Rd Sharma 2021 Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions are provided here with simple step-by-step explanations. These solutions for Arithmetic Progressions are extremely popular among Class 10 students for Maths Arithmetic Progressions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2021 Book of Class 10 Maths Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2021 Solutions. All Rd Sharma 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

#### Page No 5.11:

#### Question 1:

For the following arithmetic progressions write the first term *a* and the common difference* d:*

(i) −5, −1, 3, 7, ...

(ii) $\frac{1}{5},\frac{3}{5},\frac{5}{5},\frac{7}{5}$

(iii) 0.3, 0.55, 0.80, 1.05, ...

(iv) −1.1, −3.1, −5.1, −7.1, ...

#### Answer:

In the given problem, we need to write the first term (*a*) and the common difference (*d*) of the given A.P

(i) −5, −1, 3, 7 …

Here, first term of the given A.P is (*a*) = −5

Now, we will find the difference between the two terms of the given A.P

Similarly,

Also,

As

Therefore, the first term of the given A.P is and the common difference of the given is

(ii)

Here, first term of the given A.P is (*a*) =

Now, we will find the difference between the two terms of the given A.P

Similarly,

Also,

As

Therefore, the first term of the given A.P is and the common difference is

(iii) 0.3, 0.55, 0.80, 1.05, …

Here, first term of the given A.P is (*a*) = 0.3

Now, we will find the difference between the two terms of the given A.P

Similarly,

Also,

As

Therefore, the first term of A.P is and the common difference is

(iv) −1.1, −3.1, −5.1, −7.1...

Here, first term of the given A.P is (*a*) = −1.1

Now, we will find the difference between the two terms of the given A.P

Similarly,

Also,

As

Therefore, the first term of A.P is and the common difference is

#### Page No 5.11:

#### Question 2:

Write the arithmetic progression when first term *a* and common difference *d* are as follows:

(i) *a* = 4, *d* = −3

(ii) *a* = −1, *d* = $\frac{1}{2}$

(iii) *a* = −1.5, *d* = −0.5

#### Answer:

In the given problem, we are given its first term (*a*) and common difference (*d*).

We need to find the A.P

(i) ,

Now, as

A.P would be represented by

So,

Similarly,

Also,

Further,

Therefore, A.P with and is

(ii) ,

Now, as

A.P would be represented by

So,

Similarly,

Also,

Further,

Therefore, A.P with and is

(iii) ,

Now, as

A.P would be represented by

So,

Similarly,

Also,

Further,

Therefore, A.P with and is

#### Page No 5.11:

#### Question 3:

In which of the following situations, the sequence of numbers formed will form an A.P.?

(i) The cost of digging a well for the first metre is Rs 150 and rises by Rs 20 for each succeeding metre.

(ii) The amount of air present in the cylinder when a vacuum pump removes each time $\frac{1}{4}$ of their remaining in the cylinder.

Q

#### Answer:

(i) In the given problem,

Cost of digging a well for the first meter = Rs 150

Cost of digging a well for subsequent meter is increased by Rs 20

So,

Cost of digging a well of depth one meter= Rs. 150

Cost of digging a well of depth two meters= Rs = Rs.

Cost of digging a well of depth three meters= Rs = Rs.

Cost of digging a well of depth four meters = Rs = Rs.

Thus, the costs of digging a well of different depths are

Now, for a sequence to be an A.P., the difference between adjacent terms should be equal.

Here,

Also,

Therefore,

Since the terms of the sequence are at a common difference of 20, the above sequence is an A.P. with the first term asand common difference.

(ii) Here, let us take the initial amount of air present in the cylinder as 100 units.

So,

Amount left after vacuum pump removes air for 1^{st} time=

Amount left after vacuum pump removes air for 2^{nd} time=

Amount left after vacuum pump removes air for 3^{rd} time=

Thus, the amount left in the cylinder at various stages is

Now, for a sequence to be an A.P., the difference between adjacent terms should be equal.

Here,

Also,

Since,

The sequence is not an A.P.

(iii)

Here, prinical (P) = 1000

Rate (r) = 10%

Amount compounded annually is given by

$A=P{\left(1+\frac{r}{100}\right)}^{n}$

For the first year,

${A}_{1}=1000{\left(1+\frac{10}{100}\right)}^{1}=1100$

For the second year,

${A}_{2}=1000{\left(1+\frac{10}{100}\right)}^{2}=1210$

For the third year,

${A}_{1}=1000{\left(1+\frac{10}{100}\right)}^{3}=1331$

Therefore, first three terms are 1100, 1210, 1331.

The common difference between the consecutive terms are not same.

Hence, this is not in A.P.

#### Page No 5.11:

#### Question 4:

Find the common difference and write the next four terms of each of the following arithmetic progressions:

(i) 1, −2, −5, −8, ...

(ii) 0, −3, −6, −9, ...

(iii) $-1,\frac{1}{4},\frac{3}{2},...$

(iv) $-1,\frac{5}{6},\frac{2}{3},...$

#### Answer:

In the given problem, we need to find the common difference and the next four terms of the given A.P.

(i)

Here, first term (*a*_{1}) =1

Common difference (*d*)

Now, we need to find the next four terms of the given A.P

That is we need to find

So, using the formula

Substituting in the above formula

Substituting *n *= 5*, *we get

Substituting *n *= 6*, *we get

Substituting *n *= 7*, *we get

Substituting *n *= 8*, *we get

Therefore, the common difference is and the next four terms are

(ii)

Here, first term (*a*_{1}) =0

Common difference (*d*)

Now, we need to find the next four terms of the given A.P

That is we need to find

So, using the formula

Substituting in the above formula

Substituting *n *= 5*, *we get

Substituting *n *= 6*, *we get

Substituting *n *= 7*, *we get

Substituting *n *= 8*, *we get

Therefore, the common difference is and the next four terms are

(iii)

Here, first term (*a*_{1}) =−1

Common difference (*d*)

Now, we need to find the next four terms of the given A.P

That is we need to find

So, using the formula

Substituting in the above formula

Substituting *n *= 4*, *we get

Substituting *n *= 5*, *we get

Substituting *n *= 6*, *we get

Substituting *n *= 7*, *we get

Therefore, the common difference is and the next four terms are

(iv)

Here, first term (*a*_{1}) =−1

Common difference (*d*)

Now, we need to find the next four terms of the given A.P

That is we need to find

So, using the formula

Substituting in the above formula

Substituting *n *= 4*, *we get

${a}_{4}=-1+\left(4-1\right)\left(\frac{1}{6}\right)\phantom{\rule{0ex}{0ex}}{a}_{4}=-1+\left(\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}{a}_{4}=\frac{-2+1}{2}=\frac{-1}{2}$

Substituting *n *= 5*, *we get

Substituting *n *= 6*, *we get

Substituting *n *= 7*, *we get

Therefore, the common difference is and the next four terms are

#### Page No 5.11:

#### Question 5:

Which of the following sequences are arithmetic progressions. For those which are arithmetic progressions, find out the common difference.

(i) 3, 6, 12, 24, ...

(ii) 0, −4, −8, −12, ...

(iii) $\frac{1}{2},\frac{1}{4},\frac{1}{6},\frac{1}{8},...$

(iv) 12, 2, −8, −18, ...

(v) 3, 3, 3, 3, ...

(vi) *p*, *p* + 90, *p* + 180 *p* + 270, ... where *p *= (999)^{999}

(vii) 1.0, 1.7, 2.4, 3.1, ...

(viii) −225, −425, −625, −825, ...

(ix) 10, 10 + 2^{5}, 10 + 2^{6}, 10 + 2^{7},...

(x) a + b, (a + 1) + b, (a + 1) + (b + 1), (a + 2) + (b + 1), (a + 2) + (b + 2), ...

(xi) 1^{2}, 3^{2}, 5^{2}, 7^{2}, ...

(xii) 1^{2}, 5^{2}, 7^{2}, 73, ...

#### Answer:

In the given problem, we are given various sequences.

We need to find out that the given sequences are an A.P or not and then find its common difference (*d*)

(i)

Here,

First term (*a*) = 3

Now, for the given to sequence to be an A.P,

Common difference (*d*)

Here,

Also,

Since

Hence, the given sequence is not an A.P

(ii)

Here,

First term (*a*) = 0

Now, for the given sequence to be an A.P,

Common difference (*d*)

Here

Also,

Since

Hence, the given sequence is an A.P and its common difference is

(iii)

Here,

First term (*a*) =

Now, for the given to sequence to be an A.P,

Common difference (*d*)

Here,

Also,

Since

Hence, the given sequence is not an A.P

(iv)

Here,

First term (*a*) = 12

Now, for the given to sequence to be an A.P,

Common difference (*d*)

Here,

Also,

Since

Hence, the given sequence is an A.P with the common difference

(v)

Here,

First term (*a*) = 3

Now, for the given to sequence to be an A.P,

Common difference (*d*)

Here,

Also,

Since

Hence, the given sequence is an A.P and its common difference is

(vi) Where,

Here,

First term (*a*) = *p*

Now, for the given to sequence to be an A.P,

Common difference (*d*)

Here,

Also,

Since

Hence, the given sequence is an A.P and its common difference is

(vii)

Here,

First term (*a*) = 1.0

Now, for the given to sequence to be an A.P,

Common difference (*d*)

Here,

Also,

Since

Hence, the given sequence is an A.P and its common difference is

(viii)

Here,

First term (*a*) = −225

Now, for the given to sequence to be an A.P,

Common difference (*d*)

Here,

Also,

Since

Hence, the given sequence is an A.P and its common difference is

(ix)

Here,

First term (*a*) = 10

Now, for the given to sequence to be an A.P,

Common difference (*d*) = ${a}_{1}-a={a}_{2}-{a}_{1}$

Here,

Also,

Since

Hence, the given sequence is not an A.P

(x)

Here,

First term (*a*) = *a *+* b*

Now, for the given to sequence to be an A.P,

Common difference (*d*)

Here,

Also,

Since

Hence, the given sequence is an A.P and its common difference is

(xi)

Here,

First term (*a*) =

Now, for the given to sequence to be an A.P,

Common difference (*d*)

Here,

Also,

Since

Hence, the given sequence is not an A.P

(xii)

Here,

First term (*a*) =

Now, for the given to sequence to be an A.P,

Common difference (*d*)

Here,

Also,

${a}_{3}-{a}_{2}={7}^{2}-{5}^{2}=49-25=24$

Since

Hence, the given sequence is an A.P with the common difference .

#### Page No 5.11:

#### Question 6:

Find the common difference of the A.P. and write the next two terms:

(i) 51, 59, 67, 75, ..

(ii) 75, 67, 59, 51, ...

(iii) 1.8, 2.0, 2.2, 2.4, ...

(iv) $0,\frac{1}{4},\frac{1}{2},\frac{3}{4},...\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

(v) 119, 136, 153, 170, ...

#### Answer:

In this problem, we are given different A.P. and we need to find the common difference of the A.P., along with the next two terms.

(i)

Here,

So, common difference of the A.P. (*d*) =

Also, we need to find the next two terms of A.P., which means we have to find the 5^{th} and 6^{th} term.

So, for fifth term,

Similarly, we find the sixth term,

Therefore, the common difference is and the next two terms of the A.P. are.

(ii)

Here,

So, common difference of the A.P. (*d*) =

Also, we need to find the next two terms of A.P., which means we have to find the 5^{th} and 6^{th} term.

So, for fifth term,

Similarly, we find the sixth term,

Therefore, the common difference is and the next two terms of the A.P. are.

(iii)

Here,

So, common difference of the A.P. (*d*) =

Also, we need to find the next two terms of A.P., which means we have to find the 5^{th} and 6^{th} term.

So, for fifth term,

Similarly, we find the sixth term,

Therefore, the common difference is and the next two terms of the A.P. are.

(iv)

Here,

So, common difference of the A.P. (*d*) =

^{th} and 6^{th} term.

So, for fifth term,

Similarly, we find the sixth term,

Therefore, the common difference is and the next two terms of the A.P. are.

(v)

Here,

So, common difference of the A.P. (*d*) =

^{th} and 6^{th} term.

So, for fifth term,

Similarly, we find the sixth term,

Therefore, the common difference is and the next two terms of the A.P. are.

#### Page No 5.11:

#### Question 7:

Prove that no matter what the real numbers *a* and *b *are, the sequence with *n*th term *a* + *nb* is always an A.P. What is the common difference?

#### Answer:

In the given problem, we are given the sequence with the *n*^{th} term () as where *a* and *b* are real numbers.

We need to show that this sequence is an A.P and then find its common difference (*d*)

Here,

Now, to show that it is an A.P, we will find its few terms by substituting

So,

Substituting* n *= 1*, *we get

Substituting* n *= 2*, *we get

Substituting* n *= 3*, *we get

Further, for the given to sequence to be an A.P,

Common difference (*d*)

Here,

Also,

Since

Hence, the given sequence is an A.P and its common difference is .

#### Page No 5.24:

#### Question 1:

Find:

(i) 10^{th} term of the A.P. 1, 4, 7, 10, ...

(ii) 18^{th} term of the A.P. $\sqrt{2},3\sqrt{2},5\sqrt{2}...$

(iii) *n*^{th} term of the A.P. 13, 8, 3, −2, ...

(iv) 10^{th} term of the A.P. −40, −15, 10, 35, ...

(v) 8th term of the A.P. 117, 104, 91, 78, ...

(vi) 11th term of the A.P. 10.0, 10.5, 11.0, 11.5, ...

(vii) 9th term of the A.P. $\frac{3}{4},\frac{5}{4},\frac{7}{4},\frac{9}{4},...$

#### Answer:

In this problem, we are given different A.P. and we need to find the required term of that A.P.

(i) 10^{th} term of the A.P.

Here,

First term (*a*) = 1

Common difference of the A.P. (*d*)

Now, as we know,

So, for 10^{th} term,

Therefore, the 10^{th} term of the given A.P. is.

(ii) 18^{th} term of the A.P.

Here,

First term (*a*) =

Common difference of the A.P. (*d*)

Now, as we know,

So, for 18^{th} term,

Therefore, the 18^{th} term of the given A.P. is.

(iii)* n*^{th} term of the A.P.

Here,

First term (*a*) = 13

Common difference of the A.P. (*d*)

Now, as we know,

So, for *n*^{th} term,

Therefore, the *n*^{th} term of the given A.P. is.

(iv) 10^{th} term of the A.P.

Here,

First term (*a*) = -40

Common difference of the A.P. (*d*)

Now, as we know,

So, for 10^{th} term,

Therefore, the 10^{th} term of the given A.P. is.

(v) 8^{th} term of the A.P.

Here,

First term (*a*) = 117

Common difference of the A.P. (*d*)

Now, as we know,

So, for 8^{th} term,

Therefore, the 8^{th} term of the given A.P. is.

(vi) 11^{th} term of the A.P.

Here,

First term (*a*) = 10.0

Common difference of the A.P. (*d*)

Now, as we know,

So, for 11^{th} term,

Therefore, the 11^{th} term of the given A.P. is.

(vii) 9^{th} term of the A.P.

Here,

First term (*a*) =

Common difference of the A.P. (*d*)

Now, as we know,

So, for 9^{th} term,

Therefore, the 9^{th} term of the given A.P. is.

#### Page No 5.24:

#### Question 2:

Find:

(i) Which term of the A.P. 3, 8, 13, ... is 248?

(ii) Which term of the A.P. 84, 80, 76, ... is 248?

(iii) Which term of the A.P. 4, 9, 14, ... is 254?

(iv) Which term of the A.P. 21, 42, 63, 84, ... is 420?

(v) Which term of the A.P. 121, 117, 113, ... is its first negative term?

(vi) Which term of the A.P. –7, –12, –17, –22,... will be –82? Is –100 any term of the A.P?

#### Answer:

In the given problem, we are given an A.P and the value of one of its term.

We need to find which term it is (*n*)

So here we will find the value of *n* using the formula,

(i) Here, A.P is

Now,

Common difference (*d*) =

=

= 5

Thus, using the above mentioned formula

Thus,

Therefore 248 is the of the given A.P

(ii) Here, A.P is

Now,

Common difference (*d*) =

=80-84

= -4

Thus, using the above mentioned formula

On further simplifying, we get,

Thus,

Therefore 84 is the of the given A.P

(iii) Here, A.P is

Now,

Common difference (*d*) =

= 9-4

= 5

Thus, using the above mentioned formula

Thus,

Therefore 254 is the of the given A.P

(iv) Here, A.P is

Now,

Common difference (*d*) =

= 42-21

= 21

Thus, using the above mentioned formula

Thus,

Therefore 420 is the of the given A.P

(v) Here, A.P is

We need to find first negative term of the A.P

Now,

Common difference (*d*) =

Now, we need to find the first negative term,

Further simplifying, we get,

Thus,

Therefore, the first negative term is the of the given A.P.

(vi) –7, –12, –17, –22,...$\mathrm{Given}:\phantom{\rule{0ex}{0ex}}a=-7\phantom{\rule{0ex}{0ex}}d=-12-\left(-7\right)\phantom{\rule{0ex}{0ex}}=-12+7\phantom{\rule{0ex}{0ex}}=-5\phantom{\rule{0ex}{0ex}}{a}_{n}=-82\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}{a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}\Rightarrow -82=-7+\left(n-1\right)\left(-5\right)\phantom{\rule{0ex}{0ex}}\Rightarrow -82=-7-5n+5\phantom{\rule{0ex}{0ex}}\Rightarrow -82=-2-5n\phantom{\rule{0ex}{0ex}}\Rightarrow -82+2=-5n\phantom{\rule{0ex}{0ex}}\Rightarrow -80=-5n\phantom{\rule{0ex}{0ex}}\Rightarrow n=\frac{-80}{-5}\phantom{\rule{0ex}{0ex}}\Rightarrow n=16\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},{16}^{\mathrm{th}}\mathrm{term}\mathrm{of}\mathrm{the}\mathrm{A}.\mathrm{P}.\mathrm{is}-82.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{If}{a}_{n}=-100\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}\Rightarrow -100=-7+\left(n-1\right)\left(-5\right)\phantom{\rule{0ex}{0ex}}\Rightarrow -100=-7-5n+5\phantom{\rule{0ex}{0ex}}\Rightarrow -100=-2-5n\phantom{\rule{0ex}{0ex}}\Rightarrow -100+2=-5n\phantom{\rule{0ex}{0ex}}\Rightarrow -98=-5n\phantom{\rule{0ex}{0ex}}\Rightarrow n=\frac{-98}{-5}\phantom{\rule{0ex}{0ex}}\Rightarrow n=19.6\mathrm{which}\mathrm{is}\mathrm{not}\mathrm{possible}\mathrm{since}n\mathrm{is}\mathrm{a}\mathrm{natural}\mathrm{number}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},-100\mathrm{is}\mathrm{not}\mathrm{the}\mathrm{term}\mathrm{of}\mathrm{this}\mathrm{A}.\mathrm{P}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Page No 5.24:

#### Question 3:

Find:

(i) Is 68 a term of the A.P. 7, 10, 13, ...?

(ii) Is 302 a term of the A.P. 3, 8, 13, ...?

(iii) Is − 150 a term of the A.P. 11, 8, 5, 2, ...?

#### Answer:

In the given problem, we are given an A.P and the value of one of its term.

We need to find whether it is a term of the A.P or not.

So here we will use the formula,

(i) Here, A.P is

Now,

Common difference (*d*) =

Thus, using the above mentioned formula, we get,

Since, the value of *n* is a fraction.

Thus, 68 is not the term of the given A.P

Therefore the answer is

(ii) Here, A.P is

Now,

Common difference (*d*) =

Thus, using the above mentioned formula, we get,

Since, the value of *n* is a fraction.

Thus, 302 is not the term of the given A.P

Therefore the answer is

(iii) Here, A.P is

Now,

Common difference (*d*) =

Thus, using the above mentioned formula

Since, the value of *n* is a fraction.

Thus, -150 is not the term of the given A.P

Therefore, the answer is

#### Page No 5.24:

#### Question 4:

How many terms are there in the A.P.?

(i) 7, 10, 13, ... 43.

(ii) $-1,\frac{5}{6},\frac{2}{3},\frac{1}{2},...\frac{10}{3}.$

(iii) 7, 13, 19, ..., 205.

(iv) $18,15\frac{1}{2},13,...,-47.$

#### Answer:

In the given problem, we are given an A.P.

We need to find the number of terms present in it

So here we will find the value of *n* using the formula,

(i) Here, A.P is

The first term (*a*) = 7

The last term () = 43

Now,

Common difference (*d*) =

Thus, using the above mentioned formula, we get,

Thus,

Therefore, the number of terms present in the given A.P is

(ii) Here, A.P is

The first term (*a*) = -1

The last term () =

Now,

Common difference (*d*) =

Thus, using the above mentioned formula, we get,

Further solving for *n*, we get

Thus,

Therefore, the number of terms present in the given A.P is

(iii) Here, A.P is

The first term (*a*) = 7

The last term () = 205

Now,

Common difference (*d*) =

Thus, using the above mentioned formula, we get,

Thus,

Therefore, the number of terms present in the given A.P is

(iv) Here, A.P is

The first term (*a*) = 18

The last term () = -47

Now,

Common difference (*d*) =

Thus, using the above mentioned formula, we get,

Further, solving for *n*, we get

Thus,

Therefore, the number of terms present in the given A.P is .

#### Page No 5.24:

#### Question 5:

The first term of an A.P. is 5, the common difference is 3 and the last term is 80;

#### Answer:

In the given problem, we are given an A.P whose,

First term (*a*) = 5

Last term () = 80

Common difference (*d*) = 3

We need to find the number of terms present in it (*n*)

So here we will find the value of *n* using the formula,

So, substituting the values in the above mentioned formula

Thus,

Therefore, the number of terms present in the given A.P is

#### Page No 5.24:

#### Question 6:

The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.

#### Answer:

In the given problem, we are given 6^{th} and 17^{th}^{ }term of an A.P.

We need to find the 40^{th} term

Here,

Now, we will find and using the formula

So,

Also,

So, to solve for *a* and *d*

On subtracting (1) from (2), we get

Substituting (3) in (1), we get

Thus,

Substituting the above values in the formula

Therefore,

#### Page No 5.24:

#### Question 7:

The 10^{th} and 18^{th} terms of an A.P. are 41 and 73 respectively. Find 26^{th}^{ }term.

#### Answer:

In the given problem, we are given 10^{th} and 18^{th}^{ }term of an A.P.

We need to find the 26^{th} term

Here,

Now, we will find and using the formula

So,

Also,

So, to solve for *a* and *d*

On subtracting (1) from (2), we get

Substituting *d*=4 in (1), we get

Thus,

Substituting the above values in the formula,

Therefore,

#### Page No 5.24:

#### Question 8:

If the *n*th term of the A.P. 9, 7, 5, ... is same as the *n*th term of the A.P. 15, 12, 9, ... find *n*.

#### Answer:

Here, we are given two A.P. sequences whose *n*^{th} terms are equal. We need to find *n. *

So let us first find the *n*^{th} term for both the A.P.

First A.P. is 9, 7, 5 …

Here,

First term (*a*) = 9

Common difference of the A.P. (*d*)

Now, as we know,

So, for *n*^{th} term,

Second A.P. is 15, 12, 9 …

Here,

First term (*a*) = 15

Common difference of the A.P. (*d*)

Now, as we know,

So, for *n*^{th} term,

Now, we are given that the *n*^{th} terms for both the A.P. sequences are equal, we equate (1) and (2),

Therefore,

#### Page No 5.24:

#### Question 9:

Find the 12^{th} term from the end of the following arithmetic progressions:

(i) 3, 5, 7, 9, ... 201

(ii) 3, 8, 13, ..., 253

(iii) 1, 4, 7, 10, ..., 88

#### Answer:

In the given problem, we need to find the 12^{th} term from the end for the given A.P.

(i) 3, 5, 7, 9 …201

Here, to find the 12^{th} term from the end let us first find the total number of terms. Let us take the total number of terms as *n*.

So,

First term (*a*) = 3

Last term (*a*_{n}) = 201

Common difference (*d*) =

Now, as we know,

So, for the last term,

Further simplifying,

So, the 12^{th} term from the end means the 89^{th} term from the beginning.

So, for the 89^{th} term (*n =* 89)

Therefore, the 12^{th} term from the end of the given A.P. is.

(ii) 3, 8, 13 …253

Here, to find the 12^{th} term from the end let us first find the total number of terms. Let us take the total number of terms as *n*.

So,

First term (*a*) = 3

Last term (*a*_{n}) = 253

Common difference, *d* =

Now, as we know,

So, for the last term,

Further simplifying,

So, the 12^{th} term from the end means the 40^{th} term from the beginning.

So, for the 40^{th} term (*n =* 40)

Therefore, the 12^{th} term from the end of the given A.P. is.

(iii) 1, 4, 7, 10 …88

Here, to find the 12^{th} term from the end let us first find the total number of terms. Let us take the total number of terms as *n*.

So,

First term (*a*) = 1

Last term (*a*_{n}) = 88

Common difference, *d*= =3

Now, as we know,

So, for the last term,

Further simplifying,

So, the 12^{th} term from the end means the 19^{th} term from the beginning.

So, for the 19^{th} term (*n =* 19)

Therefore, the 12^{th} term from the end of the given A.P. is.

#### Page No 5.25:

#### Question 10:

If 9th term of an A.P is zero, prove that its 29th term is double the 19th term.

#### Answer:

In the given problem, the 9^{th} term of an A.P. is zero.

Here, let us take the first term of the A.P as *a* and the common difference as *d*

So, as we know,

We get,

Now, we need to prove that 29^{th} term is double of 19^{th} term. So, let us first find the two terms.

For 19^{th} term (*n* = 19),

(Using 1)

For 29^{th} term (*n* = 29),

${a}_{29}=a+\left(29-1\right)d\phantom{\rule{0ex}{0ex}}=-8d+28d\phantom{\rule{0ex}{0ex}}=20d\phantom{\rule{0ex}{0ex}}=2\times 10d\phantom{\rule{0ex}{0ex}}=2\times {a}_{19}$ (Using 1)

Therefore, for the given A.P. the 29^{th} term is double of the 19^{th} term.

Hence proved.

#### Page No 5.25:

#### Question 11:

If 10 times the 10th term of an A.P. is equal to 15 times the 15th term, show that 25th term of the A.P. is zero.

#### Answer:

Here, let us take the first term of the A.P. as *a* and the common difference as *d*

We are given that 10 times the 10^{th} term is equal to 15 times the 15^{th} term. We need to show that 25^{th} term is zero.

So, let us first find the two terms.

So, as we know,

For 10^{th} term (*n* = 10),

For 15^{th} term (*n* = 15),

Now, we are given,

Solving this, we get,

Next, we need to prove that the 25^{th} term of the A.P. is zero. For that, let us find the 25^{th} term using *n* = 25,

Thus, the 25^{th} term of the given A.P. is zero.

Hence proved

#### Page No 5.25:

#### Question 12:

In a certain A.P. the 24^{th} term is twice the 10^{th} term. Prove that the 72nd term is twice the 34th term.

#### Answer:

Here, we are given that 24^{th} term is twice the 10^{th} term, for a certain A.P. Here, let us take the first term of the A.P. as *a* and the common difference as *d*

We have to prove that

So, let us first find the two terms.

As we know,

For 10^{th} term (*n* = 10),

For 24^{th} term (*n* = 24),

Now, we are given that

So, we get,

Further, we need to prove that the 72^{nd} term is twice of 34^{th} term. So let now find these two terms,

For 34^{th} term (*n* = 34),

(Using 1)

For 72^{nd} term (*n* = 72),

(Using 1)

Therefore,

#### Page No 5.25:

#### Question 13:

The 26^{th} , 11^{th} and last term of an A.P. are 0, 3 and $-\frac{1}{5}$, respectively. Find the common difference and the number of terms .

#### Answer:

It is given that, ${a}_{26}=0,{a}_{11}=3,{a}_{n}=-\frac{1}{5}$.

$\Rightarrow a+25d=0.....\left(i\right)\phantom{\rule{0ex}{0ex}}\Rightarrow a+10d=3.....\left(ii\right)\phantom{\rule{0ex}{0ex}}\mathrm{Subtracting}\left(ii\right)\mathrm{from}\left(i\right).\phantom{\rule{0ex}{0ex}}15d=-3\phantom{\rule{0ex}{0ex}}\Rightarrow d=-\frac{1}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow a=-25d=5\phantom{\rule{0ex}{0ex}}Now,{a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}\Rightarrow -\frac{1}{5}=5+\left(n-1\right)\left(-\frac{1}{5}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow -\frac{1}{5}-5=\left(n-1\right)\left(-\frac{1}{5}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow n=27\phantom{\rule{0ex}{0ex}}$

#### Page No 5.25:

#### Question 14:

The 4^{th} term of an A.P. is three times the first and the 7^{th} term exceeds twice the third term by 1. Find the first term and the common difference.

#### Answer:

In the given problem, let us take the first term as *a* and the common difference as *d*

Here, we are given that,

We need to find *a* and *d*

So, as we know,

For the 4^{th} term (*n = *4),

Similarly, for the 3^{rd} term (*n = *3),

Also, for the 7^{th} term (*n = *7),

Now, using the value of *a*_{3} in equation (2), we get,

Equating (3) and (4), we get,

On further simplification, we get,

Now, to find *a*,

Therefore, for the given A.P

#### Page No 5.25:

#### Question 15:

Find the second term and *n*^{th} term of an A.P. whose 6^{th} term is 12 and 8^{th} term is 22.

#### Answer:

In the given problem, we are given 6^{th} and 8^{th}^{ }term of an A.P.

We need to find the 2^{nd} and *n*^{th} term

Here, let us take the first term as *a* and the common difference as *d*

We are given,

Now, we will find and using the formula

So,

Also,

So, to solve for *a* and *d*

On subtracting (1) from (2), we get

Substituting (3) in (1), we get

Thus,

So, for the 2^{nd} term (*n *= 2),

For the *n*^{th} term,

Therefore,

#### Page No 5.25:

#### Question 16:

How many numbers of two digit are divisible by 3?

#### Answer:

In this problem, we need to find out how many numbers of two digits are divisible by 3.

So, we know that the first two digit number that is divisible by 3 is 12 and the last two digit number divisible by 3 is 99. Also, all the terms which are divisible by 3 will form an A.P. with the common difference of 3.

So here,

First term (*a*) = 12

Last term (*a*_{n}) = 99

Common difference (*d*) = 3

So, let us take the number of terms as *n*

Now, as we know,

So, for the last term,

Further simplifying,

Therefore, the number of two digit terms divisible by 3 is.

#### Page No 5.25:

#### Question 17:

An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find the 32nd term.

#### Answer:

In the given problem, we need to find the 32^{nd} term of an A.P. which contains a total of 60 terms.

Here we are given the following,

First term (*a*) = 7

Last term (*a*_{n}) = 125

Number of terms (*n*) = 60

So, let us take the common difference as *d*

Now, as we know,

So, for the last term,

Further simplifying,

So, for the 32^{nd} term (*n =* 32)

Therefore, the 32^{nd} term of the given A.P. is.

#### Page No 5.25:

#### Question 18:

The sum of 4^{th}^{ }and 8^{th} terms of an A.P. is 24 and the sum of the 6^{th} and 10^{th} terms is 34. Find the first term and the common difference of the A.P.

#### Answer:

In the given problem, the sum of 4^{th} and 8^{th} term is 24 and the sum of 6^{th} and 10^{th} term is 34.

We can write this as,

We need to find *a* and *d*

For the given A.P., let us take the first term as *a* and the common difference as *d*

As we know,

For 4^{th} term (*n* = 4),

For 8^{th} term (*n* = 8),

So, on substituting the above values in (1), we get,

Also, for 6^{th} term (*n* = 6),

For 10^{th} term (*n* = 10),

So, on substituting the above values in (2), we get,

Next we simplify (3) and (4). On subtracting (3) from (4), we get,

Further, using the value of *d* in equation (3), we get,

On further simplifying, we get,

Therefore, for the given A.P

#### Page No 5.25:

#### Question 19:

The first term of an A.P. is 5 and its 100^{th} term is −292. Find the 50^{th} term of this A.P.

#### Answer:

In the given problem, we are given 1^{st} and 100^{th}^{ }term of an A.P.

We need to find the 50^{th} term

Here,

Now, we will find *d *using the formula

So,

Also,

So, to solve for *d*

Substituting *a *= 5, we get

Thus,

Substituting the above values in the formula,

Therefore,

#### Page No 5.25:

#### Question 20:

Find a_{30} − a_{20} for the A.P.

(i) −9, −14, −19, −24, ...

(ii) a,a + d, a + 2d, a + 3d, ...

#### Answer:

In this problem, we are given different A.P. and we need to find.

(i) A.P.

Here,

First term (*a*) = -9

Common difference of the A.P. (*d*)

Now, as we know,

Here, we find and

So, for 30^{th} term,

Also, for 20^{th} term,

So,

Therefore, for the given A.P

(ii) A.P.

Here,

First term (*a*) = *a*

Common difference of the A.P. (*d*) =*d*

Now, as we know,

Here, we find *a*_{30} and *a*_{20}*.*

So, for 30^{th} term,

Also, for 20^{th} term,

So,

Therefore, for the given A.P

#### Page No 5.25:

#### Question 21:

Write the expression a_{n}- a_{k} for the A.P. a, a + d, a + 2d, ...

Hence, find the common difference of the A.P. for which

(i) 11^{th} term is 5 and 13^{th} term is 79.

(ii) a_{10} −a_{5}_{ }= 200

(iii) 20^{th} term is 10 more than the 18^{th} term.

#### Answer:

A.P: *a, a+d, a+2d …*

Here, we first need to write the expression for

Now, as we know,

So, for *n*^{th} term,

Similarly, for *k*^{th} term

So,

So,

(i) In the given problem, we are given 11^{th} and 13^{th}^{ }term of an A.P.

We need to find the common difference. Let us take the common difference as *d *and the first term as *a*.

Here,

Now, we will find and using the formula

So,

Also,

Solving for *a* and *d*

On subtracting (1) from (2), we get

Therefore, the common difference for the A.P. is.

(ii) We are given,

Here,

Let us take the first term as *a* and the common difference as *d*

Now, as we know,

Here, we find *a*_{30} and *a*_{20}*.*

So, for 10^{th} term,

Also, for 5^{th} term,

So,

Therefore, the common difference for the A.P. is.

(iii) In the given problem, the 20^{th} term is 10 more than the 18^{th} term. So, let us first find the 20^{th} term and 18^{th} term of the A.P.

Here

Let us take the first term as *a* and the common difference as *d*

Now, as we know,

So, for 20^{th} term (*n* = 20),

Also, for 18^{th} term (*n* = 18),

Now, we are given,

On substituting the values, we get,

Therefore, the common difference for the A.P. is

#### Page No 5.25:

#### Question 22:

Find n if the given value of x is the nth term of the given A.P.

(i) 25, 50, 75, 100, ...; x = 1000

(ii) −1, −3, −5, −7, ...; x = −151

(iii) $5\frac{1}{2},11,16\frac{1}{2},22,...;x=550$

(iv) $1,\frac{21}{11},\frac{31}{11},\frac{41}{11},...,x=\frac{171}{11}$

#### Answer:

In the given problem, we need to find the number of terms in an A.P

(i) 25, 50, 75, 100 …

We are given,

Let us take the total number of terms as *n*.

So,

First term (*a*) = 25

Last term (*a*_{n}) = 1000

Common difference (*d*) =

Now, as we know,

So, for the last term,

Therefore, the total number of terms of the given A.P. is.

(ii) -1, -3, -5, -7 …

We are given,

Let us take the total number of terms as *n*.

So,

First term (*a*) = −1

Last term (*a*_{n}) = −151

Common difference (*d*) =

Now, as we know,

So, for the last term,

On further simplifying, we get,

Therefore, the total number of terms of the given A.P. is.

(iii)

We are given,

Let us take the total number of terms as *n*.

So,

First term (*a*) =

Last term (*a*_{n}) = 550

Common difference (*d*) =

Now, as we know,

So, for the last term,

On further simplifying, we get,

Therefore, the total number of terms of the given A.P. is

_{(iv) }

We are given,

Let us take the total number of terms as *n*.

So,

First term (*a*) = 1

Last term (*a*_{n}) =

Common difference (*d*) =

Now, as we know,

So, for the last term,

On further simplifying, we get,

Therefore, the total number of terms of the given A.P. is.

#### Page No 5.25:

#### Question 23:

THe eighth term of an A.P. is half of its second term and the eleventh term exceeds one third of its fourth term by 1 . Find the 15^{th} term.

#### Answer:

${a}_{8}=\frac{1}{2}{a}_{2}and{a}_{11}=\frac{1}{3}{a}_{4}+1\phantom{\rule{0ex}{0ex}}\Rightarrow a+7d=\frac{\left(a+d\right)}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow a+13d=0.....\left(i\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$And,\phantom{\rule{0ex}{0ex}}{a}_{11}=\frac{1}{3}{a}_{4}+1\phantom{\rule{0ex}{0ex}}\Rightarrow a+10d=\frac{a+3d}{3}+1\phantom{\rule{0ex}{0ex}}\Rightarrow 2a+27d=1.....\left(ii\right)\phantom{\rule{0ex}{0ex}}Solve\left(i\right)and\left(ii\right),wegetd=1.\phantom{\rule{0ex}{0ex}}Therefore,a=-13\phantom{\rule{0ex}{0ex}}{a}_{15}=a+14d=-13+14=1$

#### Page No 5.25:

#### Question 24:

Find the arithmetic progression whose third term is 16 and seventh term exceeds its fifth term by 12.

#### Answer:

Here, let us take the first term of the A.P as *a* and the common difference of the A.P as *d*

Now, as we know,

So, for 3^{rd} term (*n* = 3),

Also, for 5^{th} term (*n* = 5),

For 7^{th} term (*n* = 7),

Now, we are given,

Substituting the value of *d* in (1), we get,

So, the first term is 4 and the common difference is 6.

Therefore, the A.P. is

#### Page No 5.25:

#### Question 25:

The 7^{th} term of an A.P. is 32 and its 13^{th} term is 62. Find the A.P.

#### Answer:

Here, let us take the first term of the A.P. as *a* and the common difference of the A.P as *d*

Now, as we know,

So, for 7^{th} term (*n* = 7),

Also, for 13^{th} term (*n* = 13),

Now, on subtracting (2) from (1), we get,

Substituting the value of *d* in (1), we get,

So, the first term is 2 and the common difference is 5.

Therefore, the A.P. is

#### Page No 5.25:

#### Question 26:

Which term of the A.P. 3, 10, 17, ... will be 84 more than its 13*th* term?

#### Answer:

In the given problem, let us first find the 13^{th} term of the given A.P.

A.P. is 3, 10, 17 …

Here,

First term (*a*) = 3

Common difference of the A.P. (*d*) =7

Now, as we know,

So, for 13^{th} term (*n* = 13),

Let us take the term which is 84 more than the 13^{th} term as *a*_{n}. So,

Also,

Further simplifying, we get,

Therefore, theof the given A.P. is 84 more than the 13^{th} term.

#### Page No 5.25:

#### Question 27:

Two arithmetic progression have the same common difference. The difference between their 100th terms is 100, What is the difference between their 1000th terms?

#### Answer:

Here, we are given two A.P sequences which have the same common difference. Let us take the first term of one A.P. as *a* and of other A.P. as *a’*

Also, it is given that the difference between their 100^{th} terms is 100.

We need to find the difference between their 100^{th} terms

So, let us first find the 100^{th} terms for both of them.

Now, as we know,

So, for 100^{th} term of first A.P. (*n* = 100),

Now, for 100^{th} term of second A.P. (*n* = 100),

Now, we are given,

On substituting the values, we get,

Now, we need the difference between the 1000^{th} terms of both the A.P.s

So, for 1000^{th} term of first A.P. (*n* = 1000),

Now, for 1000^{th} term of second A.P. (*n* = 1000),

So,

Therefore, the difference between the 1000^{th} terms of both the arithmetic progressions will be.

#### Page No 5.25:

#### Question 28:

For what value of *n*, the nth terms of the arithmetic progressions 63, 65, 67, ... and 3, 10, 17, ... are equal?

#### Answer:

Here, we are given two A.P. sequences. We need to find the value of *n* for which the *n*^{th }terms of both the sequences are equal. We need to find *n*

So let us first find the *n*^{th} term for both the A.P.

First A.P. is 63, 65, 67 …

Here,

First term (*a*) = 63

Common difference of the A.P. (*d*) =2

Now, as we know,

So, for *n*^{th} term,

Second A.P. is 3, 10, 17 …

Here,

First term (*a*) = 3

Common difference of the A.P. (*d*) =7

Now, as we know,

So, for *n*^{th} term,

Now, we are given that the *n*^{th} terms for both the A.P. sequences are equal, we equate (1) and (2),

Therefore,

#### Page No 5.25:

#### Question 29:

How many multiples of 4 lie between 10 and 250?

#### Answer:

In this problem, we need to find out how many multiples of 4 lie between 10 and 250.

So, we know that the first multiple of 4 after 10 is 12 and the last multiple of 4 before 250 is 248. Also, all the terms which are divisible by 4 will form an A.P. with the common difference of 4.

So here,

First term (*a*) = 12

Last term (*a*_{n}) = 248

Common difference (*d*) = 4

So, let us take the number of terms as *n*

Now, as we know,

So, for the last term,

Further simplifying,

Therefore, the number of multiples of 4 that lie between 10 and 250 is.

#### Page No 5.25:

#### Question 30:

How many three digit numbers are divisible by 7?

#### Answer:

In this problem, we need to find out how many numbers of three digits are divisible by 7.

So, we know that the first three digit number that is divisible by 7 is 105 and the last three digit number divisible by 7 is 994. Also, all the terms which are divisible by 7 will form an A.P. with the common difference of 7.

So here,

First term (*a*) = 105

Last term (*a*_{n}) = 994

Common difference (*d*) = 7

So, let us take the number of terms as *n*

Now, as we know,

So, for the last term,

Further simplifying,

Therefore, the number of three digit terms divisible by 7 is.

#### Page No 5.25:

#### Question 31:

Which term of the arithmetic progression 8, 14, 20, 26, ... will be 72 more than its 41^{st} term.

#### Answer:

In the given problem, let us first find the 41^{st} term of the given A.P.

A.P. is 8, 14, 20, 26 …

Here,

First term (*a*) = 8

Common difference of the A.P. (*d*) =6

Now, as we know,

So, for 41^{st} term (*n* = 41),

Let us take the term which is 72 more than the 41^{st} term as *a*_{n}. So,

Also,

Further simplifying, we get,

Therefore, the of the given A.P. is 72 more than the 41^{st} term.

#### Page No 5.26:

#### Question 32:

Find the term of the arithmetic progression 9, 12, 15, 18, ... which is 39 more than its 36^{th} term.

#### Answer:

In the given problem, let us first find the 36^{st} term of the given A.P.

A.P. is 9, 12, 15, 18 …

Here,

First term (*a*) = 9

Common difference of the A.P. (*d*) =3

Now, as we know,

So, for 36^{th} term (*n* = 36),

Let us take the term which is 39 more than the 36^{th} term as *a*_{n}. So,

Also,

Further simplifying, we get,

Therefore, theof the given A.P. is 39 more than the 36^{th} term

#### Page No 5.26:

#### Question 33:

Find the 8^{th}^{ }term from the end of the A.P. 7, 10, 13, ..., 184.

#### Answer:

In the given problem, we need to find the 8^{th} term from the end for the given A.P.

We have the A.P as 7, 10, 13 …184

Here, to find the 8^{th} term from the end let us first find the total number of terms. Let us take the total number of terms as *n*.

So,

First term (*a*) = 7

Last term (*a*_{n}) = 184

Common difference (*d*) = =3

Now, as we know,

So, for the last term,

Further simplifying,

So, the 8^{th} term from the end means the 53^{rd} term from the beginning.

So, for the 53^{rd} term (*n =* 53)

Therefore, the 8^{th} term from the end of the given A.P. is.

#### Page No 5.26:

#### Question 34:

Find the 10^{th} term from the end of the A.P. 8, 10, 12, ..., 126.

#### Answer:

In the given problem, we need to find the 10^{th} term from the end for the given A.P.

We have the A.P as 8, 10, 12 …126

Here, to find the 10^{th} term from the end let us first find the total number of terms. Let us take the total number of terms as *n*.

So,

First term (*a*) = 8

Last term (*a*_{n}) = 126

Common difference (*d*) = =2

Now, as we know,

So, for the last term,

Further simplifying,

So, the 10^{th} term from the end means the 51^{st} term from the beginning.

So, for the 51^{st} term (*n =* 51)

Therefore, the 10^{th} term from the end of the given A.P. is.

#### Page No 5.26:

#### Question 35:

The sum 4^{th} and 8^{th} terms of an A.P. is 24 and the sum of 6^{th} and 10^{th} terms is 44. Find the A.P.

#### Answer:

In the given problem, the sum of 4^{th} and 8^{th} term is 24 and the sum of 6^{th} and 10^{th} term is 44. We have to find the A.P

We can write this as,

We need to find the A.P

For the given A.P., let us take the first term as *a* and the common difference as *d*

As we know,

For 4^{th} term (*n* = 4),

For 8^{th} term (*n* = 8),

So, on substituting the above values in (1), we get,

Also, for 6^{th} term (*n* = 6),

For 10^{th} term (*n* = 10),

So, on substituting the above values in (2), we get,

Next we simplify (3) and (4). On subtracting (3) from (4), we get,

Further, using the value of *d* in equation (3), we get,

So here,

Therefore, the A.P. is.

#### Page No 5.26:

#### Question 36:

Which term of the A.P. 3, 15, 27, 39, ... will be 120 more than its 21^{st} term?

#### Answer:

In the given problem, let us first find the 21^{st} term of the given A.P.

A.P. is 3, 15, 27, 39 …

Here,

First term (*a*) = 3

Common difference of the A.P. (*d*) =12

Now, as we know,

So, for 21^{st} term (*n* = 21),

Let us take the term which is 120 more than the 21^{st} term as *a*_{n}. So,

Also,

Further simplifying, we get,

Therefore, theof the given A.P. is 120 more than the 21^{st} term.

#### Page No 5.26:

#### Question 37:

The 17^{th} term of an A.P. is 5 more than twice its 8^{th}^{ }term. If the 11^{th} term of the A.P. is 43. find the *n*^{th} term.

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, *n*^{th }term = *a*_{n }= *a* + (*n* − 1)*d*

According to the question,

*a*_{17} = 5 + 2*a*_{8}

⇒ *a* + (17 − 1)*d* = 5 + 2(*a* + (8 − 1)*d*)

⇒ *a* + 16*d* = 5 + 2*a* + 14*d*

⇒ 16*d* − 14*d *= 5 + 2*a* − *a*

⇒ 2*d *= 5 + *a*

⇒ *a = *2*d *− 5 * ....* (1)

Also, *a*_{11} = 43

⇒ *a* + (11 − 1)*d* = 43

⇒ *a* + 10*d* = 43 ....(2)

On substituting the values of (1) in (2), we get

2*d *− 5 + 10*d* = 43

⇒ 12*d *= 5 + 43

⇒ 12*d *= 48

⇒ *d *= 4

⇒ *a* = 2 × 4 − 5 [From (1)]

⇒ *a* = 3

∴ *a*_{n }= *a* + (*n* − 1)*d
=* 3 + (

*n*− 1)4

= 3 + 4

*n*− 4

= 4

*n*− 1

Thus, the

*n*

^{th}term of the given A.P. is 4

*n*− 1.

#### Page No 5.26:

#### Question 38:

Find the number of all three digit natural numbers which are divisible by 9.

#### Answer:

First three-digit number that is divisible by 9 is 108.

Next number is 108 + 9 = 117.

And the last three-digit number that is divisible by 9 is 999.

Thus, the progression will be 108, 117, .... , 999.

All are three digit numbers which are divisible by 9, and thus forms an A.P. having first term a 108 and the common difference as 9.

We know that, *n*^{th }term = *a*_{n }= *a* + (*n* − 1)*d*

According to the question,

999 = 108 + (*n* − 1)9

⇒ 108 + 9*n* − 9 = 999

⇒ 99 + 9*n* = 999

⇒ 9*n* = 999 − 99

⇒ 9*n* = 900

⇒ *n* = 100

Thus, the number of all three digit natural numbers which are divisible by 9 is 100.

#### Page No 5.26:

#### Question 39:

The 19^{th} term of an A.P. is equal to three times its sixth term. If its 9^{th} term is 19, find the A.P.

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, *n*^{th }term = *a*_{n }= *a* + (*n* − 1)*d*

According to the question,

*a*_{19} = 3*a*_{6}

⇒ *a* + (19 − 1)*d* = 3(*a* + (6 − 1)*d*)

⇒ *a* + 18*d* = 3*a* + 15*d*

⇒ 18*d* − 15*d *= 3*a* − *a*

⇒ 3*d *= 2*a*

⇒ *a = $\frac{3}{2}$**d* * ....* (1)

Also, *a*_{9} = 19

⇒ *a* + (9 − 1)*d* = 19

⇒ *a* + 8*d* = 19 ....(2)

On substituting the values of (1) in (2), we get

*$\frac{3}{2}$d *+ 8*d* = 19

⇒ 3*d *+ 16*d *= 19 × 2

⇒ 19*d *= 38

⇒ *d *= 2

⇒ *a* = $\frac{3}{2}\times 2$ [From (1)]

⇒ *a* = 3

Thus, the A.P. is 3, 5, 7, 9, .... .

#### Page No 5.26:

#### Question 40:

The 9^{th} term of an A.P. is equal to 6 times its second term. If its 5^{th} term is 22, find the A.P.

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, *n*^{th }term = *a*_{n }= *a* + (*n* − 1)*d*

According to the question,

*a*_{9} = 6*a*_{2}

⇒ *a* + (9 − 1)*d* = 6(*a* + (2 − 1)*d*)

⇒ *a* + 8*d* = 6*a* + 6*d*

⇒ 8*d* − 6*d *= 6*a* − *a*

⇒ 2*d *= 5*a*

⇒ *a = $\frac{2}{5}$**d* * ....* (1)

Also, *a*_{5} = 22

⇒ *a* + (5 − 1)*d* = 22

⇒ *a* + 4*d* = 22 ....(2)

On substituting the values of (1) in (2), we get

*$\frac{2}{5}$d *+ 4*d* = 22

⇒ 2*d *+ 20*d *= 22 × 5

⇒ 22*d *= 110

⇒ *d *= 5

⇒ *a* = $\frac{2}{5}\times 5$ [From (1)]

⇒ *a* = 2

Thus, the A.P. is 2, 7, 12, 17, .... .

#### Page No 5.26:

#### Question 41:

The 24^{th} term of an A.P. is twice its 10^{th} term. Show that its 72^{nd} term is 4 times its 15^{th}^{ }term.

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, *n*^{th }term = *a*_{n }= *a* + (*n* − 1)*d*

According to the question,

*a*_{24} = 2*a*_{10}

⇒ *a* + (24 − 1)*d* = 2(*a* + (10 − 1)*d*)

⇒ *a* + 23*d* = 2*a* + 18*d*

⇒ 23*d* − 18*d *= 2*a* − *a*

⇒ 5*d *= *a*

⇒ *a = *5*d* * ....* (1)

Also,

*a*_{72} = *a* + (72 − 1)*d*

= 5*d* + 71*d * [From (1)]

= 76*d * ..... (2)

and

*a*_{15 }= *a* + (15 − 1)*d*

= 5*d* + 14*d * [From (1)]

= 19*d * ..... (3)

On comparing (2) and (3), we get

76*d *= 4 × 19*d*

⇒ *a*_{72} = 4 × *a*_{15}

Thus, 72^{nd} term of the given A.P. is 4 times its 15^{th}^{ }term.

#### Page No 5.26:

#### Question 42:

Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.

#### Answer:

Since, the number is divisible by both 2 and 5, means it must be divisible by 10.

In the given numbers, first number that is divisible by 10 is 110.

Next number is 110 + 10 = 120.

The last number that is divisible by 10 is 990.

Thus, the progression will be 110, 120, ..., 990.

All the terms are divisible by 10, and thus forms an A.P. having first term as 110 and the common difference as 10.

We know that, *n*^{th }term = *a*_{n }= *a* + (*n* − 1)*d*

According to the question,

990 = 110 + (*n* − 1)10

⇒ 990 = 110 + 10*n* − 10

⇒ 10*n* = 990 − 100

⇒ 10*n *= 890

⇒ *n = *89

Thus, the number of natural numbers between 101 and 999 which are divisible by both 2 and 5 is 89.

#### Page No 5.26:

#### Question 43:

If the seventh term of an A.P. is $\frac{1}{9}$ and its ninth term is $\frac{1}{7}$, find its (63)^{rd} term.

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, *n*^{th }term = *a*_{n }= *a* + (*n* − 1)*d*

According to the question,

*a*_{7} = $\frac{1}{9}$

⇒ *a* + (7 − 1)*d* = $\frac{1}{9}$

⇒ *a* + 6*d* = $\frac{1}{9}$ * ....* (1)

Also, *a*_{9} = $\frac{1}{7}$

⇒ *a* + (9 − 1)*d* = $\frac{1}{7}$

⇒ *a* + 8*d* = $\frac{1}{7}$ ....(2)

On Subtracting (1) from (2), we get

8*d *− 6*d* = $\frac{1}{7}-\frac{1}{9}$

⇒ 2*d *= $\frac{9-7}{63}$

⇒ 2*d *= $\frac{2}{63}$

⇒ *d *= $\frac{1}{63}$

⇒ *a* = $\frac{1}{9}-\frac{6}{63}$ [From (1)]

⇒ *a* = $\frac{7-6}{63}$

⇒ *a* = $\frac{1}{63}$

∴ *a*_{63 }= *a* + (63 − 1)*d
=* $\frac{1}{63}+\frac{62}{63}$

= $\frac{63}{63}$

= 1

Thus, (63)

^{rd}term of the given A.P. is 1.

#### Page No 5.26:

#### Question 44:

The sum of 5^{th }and 9^{th }terms of an A.P. is 30. If its 25^{th} term is three times its 8^{th} term, find the A.P.

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, *n*^{th }term = *a*_{n }= *a* + (*n* − 1)*d*

According to the question,

*a*_{5} + *a*_{9} = 30

⇒ *a* + (5 − 1)*d* + *a* + (9 − 1)*d *= 30

⇒ *a* + 4*d* + *a* + 8*d* = 30

⇒ 2*a* + 12*d* = 30

⇒ *a* + 6*d* = 15 * ....* (1)

Also, *a*_{25} = 3(*a*_{8})

⇒ *a* + (25 − 1)*d* = 3[*a* + (8 − 1)*d*]

⇒ *a* + 24*d* = 3*a* + 21*d*

⇒ 3*a* − *a* = 24*d* − 21*d*

⇒ 2*a* = 3*d*

⇒ *a* = $\frac{3}{2}$*d* ....(2)

Substituting the value of (2) in (1), we get

*$\frac{3}{2}$d *+ 6*d* = 15

⇒ 3*d *+ 12*d *= 15 × 2

⇒ 15*d *= 30

⇒ *d *= 2

⇒ *a* = $\frac{3}{2}\times 2$ [From (1)]

⇒ *a* = 3

Thus, the A.P. is 3, 5, 7, 9, .... .

#### Page No 5.26:

#### Question 45:

Find where 0 (zero) is a term of the A.P. 40, 37, 34, 31, ..... .

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, *n*^{th }term = *a*_{n }= *a* + (*n* − 1)*d*

It is given that *a* = 40, *d* = −3 and *a _{n}* = 0

According to the question,

⇒ 0 = 40 + (

*n*− 1)(−3)

⇒ 0 = 40 − 3

*n*+ 3

⇒ 3

*n*= 43

⇒

*n*= $\frac{43}{3}$

*....*(1)

Here,

*n*is the number of terms, so must be an integer.

Thus, there is no term where 0 (zero) is a term of the A.P. 40, 37, 34, 31, ..... .

#### Page No 5.26:

#### Question 46:

Find the middle term of the A.P. 213, 205, 197, ...., 37.

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, *n*^{th }term = *a*_{n }= *a* + (*n* − 1)*d*

It is given that *a* = 213, *d* = −8 and *a _{n}* = 37

According to the question,

⇒ 37 = 213 + (

*n*− 1)(−8)

⇒ 37 = 213 − 8

*n*+ 8

⇒ 8

*n*= 221 − 37

⇒ 8

*n*= 184

⇒

*n*= 23

*....*(1)

Therefore, total

*number of terms is 23.*

Since, there are odd number of terms.

So, Middle term will be ${\left(\frac{23+1}{2}\right)}^{\mathrm{th}}$ term, i.e., the 12

^{th}term.

∴

*a*

_{12 }= 213 + (12 − 1)(−8)

= 213 − 88

= 125

Thus, the middle term of the A.P. 213, 205, 197, ...., 37 is 125.

#### Page No 5.26:

#### Question 47:

If the 5^{th }term of an A.P. is 31 and 25^{th} term is 140 more than the 5^{th} term, find the A.P.

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, *n*^{th }term = *a*_{n }= *a* + (*n* − 1)*d*

According to the question,

*a*_{5} = 31

⇒ *a* + (5 − 1)*d* = 31

⇒ *a* + 4*d* = 31

⇒ *a = *31 − 4*d ** ....* (1)

Also, *a*_{25} = 140 + *a*_{5}

⇒ *a* + (25 − 1)*d* = 140 + 31

⇒ *a* + 24*d* = 171 .... (2)

On substituting the values of (1) in (2), we get

31 − 4*d* + 24*d* = 171

⇒ 20*d *= 171 − 31

⇒ 20*d *= 140

⇒ *d *= 7

⇒ *a* = 31 − 4 × 7 [From (1)]

⇒ *a* = 3

Thus, the A.P. is 3, 10, 17, 24, .... .

#### Page No 5.26:

#### Question 48:

Find the sum of two middle terms of the A.P. : $-\frac{4}{3},-1,\frac{-2}{3},\frac{-1}{3},.......,4\frac{1}{3}$.

#### Answer:

$a=-\frac{4}{3},d=\frac{1}{3},{a}_{n}=\frac{13}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow a+\left(n-1\right)d=\frac{13}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \left(-\frac{4}{3}\right)+\left(n-1\right)\left(\frac{1}{3}\right)=\frac{13}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow 13=-4+n-1\phantom{\rule{0ex}{0ex}}\Rightarrow n=18\phantom{\rule{0ex}{0ex}}\mathrm{Midlle}\mathrm{terms}\mathrm{are}{\frac{\mathrm{n}}{2}}^{\mathrm{th}}\mathrm{and}\frac{\mathrm{n}}{2}+{1}^{\mathrm{th}},\mathrm{i}.\mathrm{e}9\mathrm{th}\mathrm{and}10\mathrm{th}\mathrm{terms}.\phantom{\rule{0ex}{0ex}}{a}_{9}=a+8d=-\frac{4}{3}+\frac{8}{3}=\frac{4}{3}\phantom{\rule{0ex}{0ex}}{a}_{10}=a+9d=-\frac{4}{3}+\frac{9}{3}=\frac{5}{3}\phantom{\rule{0ex}{0ex}}\therefore {a}_{9}+{a}_{10}=\frac{9}{3}=3\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Page No 5.26:

#### Question 49:

If (m + 1)^{th} term of an A.P is twice the (n + 1)^{th} term, prove that (3m + 1)^{th} term is twice the (m + n + 1)^{th} term.

#### Answer:

Here, we are given that (*m+*1)^{th} term is twice the (*n+*1)^{th} term, for a certain A.P. Here, let us take the first term of the A.P. as *a* and the common difference as *d*

We need to prove that

So, let us first find the two terms.

As we know,

For (*m+*1)^{th} term (*n’* = *m*+1)

For (*n+*1)^{th} term (*n’* = *n+1*),

Now, we are given that

So, we get,

Further, we need to prove that the (3*m*+1)^{th }term is twice of (*m+n+*1)^{th} term. So let us now find these two terms,

For (*m+n+*1)^{th} term (*n’* = *m+n+*1 ),

(Using 1)

For (3*m*+1)^{th} term (*n’* = 3*m**+*1),

Therefore,

Hence proved

#### Page No 5.26:

#### Question 50:

If an A.P. consists of *n* terms with first term a and *n*^{th} term *l *show that the sum of the m^{th} term from the beginning and the m^{th} term from the end is (a + l).

#### Answer:

In the given problem, we have an A.P. which consists of *n* terms.

Here,

The first term (*a*) = *a*

The last term (*a*_{n}) = *l*

Now, as we know,

So, for the *m*^{th}^{ }term from the beginning, we take (*n = m*),

Similarly, for the *m*^{th} term from the end, we can take *l* as the first term.

So, we get,

Now, we need to prove

So, adding (1) and (2), we get,

Therefore,

Hence proved

#### Page No 5.26:

#### Question 51:

How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3 ?

#### Answer:

The given A.P is 11, 15, 19, .....,299.

$a=11\phantom{\rule{0ex}{0ex}}d=4\phantom{\rule{0ex}{0ex}}{a}_{n}=299\phantom{\rule{0ex}{0ex}}\Rightarrow a+\left(n-1\right)d=299\phantom{\rule{0ex}{0ex}}\Rightarrow 11+\left(n-1\right)4=299\phantom{\rule{0ex}{0ex}}\Rightarrow \left(n-1\right)4=288\phantom{\rule{0ex}{0ex}}\Rightarrow n=73$

#### Page No 5.26:

#### Question 52:

Find the 12^{th} term from the end of the A.P. $-2,-4,-6,........-100$ .

#### Answer:

Consider the A.P −2, −4, −6, ...., −100.

$a=-2,d=-2,{a}_{n}=-100\phantom{\rule{0ex}{0ex}}\Rightarrow a+\left(n-1\right)d=-100\phantom{\rule{0ex}{0ex}}\Rightarrow \left(-2\right)+\left(n-1\right)\left(-2\right)=-100\phantom{\rule{0ex}{0ex}}\Rightarrow 2n=100\phantom{\rule{0ex}{0ex}}\Rightarrow n=50\phantom{\rule{0ex}{0ex}}\mathrm{The}12\mathrm{th}\mathrm{term}\mathrm{from}\mathrm{the}\mathrm{end}\mathrm{will}\mathrm{be}\mathrm{the}39\mathrm{th}\mathrm{term}\mathrm{from}\mathrm{the}\mathrm{starting}.\phantom{\rule{0ex}{0ex}}\therefore {a}_{39}=a+38d=-2+38\left(-2\right)=-78$

#### Page No 5.26:

#### Question 53:

For the A.P. : $-3,-7,-11,.....,$can we find ${a}_{30}-{a}_{20}$ withoput actually finding ${a}_{30}\mathrm{and}{a}_{20}$ ? Give reasons for your answer.

#### Answer:

Consider the A.P −3, −7, −11, ....

$a=-3,d=-4,\phantom{\rule{0ex}{0ex}}{a}_{30}-{a}_{20}=\left[a+\left(30-1\right)d\right]-\left[a+\left(20-1\right)d\right]\phantom{\rule{0ex}{0ex}}{a}_{30}-{a}_{20}=a+29d-a-19d\phantom{\rule{0ex}{0ex}}\therefore {a}_{30}-{a}_{20}=10d\phantom{\rule{0ex}{0ex}}=10\left(-4\right)\phantom{\rule{0ex}{0ex}}=-40\phantom{\rule{0ex}{0ex}}$

#### Page No 5.26:

#### Question 54:

Two A.P.s have the same common difference . The first term of one A.P. is 2 and that of the other is 7 . The difference between their 10^{th} terms is the same as the difference between their 21^{st }terms , which is the same as the difference between any two corresponding terms . Why ?

#### Answer:

First term of 1st A.P is 2.

First term of 2nd A.P is 7.

Consider the difference of their 10th terms,

${a}_{10}-a{\text{'}}_{10}=a+9d-a\text{'}-9d\text{'}\phantom{\rule{0ex}{0ex}}=a-a\text{'}+9d-9d\text{'}\phantom{\rule{0ex}{0ex}}=2-7+0\left[d=d\text{'}\right]\phantom{\rule{0ex}{0ex}}=-5\phantom{\rule{0ex}{0ex}}{a}_{21}-a{\text{'}}_{21}=a+20d-a\text{'}-20d\text{'}\phantom{\rule{0ex}{0ex}}=a-a\text{'}+20d-20d\text{'}\phantom{\rule{0ex}{0ex}}=2-7+0\left[d=d\text{'}\right]\phantom{\rule{0ex}{0ex}}=-5\phantom{\rule{0ex}{0ex}}Therefore,{a}_{10}-a{\text{'}}_{10}={a}_{21}-a{\text{'}}_{21}.\phantom{\rule{0ex}{0ex}}$

The difference between any two corresponding terms of A.P's is same as the difference between their terms.

#### Page No 5.30:

#### Question 1:

Find the value of x for which (8x + 4), (6x − 2) and (2x + 7) are in A.P.

#### Answer:

Here, we are given three terms,

First term (*a*_{1}) =

Second term (*a*_{2}) =

Third term (*a*_{3}) =

We need to find the value of *x* for which these terms are in A.P. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,

…… (1)

Also,

…… (2)

Now, on equating (1) and (2), we get,

Therefore, for, these three terms will form an A.P.

#### Page No 5.30:

#### Question 2:

If x + 1, 3x and 4x + 2 are in A.P., find the value of x.

#### Answer:

Here, we are given three terms which are in A.P.,

First term (*a*_{1}) =

Second term (*a*_{2}) =

Third term (*a*_{3}) =

We need to find the value of *x*. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,

…… (1)

Also,

…… (2)

Now, on equating (1) and (2), we get,

Therefore, for, these three terms will form an A.P.

#### Page No 5.30:

#### Question 3:

Show that (a − b)^{2}, (a^{2} + b^{2}) and (a + b)^{2} are in A.P.

#### Answer:

Here, we are given three terms and we need to show that they are in A.P.,

First term (*a*_{1}) =

Second term (*a*_{2}) =

Third term (*a*_{3}) =

So, in an A.P. the difference of two adjacent terms is always constant. So, to prove that these terms are in A.P. we find the common difference, we get,

…… (1)

Also,

…… (2)

Now, since in equations (1) and (2) the values of *d *are equal, we can say that these terms are in A.P. with 2*ab* as the common difference.

Hence proved

#### Page No 5.30:

#### Question 4:

Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers.

#### Answer:

In the given problem, the sum of three terms of an A.P is 27 and the product of the three terms is 648. We need to find the three terms.

Here,

Let the three terms be where *a* is the first term and *d* is the common difference of the A.P

So,

…… (1)

Also,

Further solving for *d*,

…… (2)

Now, substituting (1) and (2) in three terms

First term =

So,

Also,

Second term = *a*

So,

Also,

Third term =

So,

Therefore, the three terms are .

#### Page No 5.30:

#### Question 5:

The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.

#### Answer:

In the given problem, the sum of three terms of an A.P is 21 and the product of the first and the third term exceeds the second term by 6.

We need to find the three terms.

Here,

Let the three terms be where, *a* is the first term and *d* is the common difference of the A.P

So,

…… (1)

Also,

(Using)

(Using 1)

Further solving for *d*,

…… (2)

Now, using the values of *a *and *d* in the expressions of the three terms, we get,

First term =

So,

Second term = *a*

So,

Also,

Third term =

So,

Therefore, the three terms are .

#### Page No 5.30:

#### Question 6:

Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.

#### Answer:

Here, we are given that four number are in A.P., such that there sum is 50 and the greatest number is 4 times the smallest.

So, let us take the four terms as.

Now, we are given that sum of these numbers is 50, so we get,

…… (1)

Also, the greatest number is 4 times the smallest, so we get,

…… (2)

Now, using (2) in (1), we get,

Now, using the value of *a* in (2), we get

So, first term is given by,

Second term is given by,

Third term is given by,

Fourth term is given by,

Therefore, the four terms are .

#### Page No 5.30:

#### Question 7:

The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the numbers.

#### Answer:

In the given problem, the sum of three terms of an A.P is 12 and the sum of their cubes is 288.

We need to find the three terms.

Here,

Let the three terms be where, *a* is the first term and *d* is the common difference of the A.P

So,

Also, it is given that

So, using the properties:

We get,

Further solving for *d *by substituting the value of *a*, we get,

On further simplification, we get,

Now, here *d* can have two values +2 and -2.

So, on substituting the values of* a* = 4 and *d = *2 in three terms, we get,

First term =

So,

Second term = *a*

So,

Third term =

So,

Also, on substituting the values of* a* = 4 and in three terms, we get,

First term =

So,

Second term = *a*

So,

Third term =

So,

Therefore, the three terms are .

#### Page No 5.30:

#### Question 8:

Divide 56 in four parts in A.P. such that the ratio of the product of their extremes to the product of their means is 5 : 6.

#### Answer:

Let the four terms of the AP be *a* − 3*d*, *a* − *d*, *a* + *d *and *a* + 3*d*.

Given:

(*a *− 3*d*) + (*a − d*) + (*a + d*) + (*a* + 3*d*) = 56

$\Rightarrow $ 4*a* = 56

$\Rightarrow $ *a* = 14

$\mathrm{Also},\phantom{\rule{0ex}{0ex}}\frac{\left(a-3d\right)\left(a+3d\right)}{\left(a-d\right)\left(a+d\right)}=\frac{5}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{a}^{2}-9{d}^{2}}{{a}^{2}-{d}^{2}}=\frac{5}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\left(14\right)}^{2}-9{d}^{2}}{{\left(14\right)}^{2}-{d}^{2}}=\frac{5}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{196-9{d}^{2}}{196-{d}^{2}}=\frac{5}{6}$

$\Rightarrow 1176-54{d}^{2}=980-5{d}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 196=49{d}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{2}=4\phantom{\rule{0ex}{0ex}}\Rightarrow d=\pm 2$

When *d* = 2, the terms of the AP are 8, 12, 16, 20. When *d* = −2, the terms of the AP are 20, 18, 12, 8.

#### Page No 5.30:

#### Question 9:

The sum of the first three numbers in an arithmetic progression is 18. If the product of the first and third term is 5 times the common difference, find the three numbers.

#### Answer:

Let the first three numbers in an arithmetic progression be *a − d*,* a*,* a + d*.

The sum of the first three numbers in an arithmetic progression is 18.

*a − d* +* a* +* a + d* = 18

⇒ 3*a* = 18

⇒ *a* = 6

The product of the first and third term is 5 times the common difference.

$\left(a-d\right)\left(a+d\right)=5d\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{2}-{d}^{2}=5d\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(6\right)}^{2}-{d}^{2}=5d\phantom{\rule{0ex}{0ex}}\Rightarrow 36-{d}^{2}=5d\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{2}+5d-36=0\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{2}+9d-4d-36=0\phantom{\rule{0ex}{0ex}}\Rightarrow d\left(d+9\right)-4\left(d+9\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(d+9\right)\left(d-4\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow d=4,-9\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{For}d=4,\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{numbers}\mathrm{are}6-4,6,6+4\phantom{\rule{0ex}{0ex}}\mathrm{i}.\mathrm{e}.2,6,10\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{For}d=-9,\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{numbers}\mathrm{are}6+9,6,6-9\phantom{\rule{0ex}{0ex}}\mathrm{i}.\mathrm{e}.15,6,-3$

Hence, the three numbers are 2, 6, 10 or 15, 6, −3.

#### Page No 5.30:

#### Question 10:

The angles of a quadrilateral are in A.P. whose common difference is 10°. Find the numbers.

#### Answer:

Here, we are given that the angles of a quadrilateral are in A.P, such that the common difference is 10°.

So, let us take the angles as

Now, we know that the sum of all angles of a quadrilateral is 360°. So, we get,

On further simplifying for *a*, we get,

So, the first angle is given by,

Second angle is given by,

Third angle is given by,

Fourth angle is given by,

Therefore, the four angles of the quadrilateral are .

#### Page No 5.30:

#### Question 11:

Split 207 into three parts such that these are in A.P. and the product of the two smaller parts is 4623.

#### Answer:

Suppose three parts of 207 are (*a *− *d*), *a* , (*a* + *d*) such that , (*a* + *d*) >*a > (a − d).*

$a-d+a+a+d=207\phantom{\rule{0ex}{0ex}}\Rightarrow 3a=207\phantom{\rule{0ex}{0ex}}\Rightarrow a=69\phantom{\rule{0ex}{0ex}}\mathrm{Now},\left(a-d\right)\times a=4623\phantom{\rule{0ex}{0ex}}\Rightarrow 69\left(69-d\right)=4623\phantom{\rule{0ex}{0ex}}\Rightarrow \left(69-d\right)=67\phantom{\rule{0ex}{0ex}}\Rightarrow d=2\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{the}\mathrm{three}\mathrm{required}\mathrm{parts}\mathrm{are}67,69\mathrm{and}71.$

#### Page No 5.30:

#### Question 12:

The angles of a triangle are in A.P. The greatest angle is twice the least . Find all the angles.

#### Answer:

Suppose the angles of a triangle are (*a *− *d*), *a* , (*a* + *d*) such that , (*a* + *d*) >*a > (a − d).*

$a-d+a+a+d=180\left[\mathrm{angle}\mathrm{sum}\mathrm{property}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 3a=180\phantom{\rule{0ex}{0ex}}\Rightarrow a=60\phantom{\rule{0ex}{0ex}}\mathrm{Now},\left(a+d\right)=2\left(a-d\right)\phantom{\rule{0ex}{0ex}}\Rightarrow a+d=2a-2d\phantom{\rule{0ex}{0ex}}\Rightarrow a=3d\phantom{\rule{0ex}{0ex}}\Rightarrow d=\frac{60}{3}=20\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{the}\mathrm{three}anglesofatriangle\mathrm{are}40,60,80.$

#### Page No 5.30:

#### Question 13:

The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 : 15 . Find the number .

#### Answer:

Let the four terms of the AP be *a* − 3*d*, *a* − *d*, *a* + *d *and *a* + 3*d*.

Given:

(*a *− 3*d*) + (*a − d*) + (*a + d*) + (*a* + 3*d*) = 32

$\Rightarrow $ 4*a* = 32

$\Rightarrow $ *a* = 8

$\mathrm{Also},\phantom{\rule{0ex}{0ex}}\frac{\left(a-3d\right)\left(a+3d\right)}{\left(a-d\right)\left(a+d\right)}=\frac{7}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{a}^{2}-9{d}^{2}}{{a}^{2}-{d}^{2}}=\frac{7}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\left(8\right)}^{2}-9{d}^{2}}{{\left(8\right)}^{2}-{d}^{2}}=\frac{7}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{64-9{d}^{2}}{64-{d}^{2}}=\frac{7}{15}$

$\Rightarrow 960-135{d}^{2}=448-7{d}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 512=128{d}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {d}^{2}=4\phantom{\rule{0ex}{0ex}}\Rightarrow d=\pm 2$

When a = 8 and d= 2, then the terms are 2, 6, 10, 14.

When a = 8 and d= −2, then the terms are 14, 10, 6, 2.

#### Page No 5.50:

#### Question 1:

Find the sum of the following arithmetic progressions:

(i) 50, 46, 42, ... to 10 terms

(ii) 1, 3, 5, 7, ... to 12 terms

(iii) 3, 9/2, 6, 15/2, ... to 25 terms

(iv) 41, 36, 31, ... to 12 terms

(v) a + b, a − b, a − 3b, ... to 22 terms

(vi) (x − y)^{2}, (x^{2} + y^{2}), (x + y)^{2}, ..., to *n* terms

(vii) $\frac{x-y}{x+y}\frac{3x-2y}{x+y}\frac{5x-3y}{x+y},...$to n terms

(viii) −26, −24, −22, ... to 36 terms.

#### Answer:

In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of *n* terms of an A.P.,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

(i) To 10 terms

Common difference of the A.P. (*d*)

=

Number of terms* *(*n*) = 10

First term for the given A.P. (*a*) = 50

So, using the formula we get,

Therefore, the sum of first 10 terms for the given A.P. is.

(ii) To 12 terms.

Common difference of the A.P. (*d*)

=

Number of terms* *(*n*) = 12

First term for the given A.P. (*a*) = 1

So, using the formula we get,

Therefore, the sum of first 12 terms for the given A.P. is.

(iii) To 25 terms.

Common difference of the A.P. (*d*) =

Number of terms* *(*n*) = 25

First term for the given A.P. (*a*) = 3

So, using the formula we get,

On further simplifying, we get,

Therefore, the sum of first 25 terms for the given A.P. is.

(iv) To 12 terms.

Common difference of the A.P. (*d*) =

Number of terms* *(*n*) = 12

First term for the given A.P. (*a*) = 41

So, using the formula we get,

Therefore, the sum of first 12 terms for the given A.P. is.

(v) To 22 terms.

Common difference of the A.P. (*d*) =

Number of terms* *(*n*) = 22

First term for the given A.P. (*a*) =

So, using the formula we get,

Therefore, the sum of first 22 terms for the given A.P. is.

(vi) To *n* terms.

Common difference of the A.P. (*d*) =

Number of terms* *(*n*) = *n*

First term for the given A.P. (*a*) =

So, using the formula we get,

Now, taking 2 common from both the terms inside the bracket we get,

Therefore, the sum of first *n* terms for the given A.P. is

(vii) To *n* terms.

Number of terms* *(*n*) = *n*

First term for the given A.P. (*a*) =

Common difference of the A.P. (*d*) =

So, using the formula we get,

Now, on further solving the above equation we get,

Therefore, the sum of first *n* terms for the given A.P. is.

(viii) To 36 terms.

Common difference of the A.P. (*d*) =

Number of terms* *(*n*) = 36

First term for the given A.P. (*a*) = −26

So, using the formula we get,

Therefore, the sum of first 36 terms for the given A.P. is.

#### Page No 5.5:

#### Question 1:

Write the first five terms of each of the following sequences whose *n*th terms are:

(a) a_{n} = 3n + 2

(b) ${a}_{n}=\frac{n-3}{3}$

(c) a_{n} = 3^{n}

(d) ${a}_{n}=\frac{3n-2}{5}$

(e) a_{n} = (−1)^{n} 2^{n}

(f) ${a}_{n}=\frac{n(n-2)}{2}$

(g) an =n^{2} − n + 1

(h) a_{n} = 2n^{2} − 3n + 1

(i) ${a}_{n}=\frac{2n-3}{6}$

#### Answer:

Here, we are given the *n*^{th }term for various sequences. We need to find the first five terms of the sequence.

(i)

Here, the *n*^{th} term is given by the above expression. So, to find the first term we use, we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(ii)

Here, the *n*^{th} term is given by the above expression. So, to find the first term we use, , we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(iii)

Here, the *n*^{th} term is given by the above expression. So, to find the first term we use, we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(iv)

Here, the *n*^{th} term is given by the above expression. So, to find the first term we use, , we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(v)

Here, the *n*^{th} term is given by the above expression. So, to find the first term we use, we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms of the given A.P are.

(vi)

Here, the *n*^{th} term is given by the above expression. So, to find the first term we use , we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(vii)

Here, the *n*^{th} term is given by the above expression. So, to find the first term we use, we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(viii)

Here, the *n*^{th} term is given by the above expression. So, to find the first term we use, we get,

Similarly, we find the other four terms,

Second term (),

Third term,

Fourth term,

Fifth term,

Therefore, the first five terms for the given sequence are.

(ix)

Here, the *n*^{th} term is given by the above expression. So, to find the first term we use, we get,

Similarly, we find the other four terms,

Second term (),

Third term (),

Fourth term (),

Fifth term (),

Therefore, the first five terms of the given A.P are

#### Page No 5.5:

#### Question 2:

Find the indicated terms in each of the following sequences whose nth terms are:

(a) a_{n} = 5n − 4; a_{12}_{ }and a_{15}

(b) ${a}_{n}=\frac{3n-2}{4n+5};{a}_{7}and{a}_{8}$

(c) a_{n} = n (n −1) (n − 2); a_{5} and a_{8}

(d) a_{n} = (n − 1) (2 − n) (3 + n); a_{1}, a_{2}, a_{3}

(e) a_{n} = (−1)^{n} n ; a_{3}, a_{5}, a_{8}

#### Answer:

Here, we are given the *n*^{th }term for various sequences. We need to find the indicated terms of the A.P.

(i)

We need to find and

Now, to find term we use, we get,

Also, to find term we use, we get,

${a}_{15}=5\left(15\right)-4\phantom{\rule{0ex}{0ex}}=75-4\phantom{\rule{0ex}{0ex}}=71$

Thus,

(ii)

We need to find and

Now, to find term we use, we get,

Also, to find term we use, we get,

Thus,

(iii)

We need to find and

Now, to find term we use, we get,

Also, to find term we use, we get,

Thus,

(iv)

We need to find, and

Now, to find term we use, we get,

Also, to find term we use, we get,

Similarly, to find term we use, we get,

Thus,

(v)

We need to find, and

Now, to find term we use, we get,

Also, to find term we use, we get,

Similarly, to find term we use, we get,

Thus,

#### Page No 5.5:

#### Question 3:

Find the next five terms of each of the following sequences given by:

(i) *a*_{1} = 1, *a _{n} = a*

_{n−1}+2,

*n*≥ 2

(ii)

*a*

_{1}=

*a*

_{2}= 2,

*a*=

_{n}*a*

_{n−1}− 3,

*n*> 2

(iii) ${a}_{1}=-1,{a}_{n}=\frac{{a}_{n}-1}{n},n\ge 2$

(iv)

*a*

_{1}= 4,

*a*= 4

_{n}*a*

_{n−1}+ 3,

*n*> 1.

#### Answer:

In the given problem, we are given the first, second term and* *the *n*^{th }term of an A.P.

We need to find its next five terms

(i) , ,

Here, we are given that

So, the next five terms of this A.P would be,, , and

Now …… (1)

So, to find theterm we use, we get,

(Using 1)

…… (2)

For, using, we get,

(Using 2)

…… (3)

For, using, we get,

(Using 3)

…… (4)

For, using, we get,

(Using 4)

……. (5)

For, using, we get,

(Using 5)

Therefore, the next five terms, of the given A.P are

(ii) , ,

Here, we are given that

So, the next five terms of this A.P would be,,,and

Now …… (1)

So, to find theterm we use, we get,

(Using 1)

…… (2)

For, using, we get,

(Using 2)

…… (3)

For, using, we get,

(Using 3)

…… (4)

For, using, we get,

(Using 4)

…… (5)

For, using, we get,

(Using 5)

Therefore, the next five terms, of the given A.P are

(iii) , ,

Here, we are given that

So, the next five terms of this A.P would be, ,, and

Now …… (1)

So, to find theterm we use, we get,

(Using 1)

…… (2)

For, using, we get,

(Using 2)

…… (3)

For, using, we get,

(Using 3)

…… (4)

For, using, we get,

(Using 4)

…… (5)

For, using, we get,

(Using 5)

Therefore, the next five terms, of the given A.P are

(iv) , ,

Here, we are given that *n* > 1.

So, the next five terms of this A.P would be, , , and

Now …… (1)

So, to find theterm we use, we get,

(Using 1)

…… (2)

For, using, we get,

(Using 2)

…… (3)

For, using, we get,

(Using 3)

…… (4)

For, using, we get,

(Using 4)

…… (5)

For, using, we get,

(Using 5)

Therefore, the next five terms, of the given A.P are

#### Page No 5.51:

#### Question 2:

Find the sum to n term of the A.P. 5, 2, −1, −4, −7, ...,

#### Answer:

In the given problem, we need to find the sum of the *n* terms of the given A.P. “”.

So, here we use the following formula for the sum of *n* terms of an A.P.,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

For the given A.P. (),

Common difference of the A.P. (*d*) =

Number of terms* *(*n*) = *n*

First term for the given A.P. (*a*) = 5

So, using the formula we get,

Therefore, the sum of first *n* terms for the given A.P. is.

#### Page No 5.51:

#### Question 3:

Find the sum of *n* terms of an A.P. whose *nth *terms is given by an = 5 − 6n.

#### Answer:

Here, we are given an A.P., whose *n*^{th} term is given by the following expression,

So, here we can find the sum of the *n* terms of the given A.P., using the formula,

Where, *a* = the first term

*l* = the last term

So, for the given A.P,

The first term (*a*) will be calculated using in the given equation for *n*^{th} term of A.P.

Now, the last term (*l*) or the *n*^{th} term is given

So, on substituting the values in the formula for the sum of *n* terms of an A.P., we get,

Therefore, the sum of the *n* terms of the given A.P. is.

#### Page No 5.51:

#### Question 4:

Find the sum of last ten terms of the A.P. : 8, 10, 12, 14,...., 126.

#### Answer:

The given A.P is 8, 10, 12, 14,...., 126.

a = 8 and d = 2.

When this A.P is reversed, we get the A.P.

126, 124, 122, 120,....

So, first term becomes 126 and common difference −2.

The sum of first 10 terms of this A.P is as follows:

${S}_{10}=\frac{10}{2}\left[2\times 126+9\left(-2\right)\right]\phantom{\rule{0ex}{0ex}}=5\left[234\right]\phantom{\rule{0ex}{0ex}}=1170$

#### Page No 5.51:

#### Question 5:

Find the sum of the first 15 terms of each of the following sequences having nth term as

(i) *a*_{n }= 3 + 4*n*

(ii) *b*_{n} = 5 + 2*n*

(iii) *x*_{n} = 6 − *n*

(iv) *y*_{n} = 9 − 5*n*

#### Answer:

(i) Here, we are given an A.P. whose *n*^{th} term is given by the following expression, . We need to find the sum of first 15 terms.

So, here we can find the sum of the *n* terms of the given A.P., using the formula,

Where, *a* = the first term

*l* = the last term

So, for the given A.P,

The first term (*a*) will be calculated using in the given equation for *n*^{th} term of A.P.

Now, the last term (*l*) or the *n*^{th} term is given

So, on substituting the values in the formula for the sum of *n* terms of an A.P., we get,

Therefore, the sum of the 15 terms of the given A.P. is.

(ii) Here, we are given an A.P. whose *n*^{th} term is given by the following expression

We need to find the sum of first 15 terms.

So, here we can find the sum of the *n* terms of the given A.P., using the formula,

Where, *a* = the first term

*l* = the last term

So, for the given A.P,

The first term (*a*) will be calculated using in the given equation for *n*^{th} term of A.P.

Now, the last term (*l*) or the *n*^{th} term is given

So, on substituting the values in the formula for the sum of *n* terms of an A.P., we get,

Therefore, the sum of the 15 terms of the given A.P. is.

(iii) Here, we are given an A.P. whose *n*^{th} term is given by the following expression, . We need to find the sum of first 15 terms.

So, here we can find the sum of the *n* terms of the given A.P., using the formula,

Where, *a* = the first term

*l* = the last term

So, for the given A.P,

The first term (*a*) will be calculated using in the given equation for *n*^{th} term of A.P.

Now, the last term (*l*) or the *n*^{th} term is given

So, on substituting the values in the formula for the sum of *n* terms of an A.P., we get,

Therefore, the sum of the 15 terms of the given A.P. is.

(iv) Here, we are given an A.P. whose *n*^{th} term is given by the following expression, . We need to find the sum of first 15 terms.

So, here we can find the sum of the *n* terms of the given A.P., using the formula,

Where, *a* = the first term

*l* = the last term

So, for the given A.P,

The first term (*a*) will be calculated using in the given equation for *n*^{th} term of A.P.

Now, the last term (*l*) or the *n*^{th} term is given

So, on substituting the values in the formula for the sum of *n* terms of an A.P., we get,

Therefore, the sum of the 15 terms of the given A.P. is.

#### Page No 5.51:

#### Question 6:

Find the sum of first 20 terms of the sequence whose *n*th term is *a*_{n} = An + B.

#### Answer:

Here, we are given an A.P. whose *n*^{th} term is given by the following expression. We need to find the sum of first 20 terms.

So, here we can find the sum of the *n* terms of the given A.P., using the formula,

Where, *a* = the first term

*l* = the last term

So, for the given A.P,

The first term (*a*) will be calculated using in the given equation for *n*^{th} term of A.P.

Now, the last term (*l*) or the *n*^{th} term is given

So, on substituting the values in the formula for the sum of *n* terms of an A.P., we get,

Therefore, the sum of the first 20 terms of the given A.P. is.

#### Page No 5.51:

#### Question 7:

Find the sum of the first 25 terms of an A.P. whose *n*th term is given by a_{n} = 2 − 3*n*.

#### Answer:

Here, we are given an A.P. whose *n*^{th} term is given by the following expression, . We need to find the sum of first 25 terms.

So, here we can find the sum of the *n* terms of the given A.P., using the formula,

Where, *a* = the first term

*l* = the last term

So, for the given A.P,

The first term (*a*) will be calculated using in the given equation for *n*^{th} term of A.P.

Now, the last term (*l*) or the *n*^{th} term is given

So, on substituting the values in the formula for the sum of *n* terms of an A.P., we get,

Therefore, the sum of the 25 terms of the given A.P. is.

#### Page No 5.51:

#### Question 8:

Find the sum of the first 25 terms of an A.P. whose *n*th term is given by a_{n }= 7 − 3*n*.

#### Answer:

Here, we are given an A.P. whose *n*^{th} term is given by the following expression, . We need to find the sum of first 25 terms.

So, here we can find the sum of the *n* terms of the given A.P., using the formula,

Where, *a* = the first term

*l* = the last term

So, for the given A.P,

The first term (*a*) will be calculated using in the given equation for *n*^{th} term of A.P.

Now, the last term (*l*) or the *n*^{th} term is given

So, on substituting the values in the formula for the sum of *n* terms of an A.P., we get,

Therefore, the sum of the 25 terms of the given A.P. is.

#### Page No 5.51:

#### Question 9:

If the sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, ..., is 116. Find the last term.

#### Answer:

In the given problem, we have the sum of the certain number of terms of an A.P. and we need to find the last term for that A.P.

So here, let us first find the number of terms whose sum is 116. For that, we will use the formula,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

So for the given A.P

The first term (*a*) = 25

The sum of* n* terms

Common difference of the A.P. (*d*) =

So, on substituting the values in the formula for the sum of *n* terms of an A.P., we get,

So, we get the following quadratic equation,

$3{n}^{2}-53n+232=0$

On solving by splitting the middle term, we get,

$3{n}^{2}-24n-29n+232=0\phantom{\rule{0ex}{0ex}}3n\left(n-8\right)-29\left(n-8\right)=0\phantom{\rule{0ex}{0ex}}\left(3n-29\right)\left(n-8\right)=0$

Further,

Also,

Now, since *n* cannot be a fraction, so the number of terms is 8.

So, the term is *a*_{8}

Therefore, the last term of the given A.P. such that the sum of the terms is 116 is.

#### Page No 5.51:

#### Question 10:

(i) How many terms of the sequence 18, 16, 14, ... should be taken so that their sum is zero?

(ii) How many terms are there in the A.P. whose first and fifth terms are −14 and 2 respectively and the sum of the terms is 40?

(iii) How many terms of the A.P. 9, 17, 25,... must be taken so that their sum is 636?

(iv) How many terms of the A.P. 63, 60, 57, ... must be taken so that their sum is 693?

(v) How many terms of the A.P 27, 24, 21... should be taken so that their sum is zero?

(vi) How many terms of the A.P 45, 39, 33... must be taken so that their sum is 180?

Explain the double answer.

#### Answer:

In the given problem, we have the sum of the certain number of terms of an A.P. and we need to find the number of terms.

(i) A.P. is

So here, let us find the number of terms whose sum is 0. For that, we will use the formula,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

The first term (*a*) = 18

The sum of* n* terms (*S*_{n}) = 0

Common difference of the A.P. (*d*) =

So, on substituting the values in the formula for the sum of *n* terms of an A.P., we get,

Further,

Or,

Since, the number of terms cannot be zero, the number of terms (*n*) is.

(ii) Here, let us take the common difference as *d.*

So, we are given,

First term (*a*_{1}) = −14

Fifth term (*a*_{5}) = 2

Sum of terms (*s*_{n}) = 40

Now,

Further, let us find the number of terms whose sum is 40. For that, we will use the formula,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

The first term (*a*_{1}) = −14

The sum of* n* terms (*S*_{n}) = 40

Common difference of the A.P. (*d*) = 4

So, on substituting the values in the formula for the sum of *n* terms of an A.P., we get,

So, we get the following quadratic equation,

On solving by splitting the middle term, we get,

Further,

Or,

Since the number of terms cannot be negative. Therefore, the number of terms (*n*) is

(iii) A.P. is

So here, let us find the number of terms whose sum is 636. For that, we will use the formula,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

The first term (*a*) = 9

The sum of* n* terms (*S*_{n}) = 636

Common difference of the A.P. (*d*) =

So, on substituting the values in the formula for the sum of *n* terms of an A.P., we get,

So, we get the following quadratic equation,

On solving by splitting the middle term, we get,

Further,

Or,

Since, the number of terms cannot be a fraction, the number of terms (*n*) is.

(iv) A.P. is

So here, let us find the number of terms whose sum is 693. For that, we will use the formula,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

The first term (*a*) = 63

The sum of* n* terms (*S*_{n}) = 693

Common difference of the A.P. (*d*) =

So, on substituting the values in the formula for the sum of *n* terms of an A.P., we get,

So, we get the following quadratic equation,

On solving by splitting the middle term, we get,

Further,

Or,

Here, 22^{nd} term will be

So, the sum of 22 as well as 21 terms is 693. Therefore, the number of terms (*n*) is.

(v)

The given AP is 27, 24, 21, ...

First term of the AP = 27

Common difference = 24 − 27 = −3

Let the sum of the first *x* terms of the AP be 0.

Sum of first *x* terms = $\frac{x}{2}\left[2\times 27+\left(x-1\right)\left(-3\right)\right]=0$

$\Rightarrow \frac{x}{2}\left[54+\left(-3x+3\right)\right]=0\phantom{\rule{0ex}{0ex}}\Rightarrow x\left(54-3x+3\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x\left(57-3x\right)=0$

Now, either *x* = 0 or 57 − 3*x* = 0.

Since the number of terms cannot be 0, $x\ne 0$.

∴ 57 − 3*x* = 0

⇒ 57 = 3*x*

⇒ *x* = 19

Thus, the sum of the first 19 terms of the AP is 0.

$\mathrm{Given}:\phantom{\rule{0ex}{0ex}}a=27\phantom{\rule{0ex}{0ex}}d=24-27\phantom{\rule{0ex}{0ex}}=-3\phantom{\rule{0ex}{0ex}}{S}_{n}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}{S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 0=\frac{n}{2}\left[2\left(27\right)+\left(n-1\right)\left(-3\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 0=n\left[2\left(27\right)+\left(n-1\right)\left(-3\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow n\left[54-3n+3\right]=0\phantom{\rule{0ex}{0ex}}\Rightarrow n\left(57-3n\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow n=0\mathrm{or}3n=57\phantom{\rule{0ex}{0ex}}\mathrm{But}n\mathrm{is}\mathrm{a}\mathrm{natural}\mathrm{number},\therefore n\ne 0\phantom{\rule{0ex}{0ex}}\Rightarrow n=19$

Hence, 19 terms of the A.P 27, 24, 21... should be taken so that their sum is zero.

#### Page No 5.51:

#### Question 11:

Find the sum of the first

(i) 11 terms of the A.P.2, 6, 10. 14

(ii) 13 terms of the A.P. −6, 0, 6, 12, ...

(iii) 51 terms of the A.P. : whose second term is 2 and fourth term is 8.

#### Answer:

In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of *n* terms of an A.P.,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

(i) To 11 terms.

Common difference of the A.P. (*d*) =

= 6 − 2

= 4

Number of terms* *(*n*) = 11

First term for the given A.P. (*a*) = 2

So, using the formula we get,

Therefore, the sum of first 11 terms for the given A.P. is.

(ii) To 13 terms.

Common difference of the A.P. (*d*) =

= 0 − (−6)

= 6

Number of terms* *(*n*) = 13

First term for the given A.P. (*a*) = −6

So, using the formula we get,

Therefore, the sum of first 13 terms for the given A.P. is.

(iii) 51 terms of an A.P whose and

Now,

Also,

${a}_{4}=a+3d\phantom{\rule{0ex}{0ex}}8=a+3d...\left(2\right)$

Subtracting (1) from (2), we get

Further substituting in (1), we get

Number of terms* *(*n*) = 51

First term for the given A.P. (*a*) = −1

So, using the formula we get,

Therefore, the sum of first 51 terms for the given A.P. is.

#### Page No 5.51:

#### Question 12:

Find the sum of

(i) the first 15 multiples of 8

(ii) the first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.

(iii) all 3 − digit natural numbers which are divisible by 13.

(iv) all 3 − digit natural numbers, which are multiples of 11.

(v) all 2 − digit natural numbers divisible by 4.

(vi) first 8 multiples of 3.

#### Answer:

In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of *n* terms of an A.P.,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

(i) First 15 multiples of 8.

So, we know that the first multiple of 8 is 8 and the last multiple of 8 is 120.

Also, all these terms will form an A.P. with the common difference of 8.

So here,

First term (*a*) = 8

Number of terms (*n*) = 15

Common difference (*d*) = 8

Now, using the formula for the sum of *n* terms, we get

Therefore, the sum of the first 15 multiples of 8 is

(ii) (a) First 40 positive integers divisible by 3

So, we know that the first multiple of 3 is 3 and the last multiple of 3 is 120.

Also, all these terms will form an A.P. with the common difference of 3.

So here,

First term (*a*) = 3

Number of terms (*n*) = 40

Common difference (*d*) = 3

Now, using the formula for the sum of *n* terms, we get

Therefore, the sum of first 40 multiples of 3 is

(b) First 40 positive integers divisible by 5

So, we know that the first multiple of 5 is 5 and the last multiple of 5 is 200.

Also, all these terms will form an A.P. with the common difference of 5.

So here,

First term (*a*) = 5

Number of terms (*n*) = 40

Common difference (*d*) = 5

Now, using the formula for the sum of *n* terms, we get

Therefore, the sum of first 40 multiples of 3 is

(c) First 40 positive integers divisible by 6

So, we know that the first multiple of 6 is 6 and the last multiple of 6 is 240.

Also, all these terms will form an A.P. with the common difference of 6.

So here,

First term (*a*) = 6

Number of terms (*n*) = 40

Common difference (*d*) = 6

Now, using the formula for the sum of *n* terms, we get

Therefore, the sum of first 40 multiples of 3 is

(iii) All 3 digit natural number which are divisible by 13

So, we know that the first 3 digit multiple of 13 is 104 and the last 3 digit multiple of 13 is 988.

Also, all these terms will form an A.P. with the common difference of 13.

So here,

First term (*a*) = 104

Last term (*l*) = 988

Common difference (*d*) = 13

So, here the first step is to find the total number of terms. Let us take the number of terms as *n.*

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of *n* terms, we get

On further simplification, we get,

Therefore, the sum of all the 3 digit multiples of 13 is.

(iv) all 3-digit natural numbers, which are multiples of 11.

We know that the first 3 digit number multiple of 11 will be 110.

Last 3 digit number multiple of 11 will be 990.

So here,

First term (*a*) = 110

Last term (*l*) = 990

Common difference (*d*) = 11

So, here the first step is to find the total number of terms. Let us take the number of terms as *n.*

Now, as we know,

So, for the last term,

$990=110+(n-1)11\phantom{\rule{0ex}{0ex}}990=110+11n-11\phantom{\rule{0ex}{0ex}}990=99+11n\phantom{\rule{0ex}{0ex}}891=11n\phantom{\rule{0ex}{0ex}}81=n$

Now, using the formula for the sum of *n* terms, we get

${S}_{n}=\frac{81}{2}\left[2\left(110\right)+\left(81-1\right)11\right]\phantom{\rule{0ex}{0ex}}{S}_{n}=\frac{81}{2}\left[220+80\times 11\right]\phantom{\rule{0ex}{0ex}}{S}_{n}=\frac{81}{2}\times 1100\phantom{\rule{0ex}{0ex}}{S}_{n}=81\times 550\phantom{\rule{0ex}{0ex}}{S}_{n}=44550$

Therefore, the sum of all the 3 digit multiples of 11 is 44550.

(v) 2-digit no. divisible by 4 are 12,16,20,........,96

We can see it forms an AP as the common difference is 4 and the first term is 4.

To find no. of terms n,

we know that

$96=12+(n-1)4\phantom{\rule{0ex}{0ex}}84=(n-1)4\phantom{\rule{0ex}{0ex}}21=n-1\phantom{\rule{0ex}{0ex}}22=n$

Now,

First term (*a*) = 12

Number of terms (*n*) = 22

Common difference (*d*) =4

Now, using the formula for the sum of *n* terms, we get

${S}_{22}=\frac{22}{2}\left\{2\left(12\right)+(22-1)4\right\}\phantom{\rule{0ex}{0ex}}{S}_{22}=11\left\{24+84\right\}\phantom{\rule{0ex}{0ex}}{S}_{22}=1188$

Hence, the sum of 22 terms is 1188 which are divisible by 4.

(vi)

First 8 multiples of 3 are { 3, 6, 9...,24}

We can observe they are in AP with first term (*a*) = 3 and last term (*l*) = 24 and number of terms are 8.

${S}_{n}=\frac{n}{2}\left(a+l\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {S}_{n}=\frac{8}{2}\left(3+24\right)\phantom{\rule{0ex}{0ex}}{S}_{8}=4\times \left(3+24\right)=108$

Hence, the sum of the first 8 multiples of 3 is 108.

#### Page No 5.51:

#### Question 13:

Find the sum:

(i) 2 + 4 + 6 ... + 200

(ii) 3 + 11 + 19 + ... + 803

(iii) (−5) + (−8)+ (−11) + ... + (−230)

(iv) 1 + 3 + 5 + 7 + ... + 199

(v) $7+10\frac{1}{2}+14+...+84$

(vi) 34 + 32 + 30 + ... + 10

(vii) 25 + 28 + 31 + ... + 100

(viii) $18+15\frac{1}{2}+13+...+\left(-49\frac{1}{2}\right)$

#### Answer:

*n* terms of an A.P.,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

(i)

Common difference of the A.P. (*d*) =

So here,

First term (*a*) = 2

Last term (*l*) = 200

Common difference (*d*) = 2

So, here the first step is to find the total number of terms. Let us take the number of terms as *n.*

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of *n* terms, we get

On further simplification, we get,

Therefore, the sum of the A.P is

(ii)

Common difference of the A.P. (*d*) =

So here,

First term (*a*) = 3

Last term (*l*) = 803

Common difference (*d*) = 8

So, here the first step is to find the total number of terms. Let us take the number of terms as *n.*

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of *n* terms, we get

Therefore, the sum of the A.P is

(iii)

Common difference of the A.P. (*d*) =

So here,

First term (*a*) = −5

Last term (*l*) = −230

Common difference (*d*) = −3

So, here the first step is to find the total number of terms. Let us take the number of terms as *n.*

Now, as we know,

So, for the last term,

Now, using the formula for the sum of *n* terms, we get

Therefore, the sum of the A.P is

(iv)

Common difference of the A.P. (*d*) =

So here,

First term (*a*) = 1

Last term (*l*) = 199

Common difference (*d*) = 2

So, here the first step is to find the total number of terms. Let us take the number of terms as *n.*

Now, as we know,

So, for the last term,

Now, using the formula for the sum of *n* terms, we get

On further simplification, we get,

Therefore, the sum of the A.P is

(v)

Common difference of the A.P is

(*d*) =

So here,

First term (*a*) = 7

Last term (*l*) = 84

Common difference (*d*) =

So, here the first step is to find the total number of terms. Let us take the number of terms as *n.*

Now, as we know,

So, for the last term,

Further solving for *n,*

Now, using the formula for the sum of *n* terms, we get

On further simplification, we get,

Therefore, the sum of the A.P is

(vi)

Common difference of the A.P. (*d*) =

So here,

First term (*a*) = 34

Last term (*l*) = 10

Common difference (*d*) = −2

So, here the first step is to find the total number of terms. Let us take the number of terms as *n.*

Now, as we know,

So, for the last term,

Further solving for *n,*

Now, using the formula for the sum of *n* terms, we get

On further simplification, we get,

Therefore, the sum of the A.P is

(vii)

Common difference of the A.P. (*d*) =

So here,

First term (*a*) = 25

Last term (*l*) = 100

Common difference (*d*) = 3

So, here the first step is to find the total number of terms. Let us take the number of terms as *n.*

Now, as we know,

So, for the last term,

Further solving for *n,*

Now, using the formula for the sum of *n* terms, we get

On further simplification, we get,

Therefore, the sum of the A.P is.

(viii) $18+15\frac{1}{2}+13+...+\left(-49\frac{1}{2}\right)$

Common difference of the A.P. (*d*) =

= $=15\frac{1}{2}-18\phantom{\rule{0ex}{0ex}}=\frac{31}{2}-18\phantom{\rule{0ex}{0ex}}=\frac{31-36}{2}\phantom{\rule{0ex}{0ex}}=\frac{-5}{2}$

So here,

First term (*a*) = 18

Last term (*l*) = $-49\frac{1}{2}=\frac{-99}{2}$

Common difference (*d*) = $\frac{-5}{2}$

So, here the first step is to find the total number of terms. Let us take the number of terms as *n.*

Now, as we know,

So, for the last term,

$\frac{-99}{2}=18+\left(n-1\right)\frac{-5}{2}\phantom{\rule{0ex}{0ex}}\frac{-99}{2}=18+\left(\frac{-5}{2}\right)n+\frac{5}{2}\phantom{\rule{0ex}{0ex}}\frac{5}{2}n=18+\frac{5}{2}+\frac{99}{2}\phantom{\rule{0ex}{0ex}}\frac{5}{2}n=18+\frac{104}{2}\phantom{\rule{0ex}{0ex}}n=28$

Now, using the formula for the sum of *n* terms, we get

${S}_{n}=\frac{28}{2}\left[2\times 18+\left(28-1\right)\left(\frac{-5}{2}\right)\right]\phantom{\rule{0ex}{0ex}}{S}_{n}=14\left[36+27\left(\frac{-5}{2}\right)\right]\phantom{\rule{0ex}{0ex}}{S}_{n}=-441\phantom{\rule{0ex}{0ex}}$

Therefore, the sum of the A.P is ${S}_{n}=-441$.

#### Page No 5.52:

#### Question 14:

The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

#### Answer:

In the given problem, we have the first and the last term of an A.P. along with the common difference of the A.P. Here, we need to find the number of terms of the A.P. and the sum of all the terms.

Here,

The first term of the A.P (*a*) = 17

The last term of the A.P (*l*) = 350

The common difference of the A.P. = 9

Let the number of terms be *n*.

So, as we know that,

We get,

Further solving this,

Using the above values in the formula,

Therefore, the number of terms is and the sum.

#### Page No 5.52:

#### Question 15:

The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. Find the first term, the common difference and the sum of first 20 terms.

#### Answer:

In the given problem, let us take the first term as *a* and the common difference as *d*.

Here, we are given that,

So, using (1) in (2), we get,

Also, we know,

For the 3^{th} term (*n = *3),

Similarly, for the 7^{th} term (*n = *7),

Subtracting (4) from (5), we get,

Now, to find *a*, we substitute the value of *d* in (4),

So, for the given A.P

So, to find the sum of first 20 terms of this A.P., we use the following formula for the sum of *n* terms of an A.P.,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

So, using the formula for *n* = 20, we get,

Therefore, the sum of first 20 terms for the given A.P. is .

#### Page No 5.52:

#### Question 16:

The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference.

#### Answer:

In the given problem, we have the first and the last term of an A.P. along with the sum of all the terms of A.P. Here, we need to find the common difference of the A.P.

Here,

The first term of the A.P (*a*) = 2

The last term of the A.P (*l*) = 50

Sum of all the terms

Let the common difference of the A.P. be *d*.

So, let us first find the number of the terms (*n*) using the formula,

Now, to find the common difference of the A.P. we use the following formula,

We get,

Further, solving for *d,*

Therefore the common difference of the A.P.

#### Page No 5.52:

#### Question 17:

If 12th term of an A.P. is −13 and the sum of the first four terms is 24, what is the sum of first 10 terms.

#### Answer:

In the given problem, we need to find the sum of first 10 terms of an A.P. Let us take the first term *a* and the common difference as *d*

Here, we are given that,

Also, we know,

For the 12^{th} term (*n = *12),

So, as we know the formula for the sum of *n* terms of an A.P. is given by,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

So, using the formula for *n* = 4, we get,

Subtracting (1) from (2), we get,

On further simplifying for *d, *we get,

Now, to find *a*, we substitute the value of *d* in (1),

Now, using the formula for the sum of *n* terms of an A.P. for *n *= 10, we get,

Therefore, the sum of first 10 terms for the given A.P. is .

#### Page No 5.52:

#### Question 18:

Find the sum of *n* terms of the series $\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+..........$

#### Answer:

Let the given series be S = $\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+..........$

$=\left[4+4+4+...\right]-\left[\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+...\right]\phantom{\rule{0ex}{0ex}}=4\left[1+1+1+...\right]-\frac{1}{n}\left[1+2+3+...\right]\phantom{\rule{0ex}{0ex}}={S}_{1}-{S}_{2}$

${S}_{1}=4\left[1+1+1+...\right]\phantom{\rule{0ex}{0ex}}a=1,d=0\phantom{\rule{0ex}{0ex}}{S}_{1}=4\times \frac{n}{2}\left[2\times 1+\left(n-1\right)\times 0\right]\left({S}_{n}=\frac{n}{2}\left(2a+\left(n-1\right)d\right)\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {S}_{1}=4n$

${S}_{2}=\frac{1}{n}\left[1+2+3+...\right]\phantom{\rule{0ex}{0ex}}a=1,d=2-1=1\phantom{\rule{0ex}{0ex}}{S}_{2}=\frac{1}{n}\times \frac{n}{2}\left[2\times 1+\left(n-1\right)\times 1\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[2+n-1\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[1+n\right]$

$\mathrm{Thus},S={S}_{1}-{S}_{2}=4n-\frac{1}{2}\left[1+n\right]\phantom{\rule{0ex}{0ex}}S=\frac{8n-1-n}{2}=\frac{7n-1}{2}$

Hence, the sum of n terms of the series is $\frac{7n-1}{2}$.

#### Page No 5.52:

#### Question 19:

In an A.P., if the first term is 22, the common difference is −4 and the sum to *n *terms is 64, find *n*.

#### Answer:

In the given problem, we need to find the number of terms of an A.P. Let us take the number of terms as *n*.

Here, we are given that,

So, as we know the formula for the sum of *n* terms of an A.P. is given by,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

So, using the formula we get,

Further rearranging the terms, we get a quadratic equation,

On taking 4 common, we get,

Further, on solving the equation for *n* by splitting the middle term, we get,

So, we get,

Also,

Therefore,

#### Page No 5.52:

#### Question 20:

In an A.P., if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms?

#### Answer:

In the given problem, let us take the first term as *a* and the common difference *d*

Here, we are given that,

Also, we know,

For the 5^{th} term (*n = *5),

Similarly, for the 12^{th} term (*n = *12),

Subtracting (3) from (4), we get,

Now, to find *a*, we substitute the value of *d* in (4),

So, for the given A.P

So, to find the sum of first 20 terms of this A.P., we use the following formula for the sum of *n* terms of an A.P.,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

So, using the formula for *n* = 20, we get,

Therefore, the sum of first 20 terms for the given A.P. is.

#### Page No 5.52:

#### Question 21:

Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.

#### Answer:

In the given problem, let us take the first term as *a* and the common difference as *d*.

Here, we are given that,

Also, we know,

For the 2^{nd} term (*n = *2),

Similarly, for the 3^{rd} term (*n = *3),

Subtracting (3) from (4), we get,

Now, to find *a*, we substitute the value of *d* in (4),

So, for the given A.P

So, to find the sum of first 51 terms of this A.P., we use the following formula for the sum of *n* terms of an A.P.,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

So, using the formula for *n* = 51, we get,

Therefore, the sum of first 51 terms for the given A.P. is.

#### Page No 5.52:

#### Question 22:

(i) If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of *n* terms.

(ii) If the sum of first four terms of an A.P. is 40 and that of first 14 terms is 280. Find the sum of its first *n* terms.

#### Answer:

(i)

In the given problem, we need to find the sum of *n* terms of an A.P. Let us take the first term as *a* and the common difference as *d*.

Here, we are given that,

So, as we know the formula for the sum of *n* terms of an A.P. is given by,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

So, using the formula for *n* = 7, we get,

Further simplifying for *a, *we get,

Also, using the formula for *n* = 17, we get,

Further simplifying for *a, *we get,

Subtracting (3) from (4), we get,

Now, to find *a*, we substitute the value of *d* in (3),

Now, using the formula for the sum of *n* terms of an A.P., we get,

Therefore, the sum of first *n* terms for the given A.P. is.

${S}_{4}=40\phantom{\rule{0ex}{0ex}}{S}_{14}=280\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}{S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{S}_{4}=\frac{4}{2}\left[2a+\left(4-1\right)d\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 40=2\left[2a+3d\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 2a+3d=20...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{S}_{14}=\frac{14}{2}\left[2a+\left(14-1\right)d\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 280=7\left[2a+13d\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 2a+13d=40...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Subtracting}\left(1\right)\mathrm{from}\left(2\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}10d=20\phantom{\rule{0ex}{0ex}}\Rightarrow d=2\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}d\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}a=7\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},a=7\mathrm{and}d=2\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\phantom{\rule{0ex}{0ex}}{S}_{n}=\frac{n}{2}\left(2\times 7+\left(n-1\right)2\right)\phantom{\rule{0ex}{0ex}}=\frac{n}{2}\left(14+2n-2\right)\phantom{\rule{0ex}{0ex}}=\frac{n}{2}\left(12+2n\right)\phantom{\rule{0ex}{0ex}}=n\left(6+n\right)\phantom{\rule{0ex}{0ex}}={n}^{2}+6n$

Hence, the sum of its first

*n*terms is

*n*

^{2}+ 6

*n*.

#### Page No 5.52:

#### Question 23:

The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

#### Answer:

In the given problem, we have the first and the last term of an A.P. along with the sum of all the terms of A.P. Here, we need to find the number of terms and the common difference of the A.P.

Here,

The first term of the A.P (*a*) = 5

The last term of the A.P (*l*) = 45

Sum of all the terms

Let the common difference of the A.P. be *d*.

So, let us first find the number of the terms (*n*) using the formula,

Now, to find the common difference of the A.P. we use the following formula,

We get,

Further, solving for *d,*

Therefore, the number of terms is and the common difference of the A.P is.

#### Page No 5.52:

#### Question 24:

In an A.P. the first term is 8, *n*th term is 33 and the sum to first *n* terms is 123. Find *n* and *d*, the common differences.

#### Answer:

In the given problem, we have the first and the *n*th term of an A.P. along with the sum of the *n* terms of A.P. Here, we need to find the number of terms and the common difference of the A.P.

Here,

The first term of the A.P (*a*) = 8

The *n*th term of the A.P (*l*) = 33

Sum of all the terms

Let the common difference of the A.P. be *d*.

So, let us first find the number of the terms (*n*) using the formula,

Now, to find the common difference of the A.P. we use the following formula,

We get,

Further, solving for *d,*

Therefore, the number of terms is and the common difference of the A.P..

#### Page No 5.52:

#### Question 25:

In an A.P., the first term is 22, nth term is −11 and the sum to first n terms is 66. Find n and d, the common difference

#### Answer:

In the given problem, we have the first and the *n*th term of an A.P. along with the sum of the *n* terms of A.P. Here, we need to find the number of terms and the common difference of the A.P.

Here,

The first term of the A.P (*a*) = 22

The *n*th term of the A.P (*l*) = −11

Sum of all the terms

Let the common difference of the A.P. be *d*.

So, let us first find the number of the terms (*n*) using the formula,

Further, solving for *n*

Now, to find the common difference of the A.P. we use the following formula,

We get,

Further, solving for *d,*

Therefore, the number of terms is and the common difference of the A.P..

#### Page No 5.52:

#### Question 26:

The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find its common difference.

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, sum of first* n*^{ }terms = *S*_{n }= $\frac{n}{2}$[2*a* + (*n* − 1)*d*]

Also, *n*^{th} term = *a _{n}* =

*a*+ (

*n*− 1)

*d*

According to the question,

*a*= 7,

*a*= 49 and

_{n}*S*= 420

_{n}Now,

*a*=

_{n}*a*+ (

*n*− 1)

*d*

⇒49 = 7 + (

⇒

*n*− 1)

*d*

*⇒*42 =

*nd*−

*d*

⇒

⇒

*nd*−

*d*= 42 ....(1)

Also,

*S*

_{n }= $\frac{n}{2}$[2 × 7 + (

*n*− 1)

*d*]

⇒ 420 = $\frac{n}{2}$[14 +

*nd*−

*d*]

⇒ 840 =

*n*[14 + 42] [From (1)]

⇒ 56

*n*= 840

⇒

*n*= 15 ....(2)

On substituting (2) in (1), we get

*nd*−

*d*= 42

⇒ (15 − 1)

*d*= 42

⇒ 14

*d*= 42

⇒

*d*= 3

Thus, common difference of the given A.P. is 3.

#### Page No 5.52:

#### Question 27:

The first and the last terms of an A.P. are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, sum of first* n*^{ }terms = *S*_{n }= $\frac{n}{2}$[2*a* + (*n* − 1)*d*]

Also, *n*^{th} term = *a _{n}* =

*a*+ (

*n*− 1)

*d*

According to the question,

*a*= 5,

*a*= 45 and

_{n}*S*= 400

_{n}Now,

*a*=

_{n}*a*+ (

*n*− 1)

*d*

⇒45 = 5 + (

⇒

*n*− 1)

*d*

*⇒*40 =

*nd*−

*d*

⇒

⇒

*nd*−

*d*= 40 ....(1)

Also,

*S*

_{n }= $\frac{n}{2}$[2 × 5 + (

*n*− 1)

*d*]

⇒ 400 = $\frac{n}{2}$[10 +

*nd*−

*d*]

⇒ 800 =

*n*[10 + 40] [From (1)]

⇒ 50

*n*= 800

⇒

*n*= 16 ....(2)

On substituting (2) in (1), we get

*nd*−

*d*= 40

⇒ (16 − 1)

*d*= 40

⇒ 15

*d*= 40

⇒

*d*= $\frac{8}{3}$

Thus, common difference of the given A.P. is $\frac{8}{3}$.

#### Page No 5.52:

#### Question 28:

The sum of first 9 terms of an A.P. is 162. The ratio of its 6^{th} term to its 13^{th} term is 1 : 2. Find the first and 15^{th} term of the A.P.

**Disclaimer: **There is a misprint in the question, 'the sum of 9 terms' should be witten instead of 'the sum of *q* terms'.

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, sum of first* n*^{ }terms = *S*_{n }= $\frac{n}{2}$[2*a* + (*n* − 1)*d*]

Also, *n*^{th} term = *a _{n}* =

*a*+ (

*n*− 1)

*d*

According to the question,

*S*= 162 and $\frac{{a}_{6}}{{a}_{13}}=\frac{1}{2}$

_{q}Now,

$\frac{{a}_{6}}{{a}_{13}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a+(6-1)d}{a+(13-1)d}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a+5d}{a+12d}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 2a+10d=a+12d\phantom{\rule{0ex}{0ex}}\Rightarrow 2a-a=12d-10d\phantom{\rule{0ex}{0ex}}\Rightarrow a=2d.....\left(1\right)$

Also,

*S*

_{9 }= $\frac{9}{2}$[2

*a*+ (9 − 1)

*d*]

⇒ 162 = $\frac{9}{2}$[2(2

*d*) + 8

*d*] [From (1)]

⇒ 18 = $\frac{1}{2}$[12

*d*]

⇒ 18 = 6

*d*

⇒

*d*= 3

⇒

*a*= 2 × 3 [From (1)]

⇒

*a*= 6

Thus, the first term of the A.P. is 6.

Now,

*a*

_{15}= 6 + (15 − 1)3

= 6 + 42

= 48

Thus, 15

^{th}term of the A.P. is 48.

#### Page No 5.52:

#### Question 29:

If the 10^{th} term of an A.P. is 21 and the sum of its first 10 terms is 120, find its *n*^{th }term.

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, sum of first* n*^{ }terms = *S*_{n }= $\frac{n}{2}$[2*a* + (*n* − 1)*d*]

and *n*^{th} term = *a _{n}* =

*a*+ (

*n*− 1)

*d*

Now,

*S*

_{10 }= $\frac{10}{2}$[2

*a*+ (10 − 1)

*d*]

⇒ 120 = 5(2

*a*+ 9

*d*)

⇒ 24 = 2

*a*+ 9

*d*

⇒ 2

*a*+ 9

*d*= 24 ....(1)

Also,

*a*

_{10}=

*a*+ (10 − 1)

*d*

⇒ 21 =

*a*+ 9

*d*

⇒ 2

*a*+ 18

*d*= 42 ....(2)

Subtracting (1) from (2), we get

18

*d −*9

*d*= 42 − 24

⇒ 9

*d*= 18

⇒

*d*= 2

⇒ 2

*a*= 24 − 9

*d*[From (1)]

⇒ 2

*a*= 24 − 9 × 2

⇒ 2

*a*= 24 − 18

⇒ 2

*a*= 6

⇒

*a*= 3

Also,

*a*=

_{n}*a*+ (

*n*− 1)

*d*

=3 + (

=

*n*− 1)2

*=*3 + 2

*n*− 2

*=*1 + 2

*n*

Thus,

*n*

^{th}term of this A.P. is 1 + 2

*n.*

#### Page No 5.52:

#### Question 30:

The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28^{th} term of this A.P.

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, sum of first* n*^{ }terms = *S*_{n }= $\frac{n}{2}$[2*a* + (*n* − 1)*d*]

It is given that sum of the first 7 terms of an A.P. is 63.

And sum of next 7 terms is 161.

∴ Sum of first 14 terms = Sum of first 7 terms + sum of next 7 terms

= 63 + 161 = 224

Now,

*S*_{7 }= $\frac{7}{2}$[2*a* + (7 − 1)*d*]

⇒ 63 = $\frac{7}{2}$(2*a* + 6*d*)

⇒ 18 = 2*a* + 6*d*

⇒ 2*a* + 6*d = *18 ....(1)

Also,

*S*_{14 }= $\frac{14}{2}$[2*a* + (14 − 1)*d*]

⇒ 224 = 7(2*a* + 13*d*)

⇒ 32 = 2*a* + 13*d*

⇒ 2*a* + 13*d = * 32 ....(2)

On subtracting (1) from (2), we get

13*d − *6*d = *32 − 18

⇒ 7*d = *14

⇒ *d = *2

⇒ 2*a = *18 − 6*d *[From (1)]

⇒ 2*a = *18 − 6 × 2

⇒ 2*a = *18 − 12

⇒ 2*a = *6

⇒ *a = *3

Also, *n*^{th} term = *a _{n}* =

*a*+ (

*n*− 1)

*d*

⇒

*a*

_{28}= 3 + (28 − 1)2

= 3 + 27 × 2

= 57

Thus, 28

^{th}term of this A.P. is 57

*.*

#### Page No 5.52:

#### Question 31:

The sum of first seven terms of an A.P. is 182. If its 4^{th} and the 17^{th} terms are in the ratio 1 : 5, find the A.P.

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, sum of first* n*^{ }terms = *S*_{n }= $\frac{n}{2}$[2*a* + (*n* − 1)*d*]

According to the question,

${S}_{7}=182\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{7}{2}\left[2a+\left(7-1\right)d\right]=182\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\left(2a+6d\right)=26\phantom{\rule{0ex}{0ex}}\Rightarrow a+3d=26\phantom{\rule{0ex}{0ex}}\Rightarrow a=26-3d....\left(1\right)$

Also,

$\frac{{a}_{4}}{{a}_{17}}=\frac{1}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a+(4-1)d}{a+(17-1)d}=\frac{1}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a+3d}{a+16d}=\frac{1}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow 5(a+3d)=a+16d\phantom{\rule{0ex}{0ex}}\Rightarrow 5a+15d=a+16d\phantom{\rule{0ex}{0ex}}\Rightarrow 5a-a=16d-15d\phantom{\rule{0ex}{0ex}}\Rightarrow 4a=d....\left(2\right)$

On substituting (2) in (1), we get

$a=26-3\left(4a\right)\phantom{\rule{0ex}{0ex}}\Rightarrow a=26-12a\phantom{\rule{0ex}{0ex}}\Rightarrow 12a+a=26\phantom{\rule{0ex}{0ex}}\Rightarrow 13a=26\phantom{\rule{0ex}{0ex}}\Rightarrow a=2\phantom{\rule{0ex}{0ex}}\Rightarrow d=4\times 2\left[\mathrm{From}\left(2\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow d=8$

Thus, the A.P. is 2, 10, 18, 26, ..... .

#### Page No 5.52:

#### Question 32:

The *n*th term of an A.P is given by (−4*n* + 15), Find the sum of first 20 terms of this A.P.

#### Answer:

${a}_{n}=-4n+15\phantom{\rule{0ex}{0ex}}\Rightarrow {a}_{1}=-4+15=11\phantom{\rule{0ex}{0ex}}\mathrm{Also},{a}_{2}=-8+15=7\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Common}\mathrm{difference},d={a}_{2}-{a}_{1}=7-11=-4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}{S}_{20}=\frac{20}{2}\left[2\times 11+\left(20-1\right)\left(-4\right)\right]\phantom{\rule{0ex}{0ex}}=10\left(22-76\right)\phantom{\rule{0ex}{0ex}}=-540$

#### Page No 5.52:

#### Question 33:

In an A.P., the sum of first ten terms is −150 and the sum of its next ten terms is −550. Find the A.P.

#### Answer:

Here, we are given and sum of the next ten terms is −550.

Let us take the first term of the A.P. as *a* and the common difference as *d.*

So, let us first find *a*_{10}. For the sum of first 10 terms of this A.P,

First term = *a*

Last term = *a*_{10}

So, we know,

For the 10^{th} term (*n = *10),

So, here we can find the sum of the *n* terms of the given A.P., using the formula,

Where, *a* = the first term

*l* = the last term

So, for the given A.P,

Similarly, for the sum of next 10 terms (*S*_{10}),

First term = *a*_{11}

Last term = *a*_{20}

For the 11^{th} term (*n = *11),

For the 20^{th} term (*n = *20),

So, for the given A.P,

Now, subtracting (1) from (2),

Substituting the value of *d *in (1)

So, the A.P. is with .

#### Page No 5.52:

#### Question 34:

Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25^{th} term.

#### Answer:

First term, *a* = 10

Sum of first 14 terms, ${S}_{14}=1505$

$\Rightarrow \frac{14}{2}\left[2\times 10+\left(14-1\right)d\right]=1505\phantom{\rule{0ex}{0ex}}\Rightarrow 7\times \left(20-13d\right)=1505\phantom{\rule{0ex}{0ex}}\Rightarrow 20-13d=\frac{1505}{7}=215\phantom{\rule{0ex}{0ex}}\Rightarrow 13d=-195\phantom{\rule{0ex}{0ex}}\Rightarrow d=-15$

Now,

${a}_{25}=10+24\left(-15\right)=-350$

#### Page No 5.52:

#### Question 35:

In an A.P., the first term is 2, the last term is 29 and the sum of the terms is 155. Find the common difference of the A.P.

#### Answer:

In the given problem, we have the first and the last term of an A.P. along with the sum of all the terms of A.P. Here, we need to find the common difference of the A.P.

Here,

The first term of the A.P (*a*) = 2

The last term of the A.P (*l*) = 29

Sum of all the terms (*S*_{n}) = 155

Let the common difference of the A.P. be *d*.

So, let us first find the number of the terms (*n*) using the formula,

Now, to find the common difference of the A.P. we use the following formula,

We get,

Therefore, the common difference of the A.P. is.

#### Page No 5.53:

#### Question 36:

The first and the last term of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

#### Answer:

In the given problem, we have the first and the last term of an A.P. along with the common difference of the A.P. Here, we need to find the number of terms of the A.P. and the sum of all the terms.

Here,

The first term of the A.P (*a*) = 17

The last term of the A.P (*l*) = 350

The common difference of the A.P. = 9

Let the number of terms be *n*.

So, as we know that,

We get,

Further solving this,

Using the above values in the formula,

Therefore, the number of terms is and the sum.

#### Page No 5.53:

#### Question 37:

Find the number of terms of the A.P. −12, −9, −6, ..., 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained.

#### Answer:

First term, ${a}_{1}=-12$

Common difference, $d={a}_{2}-{a}_{1}=-9-\left(-12\right)=3$

${a}_{n}=21\phantom{\rule{0ex}{0ex}}\Rightarrow a+\left(n-1\right)d=21\phantom{\rule{0ex}{0ex}}\Rightarrow -12+\left(n-1\right)\times 3=21\phantom{\rule{0ex}{0ex}}\Rightarrow 3n=36\phantom{\rule{0ex}{0ex}}\Rightarrow n=12$

Therefore, number of terms in the given A.P. is 12.

Now, when 1 is added to each of the 12 terms, the sum will increase by 12.

So, the sum of all terms of the A.P. thus obtained

$={S}_{12}+12\phantom{\rule{0ex}{0ex}}=\frac{12}{2}\left[2\left(-12\right)+11\left(3\right)\right]+12\phantom{\rule{0ex}{0ex}}=6\times \left(9\right)+12\phantom{\rule{0ex}{0ex}}=66\phantom{\rule{0ex}{0ex}}$

#### Page No 5.53:

#### Question 38:

The sum of the first *n* terms of an A.P. is 3*n*^{2} + 6*n*. Find the *n*^{th} term of this A.P.

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, sum of first* n*^{ }terms = *S*_{n }= $\frac{n}{2}$[2*a* + (*n* − 1)*d*]

It is given that sum of the first *n* terms of an A.P. is 3*n*^{2} + 6*n*.

∴ First term = *a *= S_{1 }= 3(1)^{2} + 6(1) = 9.

Sum of first two terms = S_{2 }= 3(2)^{2} + 6(2) = 24.

∴ Second term = S_{2} − S_{1} = 24 − 9 = 15.

∴ Common difference = *d* = Second term − First term

= 15 − 9 = 6

Also, *n*^{th} term = *a _{n}* =

*a*+ (

*n*− 1)

*d*

⇒

*a*= 9 + (

_{n}*n*− 1)6

⇒

*a*= 9 + 6

_{n}*n*− 6

⇒

*a*= 3 + 6

_{n}*n*

Thus,

*n*

^{th}term of this A.P. is 3 + 6

*n.*

#### Page No 5.53:

#### Question 39:

The sum of first *n* terms of an A.P. is 5*n* − *n*^{2}. Find the *n*^{th} term of this A.P.

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, sum of first* n*^{ }terms = *S*_{n }= $\frac{n}{2}$[2*a* + (*n* − 1)*d*]

It is given that sum of the first *n* terms of an A.P. is 5*n* − *n*^{2}.

∴ First term = *a *= S_{1 }= 5(1) − (1)^{2} = 4.

Sum of first two terms = S_{2 }= 5(2) − (2)^{2} = 6.

∴ Second term = S_{2} − S_{1} = 6 − 4 = 2.

∴ Common difference = *d* = Second term − First term

= 2 − 4 = −2

Also, *n*^{th} term = *a _{n}* =

*a*+ (

*n*− 1)

*d*

⇒

*a*= 4 + (

_{n}*n*− 1)(−2)

⇒

*a*= 4 − 2

_{n}*n*+ 2

⇒

*a*= 6 − 2

_{n}*n*

Thus,

*n*

^{th}term of this A.P. is 6 − 2

*n.*

#### Page No 5.53:

#### Question 40:

The sum of the first *n* terms of an A.P. is 4*n*^{2} + 2*n*. Find the *n*^{th} term of this A.P.

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, sum of first* n*^{ }terms = *S*_{n }= $\frac{n}{2}$[2*a* + (*n* − 1)*d*]

It is given that sum of the first *n* terms of an A.P. is 4*n*^{2} + 2*n*.

∴ First term = *a *= S_{1 }= 4(1)^{2} + 2(1) = 6.

Sum of first two terms = S_{2 }= 4(2)^{2} + 2(2) = 20.

∴ Second term = S_{2} − S_{1} = 20 − 6 = 14.

∴ Common difference = *d* = Second term − First term

= 14 − 6 = 8

Also, *n*^{th} term = *a _{n}* =

*a*+ (

*n*− 1)

*d*

⇒

*a*= 6 + (

_{n}*n*− 1)(8)

⇒

*a*= 6 + 8

_{n}*n*− 8

⇒

*a*= 8

_{n}*n*− 2

Thus,

*n*

^{th}term of this A.P. is 8

*n*− 2

*.*

#### Page No 5.53:

#### Question 41:

The sum of first *n* terms of an A.P. is 3*n*^{2} + 4*n*. Find the 25th term of this A.P.

#### Answer:

${S}_{n}=3{n}^{2}+4n$

We know

${a}_{n}={S}_{n}-{S}_{n-1}\phantom{\rule{0ex}{0ex}}\therefore {a}_{n}=3{n}^{2}+4n-3{\left(n-1\right)}^{2}-4\left(n-1\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {a}_{n}=6n+1$

${a}_{25}=6\left(25\right)+1=151$

#### Page No 5.53:

#### Question 42:

The sum of first* n* terms of an A.P is 5n^{2} + 3n. If its *m*th terms is 168, find the value of *m*. Also, find the 20th term of this A.P.

#### Answer:

${S}_{n}=5{n}^{2}+3n\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{know}\phantom{\rule{0ex}{0ex}}{a}_{n}={S}_{n}-{S}_{n-1}\phantom{\rule{0ex}{0ex}}\therefore {a}_{n}=5{n}^{2}+3n-5{\left(n-1\right)}^{2}-3\left(n-1\right)\phantom{\rule{0ex}{0ex}}{a}_{n}=10n-2$

Now,

${a}_{m}=168\phantom{\rule{0ex}{0ex}}\Rightarrow 10m-2=168\phantom{\rule{0ex}{0ex}}\Rightarrow 10m=170\phantom{\rule{0ex}{0ex}}\Rightarrow m=17$

${a}_{20}=10\left(20\right)-2=198$

#### Page No 5.53:

#### Question 43:

The sum of first q terms of an A.P. is 63q − 3q^{2}. If its *p*th term is −60, find the value of *p*. Also, find the 11th term of this A.P.

#### Answer:

${S}_{q}=63q-3{q}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{know}\phantom{\rule{0ex}{0ex}}{a}_{q}={S}_{q}-{S}_{q-1}\phantom{\rule{0ex}{0ex}}\therefore {a}_{q}=63q-3{q}^{2}-63\left(q-1\right)+3{\left(q-1\right)}^{2}\phantom{\rule{0ex}{0ex}}{a}_{q}=66-6q\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},{a}_{p}=-60\phantom{\rule{0ex}{0ex}}\Rightarrow 66-6p=-60\phantom{\rule{0ex}{0ex}}\Rightarrow 126=6p\phantom{\rule{0ex}{0ex}}\Rightarrow p=21\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{a}_{11}=66-6\times 11=0$

#### Page No 5.53:

#### Question 44:

The sum of first m terms of an A.P. is 4*m*^{2} − *m*. If its nth term is 107. find the value of *n.* Also, find the 21st term of this A.P.

#### Answer:

${S}_{m}=4{m}^{2}-m\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{We}\mathrm{know}\phantom{\rule{0ex}{0ex}}{a}_{m}={S}_{m}-{S}_{m-1}\phantom{\rule{0ex}{0ex}}\therefore {a}_{m}=4{m}^{2}-m-4{\left(m-1\right)}^{2}+\left(m-1\right)\phantom{\rule{0ex}{0ex}}{a}_{m}=8m-5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}{a}_{n}=107\phantom{\rule{0ex}{0ex}}\Rightarrow 8n-5=107\phantom{\rule{0ex}{0ex}}\Rightarrow 8n=112\phantom{\rule{0ex}{0ex}}\Rightarrow n=14\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{a}_{21}=8\left(21\right)-5=163$

#### Page No 5.53:

#### Question 45:

If the sum of the first n terms of an A.P is 4n − n^{2}, What is the first term? What is the sum of first two terms? What is the second term? Similarly, find the third, the tenth and the *n*th terms.

#### Answer:

In the given problem, the sum of *n* terms of an A.P. is given by the expression,

So here, we can find the first term by substituting,

Similarly, the sum of first two terms can be given by,

Now, as we know,

So,

Now, using the same method we have to find the third, tenth and *n*^{th} term of the A.P.

So, for the third term,

Also, for the tenth term,

So, for the *n*^{th} term,

Therefore, .

#### Page No 5.53:

#### Question 46:

If the sum of first *n* terms of an A.P. is $\frac{1}{2}$(3*n*^{2} + 7*n*), then find its *n*^{th} term. Hence write its 20^{th} term.

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, sum of first* n*^{ }terms = *S*_{n }= $\frac{n}{2}$[2*a* + (*n* − 1)*d*]

It is given that sum of the first *n* terms of an A.P. is $\frac{1}{2}$(3*n*^{2} + 7*n*).

∴ First term = *a *= S_{1 }= $\frac{1}{2}$[3(1)^{2} + 7(1)] = 5.

Sum of first two terms = S_{2 }= $\frac{1}{2}$[3(2)^{2} + 7(2)] = 13.

∴ Second term = S_{2} − S_{1} = 13 − 5 = 8.

∴ Common difference = *d* = Second term − First term

= 8 − 5 = 3

Also, *n*^{th} term = *a _{n}* =

*a*+ (

*n*− 1)

*d*

⇒

*a*= 5 + (

_{n}*n*− 1)(3)

⇒

*a*= 5 + 3

_{n}*n*− 3

⇒

*a*= 3

_{n}*n*+ 2

Thus,

*n*

^{th}term of this A.P. is 3

*n*+ 2.

Now,

*a*

_{20}=

*a*+ (20 − 1)

*d*

⇒

*a*

_{20}= 5 + 19(3)

⇒

*a*

_{20}= 5 + 57

⇒

*a*

_{20}= 62

Thus, 20

^{th}term of this A.P is 62.

#### Page No 5.53:

#### Question 47:

(i) In an A.P., the sum of first n terms is $\frac{3{n}^{2}}{2}+\frac{13}{n}n.$ Find its 25^{th}^{ }term.

(ii) If the sum of first *n* terms of an A.P. is *n*^{2}, then find its 10th term.

#### Answer:

(i)

Here, the sum of first *n* terms is given by the expression,

We need to find the 25^{th} term of the A.P.

So we know that the *n*^{th}term of an A.P. is given by,

So …… (1)

So, using the expression given for the sum of *n* terms, we find the sum of 25 terms (*S*_{25}) and the sum of 24 terms (*S*_{24}). We get,

Similarly,

Now, using the above values in (1),

Therefore, .

(ii) Given:${S}_{n}={n}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{Sum}\mathrm{of}\mathrm{one}\mathrm{term}={S}_{1}={1}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}_{1}=a=1...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Sum}\mathrm{of}\mathrm{two}\mathrm{term}s={S}_{2}={2}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}_{1}+{a}_{2}=4...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Subtracting}\left(1\right)\mathrm{from}\left(2\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}{a}_{2}=3\phantom{\rule{0ex}{0ex}}\Rightarrow a+d=3\phantom{\rule{0ex}{0ex}}\Rightarrow d=2\left(\because a=1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\phantom{\rule{0ex}{0ex}}{a}_{10}=a+\left(10-1\right)d\phantom{\rule{0ex}{0ex}}=1+9\left(2\right)\phantom{\rule{0ex}{0ex}}=19\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Hence, the 10th term is 19.

#### Page No 5.53:

#### Question 48:

Find the sum of all natural numbers between 1 and 100, which are divisible by 3.

#### Answer:

In this problem, we need to find the sum of all the multiples of 3 lying between 1 and 100.

So, we know that the first multiple of 3 after 1 is 3 and the last multiple of 3 before 100 is 99.

Also, all these terms will form an A.P. with the common difference of 3.

So here,

First term (*a*) = 3

Last term (*l*) = 99

Common difference (*d*) = 3

So, here the first step is to find the total number of terms. Let us take the number of terms as *n.*

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of *n* terms,

We get,

On further simplification, we get,

Therefore, the sum of all the multiples of 3 lying between 1 and 100 is

#### Page No 5.53:

#### Question 49:

Find the sum of first *n* odd natural numbers.

#### Answer:

In this problem, we need to find the sum of first *n* odd natural numbers.

So, we know that the first odd natural number is 1. Also, all the odd terms will form an A.P. with the common difference of 2.

So here,

First term (*a*) = 1

Common difference (*d*) = 2

So, let us take the number of terms as *n*

Now, as we know,

So, for *n *terms,

Therefore, the sum of first *n* odd natural numbers is.

#### Page No 5.53:

#### Question 50:

Find the sum of all odd numbers between (i) 0 and 50 (ii) 100 and 200.

#### Answer:

(i) In this problem, we need to find the sum of all odd numbers lying between 0 and 50.

So, we know that the first odd number after 0 is 1 and the last odd number before 50 is 49.

Also, all these terms will form an A.P. with the common difference of 2.

So here,

First term (*a*) = 1

Last term (*l*) = 49

Common difference (*d*) = 2

So, here the first step is to find the total number of terms. Let us take the number of terms as *n.*

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of *n* terms,

For *n* = 25, we get,

On further simplification, we get,

Therefore, the sum of all the odd numbers lying between 0 and 50 is.

(ii) In this problem, we need to find the sum of all odd numbers lying between 100 and 200.

So, we know that the first odd number after 0 is 101 and the last odd number before 200 is 199.

Also, all these terms will form an A.P. with the common difference of 2.

So here,

First term (*a*) = 101

Last term (*l*) = 199

Common difference (*d*) = 2

So, here the first step is to find the total number of terms. Let us take the number of terms as *n.*

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of *n* terms,

For *n* = 50, we get,

Therefore, the sum of all the odd numbers lying between 100 and 200 is.

#### Page No 5.53:

#### Question 51:

Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.

#### Answer:

In this problem, we need to prove that the sum of all odd numbers lying between 1 and 1000 which are divisible by 3 is 83667.

So, we know that the first odd number after 1 which is divisible by 3 is 3, the next odd number divisible by 3 is 9 and the last odd number before 1000 is 999.

So, all these terms will form an A.P. 3, 9, 15, 21 … with the common difference of 6

So here,

First term (*a*) = 3

Last term (*l*) = 999

Common difference (*d*) = 6

So, here the first step is to find the total number of terms. Let us take the number of terms as *n.*

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of *n* terms,

For *n* = 167, we get,

On further simplification, we get,

Therefore, the sum of all the odd numbers lying between 1 and 1000 is.

Hence proved

#### Page No 5.53:

#### Question 52:

Find the sum of all integers between 84 and 719, which are multiples of 5.

#### Answer:

In this problem, we need to find the sum of all the multiples of 5 lying between 84 and 719.

So, we know that the first multiple of 5 after 84 is 85 and the last multiple of 5 before 719 is 715.

Also, all these terms will form an A.P. with the common difference of 5.

So here,

First term (*a*) = 85

Last term (*l*) = 715

Common difference (*d*) = 5

So, here the first step is to find the total number of terms. Let us take the number of terms as *n.*

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of *n* terms,

We get,

On further simplification, we get,

Therefore, the sum of all the multiples of 5 lying between 84 and 719 is.

#### Page No 5.53:

#### Question 53:

Find the sum of all integers between 50 and 500, which are divisible by 7.

#### Answer:

In this problem, we need to find the sum of all the multiples of 7 lying between 50 and 500.

So, we know that the first multiple of 7 after 50 is 56 and the last multiple of 7 before 500 is 497.

Also, all these terms will form an A.P. with the common difference of 7.

So here,

First term (*a*) = 56

Last term (*l*) = 497

Common difference (*d*) = 7

So, here the first step is to find the total number of terms. Let us take the number of terms as *n.*

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of *n* terms,

For *n* = 64, we get,

Therefore, the sum of all the multiples of 7 lying between 50 and 500 is.

#### Page No 5.53:

#### Question 54:

Find the sum of all even integers between 101 and 999.

#### Answer:

In this problem, we need to find the sum of all the even numbers lying between 101 and 999.

So, we know that the first even number after 101 is 102 and the last even number before 999 is 998.

Also, all these terms will form an A.P. with the common difference of 2.

So here,

First term (*a*) = 102

Last term (*l*) = 998

Common difference (*d*) = 2

So, here the first step is to find the total number of terms. Let us take the number of terms as *n.*

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of *n* terms,

For *n* = 64, we get,

On further simplification, we get,

Therefore, the sum of all the even numbers lying between 101 and 999 is.

#### Page No 5.53:

#### Question 55:

Find the sum of all integers between 100 and 550, which are divisible by 9.

#### Answer:

(i) In this problem, we need to find the sum of all the multiples of 9 lying between 100 and 550.

So, we know that the first multiple of 9 after 100 is 108 and the last multiple of 9 before 550 is 549.

Also, all these terms will form an A.P. with the common difference of 9.

So here,

First term (*a*) = 108

Last term (*l*) = 549

Common difference (*d*) = 9

So, here the first step is to find the total number of terms. Let us take the number of terms as *n.*

Now, as we know,

So, for the last term,

Further simplifying,

Now, using the formula for the sum of *n* terms,

We get,

Therefore, the sum of all the multiples of 9 lying between 100 and 550 is

(ii)

In this problem, we need to find the sum of all the integers lying between 100 and 550 which are not multiples of 9.

So, we know that the sum of all the multiples of 9 lying between 100 and 550 is 16425.

The sum of all the integers lying between 100 and 550 which are not multiples of 9 is

[101 + 102 + 103 + .....+ 549] − 16425

$=\left[\left(1+2+.....+549\right)-\left(1+2+.....100\right)\right]-16425\phantom{\rule{0ex}{0ex}}=\left[\left(\frac{549\times 550}{2}\right)-\left(\frac{100\times 101}{2}\right)\right]-16425\phantom{\rule{0ex}{0ex}}=150975-5050-16425\phantom{\rule{0ex}{0ex}}=129500\phantom{\rule{0ex}{0ex}}$

(iii)

Integers between 1 and 500 which are multiples of both 2 and 5 are

10, 20, 30, ....., 490.

${a}_{n}=490\phantom{\rule{0ex}{0ex}}\Rightarrow 10+\left(n-1\right)10=490\phantom{\rule{0ex}{0ex}}\Rightarrow \left(n-1\right)10=480\phantom{\rule{0ex}{0ex}}\Rightarrow n=49\phantom{\rule{0ex}{0ex}}Now,\phantom{\rule{0ex}{0ex}}{S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}{S}_{49}=\frac{49}{2}\left[20+\left(48\right)10\right]\phantom{\rule{0ex}{0ex}}{S}_{49}=\frac{49}{2}\left[500\right]\phantom{\rule{0ex}{0ex}}{S}_{49}=49\times 250=12250$

(iv)

Integers from 1 and 500 which are multiples of both 2 and 5 are

10, 20, 30, ....., 500.

${a}_{n}=500\phantom{\rule{0ex}{0ex}}\Rightarrow 10+\left(n-1\right)10=500\phantom{\rule{0ex}{0ex}}\Rightarrow \left(n-1\right)10=490\phantom{\rule{0ex}{0ex}}\Rightarrow n=50\phantom{\rule{0ex}{0ex}}Now,\phantom{\rule{0ex}{0ex}}{S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}{S}_{50}=\frac{50}{2}\left[20+\left(49\right)10\right]\phantom{\rule{0ex}{0ex}}=25\left[510\right]\phantom{\rule{0ex}{0ex}}=12750$

(v)

Integers from 1 and 500 which are multiples of 2 or 5 are

Multiples of 2 are 2, 4, 6, .....500. Therefore sum of all the multiples of 2 from 1 and 500 is

${a}_{n}=500\phantom{\rule{0ex}{0ex}}\Rightarrow 2+\left(n-1\right)2=500\phantom{\rule{0ex}{0ex}}\Rightarrow n=250\phantom{\rule{0ex}{0ex}}{S}_{250}=\frac{250}{2}\left(4+249\times 2\right)=125\times 502=62750\phantom{\rule{0ex}{0ex}}$

and

Multiples of 5 are 5, 10, 15, ....., 500. Therefore sum of all the multiples of 5 from 1 and 500 is

${a}_{n}=500\phantom{\rule{0ex}{0ex}}\Rightarrow 5+\left(n-1\right)5=500\phantom{\rule{0ex}{0ex}}\Rightarrow n=100\phantom{\rule{0ex}{0ex}}{S}_{100}=\frac{100}{2}\left(10+99\times 5\right)=50\times 505=25250\phantom{\rule{0ex}{0ex}}$

Multiples of both 2 and 5 are 10, 20, 30, ....., 500. Therefore sum of all the multiples of 2 and 5 from 1 and 500 is

${a}_{n}=500\phantom{\rule{0ex}{0ex}}\Rightarrow 10+\left(n-1\right)10=500\phantom{\rule{0ex}{0ex}}\Rightarrow \left(n-1\right)10=490\phantom{\rule{0ex}{0ex}}\Rightarrow n=50\phantom{\rule{0ex}{0ex}}Now,\phantom{\rule{0ex}{0ex}}{S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}{S}_{50}=\frac{50}{2}\left[20+\left(49\right)10\right]\phantom{\rule{0ex}{0ex}}=25\left[510\right]\phantom{\rule{0ex}{0ex}}=12750$

Hence, the sum of Integers from 1 and 500 which are multiples of 2 or 5 is

[ sum of all the multiples of 2 + sum of all the multiples of 5 ] − [ sum of all the multiples of 2 and 5 ]

= [ 62750 + 25250 ] − 12750

= 75250

#### Page No 5.53:

#### Question 56:

Let there be an A.P. with first term '*a*', common difference '*d*'. If *a*_{n} denotes in *n*^{th} term and *S*_{n} the sum of first *n* terms, find.

(i) *n* and S_{n}, if *a* = 5, *d* = 3 and *a*_{n} = 50.

(ii) n and a, if a_{n} = 4, d = 2 and S_{n}_{ }= −14.

(iii) d, if a = 3, n = 8 and S_{n} = 192.

(iv) a, if a_{n} = 28, S_{n} = 144 and n= 9.

(v) n and d, if a = 8, a_{n} = 62 and S_{n} = 210

(vi) n and a_{n}, if a= 2, d = 8 and S_{n} = 90.

(vii) $k,\mathrm{if}{S}_{n}=3{n}^{2}+5n\mathrm{and}{a}_{k}=164$.

(viii) ${S}_{22},\mathrm{if}d=22\mathrm{and}{a}_{22}=149$

#### Answer:

(i) Here, we have an A.P. whose *n*^{th} term (*a*_{n}), first term (*a*) and common difference (*d*) are given. We need to find the number of terms (*n*) and the sum of first *n* terms (*S*_{n}).

Here,

First term (*a*) = 5

Last term () = 50

Common difference (*d*) = 3

So here we will find the value of *n* using the formula,

So, substituting the values in the above mentioned formula

Further simplifying for *n*,

Now, here we can find the sum of the *n* terms of the given A.P., using the formula,

Where, *a* = the first term

*l* = the last term

So, for the given A.P, on substituting the values in the formula for the sum of *n* terms of an A.P., we get,

Therefore, for the given A.P

(ii) Here, we have an A.P. whose *n*^{th} term (*a*_{n}), sum of first *n* terms (*S*_{n}) and common difference (*d*) are given. We need to find the number of terms (*n*) and the first term (*a*).

Here,

Last term () = 4

Common difference (*d*) = 2

Sum of *n* terms (*S*_{n}) = −14

So here we will find the value of *n* using the formula,

So, substituting the values in the above mentioned formula

Now, here the sum of the *n* terms is given by the formula,

Where, *a* = the first term

*l* = the last term

So, for the given A.P, on substituting the values in the formula for the sum of *n* terms of an A.P., we get,

Equating (1) and (2), we get,

So, we get the following quadratic equation,

Further, solving it for *a* by splitting the middle term,

So, we get,

Or

Substituting, in (1),

Here, we get* n *as negative, which is not possible. So, we take,

Therefore, for the given A.P

(iii) Here, we have an A.P. whose first term (*a*), sum of first *n* terms (*S*_{n}) and the number of terms (*n*) are given. We need to find common difference (*d*).

Here,

First term () = 3

Sum of *n* terms (*S*_{n}) = 192

Number of terms (*n*) = 8

So here we will find the value of *n* using the formula,

So, to find the common difference of this A.P., we use the following formula for the sum

of *n* terms of an A.P

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

So, using the formula for *n* = 8, we get,

Further solving for *d,*

Therefore, the common difference of the given A.P. is.

(iv) Here, we have an A.P. whose *n*^{th} term (*a*_{n}), sum of first *n* terms (*S*_{n}) and the number of terms (*n*) are given. We need to find first term (*a*).

Here,

Last term () = 28

Sum of *n* terms (*S*_{n}) = 144

Number of terms (*n*) = 9

Now,

Also, using the following formula for the sum of *n* terms of an A.P

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

So, using the formula for *n* = 9, we get,

Multiplying (1) by 9, we get

Further, subtracting (3) from (2), we get

Therefore, the first term of the given A.P. is.

(v) Here, we have an A.P. whose *n*^{th} term (*a*_{n}), sum of first *n* terms (*S*_{n}) and first term (*a*) are given. We need to find the number of terms (*n*) and the common difference (*d*).

Here,

First term () = 8

Last term () = 62

Sum of *n* terms (*S*_{n}) = 210

Now, here the sum of the *n* terms is given by the formula,

Where, *a* = the first term

*l* = the last term

So, for the given A.P, on substituting the values in the formula for the sum of *n* terms of an A.P., we get,

Also, here we will find the value of *d* using the formula,

So, substituting the values in the above mentioned formula

Therefore, for the given A.P

(vi) Here, we have an A.P. whose first term (*a*), common difference (*d*) and sum of first *n* terms are given. We need to find the number of terms (*n*) and the *n*^{th} term (*a*_{n}).

Here,

First term (*a*) = 2

Sum of first *n*^{th} terms () = 90

Common difference (*d*) = 8

So, to find the number of terms (*n*) of this A.P., we use the following formula for the sum

of *n* terms of an A.P

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

So, using the formula for *n* = 8, we get,

Further solving the above quadratic equation*,*

Further solving for *n*,

Now,

Also,

Since *n *cannot be a fraction

Thus, *n* = 5

Also, we will find the value of the *n*^{th} term (*a*_{n}) using the formula,

So, substituting the values in the above mentioned formula

Therefore, for the given A.P.

(vii)

${a}_{k}={S}_{k}-{S}_{k-1}\phantom{\rule{0ex}{0ex}}\Rightarrow 164=\left(3{k}^{2}+5k\right)-\left(3{\left(k-1\right)}^{2}+5\left(k-1\right)\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 164=3{k}^{2}+5k-3{k}^{2}+6k-3-5k+5\phantom{\rule{0ex}{0ex}}\Rightarrow 164=6k+2\phantom{\rule{0ex}{0ex}}\Rightarrow 6k=162\phantom{\rule{0ex}{0ex}}\Rightarrow k=27\phantom{\rule{0ex}{0ex}}$

(viii) Given d = 22, ${a}_{22}=149$ ,n=22

We know that

$149=a+(22-1)22\phantom{\rule{0ex}{0ex}}149=a+462\phantom{\rule{0ex}{0ex}}a=-313\phantom{\rule{0ex}{0ex}}$

Now, Sum is given by

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

So, using the formula for *n* = 22, we get

${S}_{22}=\frac{22}{2}\left\{2\times \left(-313)+(22-1)\times 22\right)\right\}\phantom{\rule{0ex}{0ex}}{S}_{22}=11\left\{-626+462\right\}\phantom{\rule{0ex}{0ex}}{S}_{22}=-1804$

Hence, the sum of 22 terms is −1804.

#### Page No 5.54:

#### Question 57:

If *S _{n}* denotes the sum of first

*n*terms of an A.P., prove that

*S*

_{12}= 3(

*S*

_{8}−

*S*

_{4}).

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, sum of first* n*^{ }terms = *S*_{n }= $\frac{n}{2}$[2*a* + (*n* − 1)*d*]

Now,

*S*_{4 }= $\frac{4}{2}$[2*a* + (4 − 1)*d*]

= 2(2*a* + 3*d*)

= 4*a* + 6*d * ....(1)

*S*_{8 }= $\frac{8}{2}$[2*a* + (8 − 1)*d*]

= 4(2*a* + 7*d*)

= 8*a* + 28*d * ....(2)

*S*_{12 }= $\frac{12}{2}$[2*a* + (12 − 1)*d*]

= 6(2*a* + 11*d*)

= 12*a* + 66*d * ....(3)

On subtracting (1) from (2), we get

*S*_{8} − *S*_{4 }*= *8*a* + 28*d* − (4*a* + 6*d*)

* = *4*a* + 22*d*

Multiplying both sides by 3, we get

3(*S*_{8} − *S*_{4}) = 3(4*a* + 22*d*)

= 12*a* + 66*d
= S*

_{12}[From (3)]

Thus,

*S*

_{12}= 3(

*S*

_{8}−

*S*

_{4}).

#### Page No 5.54:

#### Question 58:

A thief, after commiting a theft runs at a uniform speed of 50 m/minute . After 2 minutes, a policeman runs to catch him. He goes 60 m in first minute and increases his speed by 5 m/minute everey suceeding minute. After how many minutes, the policeman will catch the thief.?

#### Answer:

Suppose the policeman catches the thief after *t* minutes.

Uniform speed of the thief = 50 m/min

∴ Distance covered by thief in (*t* + 2) minutes = 50 m/min × (*t* + 2) min = 50 (*t* + 2) m

The distance covered by the policeman in *t* minutes is in AP, with 60 and 5 as the first term and the common difference, respectively.

Now,

Distance covered by policeman in *t* minutes = Sum of *t* terms

$=\frac{t}{2}\left[2\times 60+\left(t-1\right)\times 5\right]\phantom{\rule{0ex}{0ex}}=\frac{t}{2}\left[115+5t\right]\mathrm{m}$

When the policeman catches the thief, we have

$\frac{t}{2}\left[115+5t\right]=50\left(t+2\right)\phantom{\rule{0ex}{0ex}}115t+5{t}^{2}=100t+200\phantom{\rule{0ex}{0ex}}\Rightarrow 5{t}^{2}+15t-200=0\phantom{\rule{0ex}{0ex}}\Rightarrow {t}^{2}+3t-40=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(t+8\right)\left(t-5\right)=0$

So, *t* = −8 or *t* = 5

∴ *t* = 5 (As *t* cannot be negative)

Thus, the policeman catches the thief after 5 min.

#### Page No 5.54:

#### Question 59:

The sums of first n terms of three A.P. are *S*_{1},*S*_{2 }and *S*_{3}. The first term of each is 5 and their common differences are 2, 4 and 6 respectively. Prove that *S*_{1 }+ *S*_{3 }= 2*S*_{2}

#### Answer:

${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}For{S}_{1},whena=1andd=1.\phantom{\rule{0ex}{0ex}}{S}_{1}=\frac{n}{2}\left[2\left(1\right)+\left(n-1\right)\right]=\frac{n}{2}\left[2+n-1\right]=\frac{n}{2}\left[n+1\right]\phantom{\rule{0ex}{0ex}}For{S}_{2},whena=1andd=2.\phantom{\rule{0ex}{0ex}}{S}_{2}=\frac{n}{2}\left[2\left(1\right)+\left(n-1\right)\left(2\right)\right]=\frac{n}{2}\left[2+2n-2\right]=\frac{n}{2}\left[2n\right]={n}^{2}\phantom{\rule{0ex}{0ex}}For{S}_{3},whena=1andd=3.\phantom{\rule{0ex}{0ex}}{S}_{3}=\frac{n}{2}\left[2\left(1\right)+\left(n-1\right)\left(3\right)\right]=\frac{n}{2}\left[2+3n-3\right]=\frac{n}{2}\left[3n-1\right]\phantom{\rule{0ex}{0ex}}Therefore,\phantom{\rule{0ex}{0ex}}{S}_{1}+{S}_{3}=\frac{n}{2}\left[n+1\right]+\frac{n}{2}\left[3n-1\right]\phantom{\rule{0ex}{0ex}}=\frac{n}{2}\left[4n\right]=2{n}^{2}\phantom{\rule{0ex}{0ex}}{S}_{1}+{S}_{3}=2{S}_{2}\phantom{\rule{0ex}{0ex}}$

Hence proved.

#### Page No 5.54:

#### Question 60:

Resham wanted to save at least ₹6500 for sending her daughter to school next year (after 12 months). She saved ₹450 in the first month and raised her savings by ₹20 every next month. How much will she able to save in next 12 months ? Will she be able to send her daughter to the school next year ?

#### Answer:

It is given that Reshma saved ₹450 in the first month and raised her savings by ₹20 every next month.

So, her savings are in an AP, with the first term (*a*) = ₹450 and the common difference (*d*) = ₹20.

We need to find her savings for 12 months, so *n* = 12.

We know that the sum of *n* terms of an AP is ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$.

Reshma's savings for 12 months:

${S}_{12}=\frac{12}{2}\left[2\times 450+\left(12-1\right)\times 20\right]\phantom{\rule{0ex}{0ex}}=6\left(900+220\right)\phantom{\rule{0ex}{0ex}}=6\times 1120\phantom{\rule{0ex}{0ex}}=6720$

So, she will save ₹6,720 in 12 months.

She needed to save at least ₹6,500 for sending her daughter to school next year.

Since ₹6,720 is greater than ₹6,500, Reshma can send her daughter to school.

The question aims to encourage personal savings and emphasise the need of female education.

#### Page No 5.54:

#### Question 61:

In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two sections, find how many trees were planted by the students.

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, sum of first* n*^{ }terms = *S*_{n }= $\frac{n}{2}$[2*a* + (*n* − 1)*d*]

It can be observed that the number of trees planted by the students forms an A.P.

2, 4, 6, 8, ... , 24

Here, *a* = 2, *d* = 2 and *n* = 12.

$\therefore {S}_{n}=\frac{12}{2}\left[2\times 2+\left(12-1\right)2\right]$

= 6(4 + 22)

= 6(26)

= 156

Therefore, trees planted by 1 section of all the classes = 156.

Number of trees planted by 2 sections of all the classes = 2 × 156 = 312

Thus, 312 trees were planted by the students.

#### Page No 5.54:

#### Question 62:

Ramkali would need ₹1800 for admission fee and books etc., for her daughter to start going to school from next year. She saved ₹50 in the first month of this year and increased her monthly saving by ₹20. After a year, how much money will she save? Will she be able to fulfil her dream of sending her daughter to school?

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, sum of first* n*^{ }terms = *S*_{n }= $\frac{n}{2}$[2*a* + (*n* − 1)*d*]

According to the question,

Saving of Ramkali in 1 year = ₹50 + ₹70 + ₹90.......

Here, *a = *50, *d = *70 − 50 = 20 and* n* = 12.

∴ *S*_{12} = $\frac{12}{2}$[2 × 50 + (12 − 1)20]

= 6[100 + 220]

= 6 × 320

= 1920

Hence, After a year, she will save ₹1920.

Since, required amount for admission is ₹1800 and her savings will be ₹1920.

Thus, yes she will be able to fulfil her dream of sending her daughter to school.

#### Page No 5.54:

#### Question 63:

A man saved Rs 16500 in ten years. In each year after the first he saved Rs 100 more than he did in the preceding year. How much did he save in the first year?

#### Answer:

Here, we are given that the total saving of a man is Rs 16500 and every year he saved Rs 100 more than the previous year.

So, let us take the first installment as *a.*

Second installment =

Third installment =

So, these installments will form an A.P. with the common difference (*d*) = 100

The sum of his savings every year

Number of years (*n*) = 10

So, to find the first installment, we use the following formula for the sum of *n* terms of an A.P.,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

So, using the formula for *n* = 10, we get,

Further solving for *a,*

Therefore, man saved in the first year.

#### Page No 5.54:

#### Question 64:

A man saved Rs. 32 during the first year, Rs 36 in the second year and in this way he increases his saving by Rs 4 every year. Find in what time his saving will be Rs. 200.

#### Answer:

Here, we are given that the total saving of a man is Rs 200. In the first year he saved Rs 32 and every year he saved Rs 4 more than the previous year.

So, the first installment = 32*.*

Second installment = 36

Third installment =

So, these installments will form an A.P. with the common difference (*d*) = 4

The sum of his savings every year

We need to find the number of years. Let us take the number of years as *n.*

So, to find the number of years, we use the following formula for the sum of *n* terms of an A.P.,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

So, using the formula for *n* = 10, we get,

We get a quadratic equation,

Further solving for *n* by splitting the middle term, we get,

So,

Or

Since number of years cannot be negative. So, in, his savings will be Rs 200.

#### Page No 5.54:

#### Question 65:

A man arranges to pay off a dept of Rs 3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of all debt unpaid, find the value of the first instalment.

#### Answer:

In the given problem,

Total amount of debt to be paid in 40 installments

After 30 installments one−third of his debt is left unpaid. This means that he paid two third of the debt in 30 installments. So,

Amount he paid in 30 installments

Let us take the first installment as *a *and common difference as *d.*

So, using the formula for the sum of *n* terms of an A.P,

Let us find *a *and *d, *for 30 installments.

Similarly, we find *a *and *d *for 40 installments.

Subtracting (1) from (2), we get,

Further solving for *d,*

Substituting the value of *d* in (1), we get.

Therefore, the first installment is.

#### Page No 5.54:

#### Question 66:

There are 25 trees at equal distances of 5 metres in a line with a well, the distance of the well from the nearest tree being 10 metres. A gardener water all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to eater all the trees.

#### Answer:

In the given problem, there are 25 trees in a line with a well such that the distance between two trees is 5 meters and the distance between the well and the first tree is 10 meters.

So, the total distance covered to water first tree

Then he goes back to the well to get water.

So,

The total distance covered to water second tree

The total distance covered to water third tree

The total distance covered to water fourth tree

So, from second tree onwards, the distance covered by the gardener forms an A.P. with the first term as 25 and common difference as 10.

So, the total distance covered for 24 trees can be calculated by using the formula for the sum of *n* terms of an A.P,

We get,

So, while watering the 24 trees he covered 3360 meters. Also, to water the first tree he covers 10 meters. So the distance covered while watering 25 trees is 3370 meters.

Now, the distance between the last tree and the well

So, to get back to the well he covers an additional 130 m. Therefore, the total distance covered by the gardener

Therefore, the total distance covered by the gardener is.

#### Page No 5.54:

#### Question 67:

A man is employed to count Rs 10710. He counts at the rate of Rs 180 per minute for half an hour. After this he counts at the rate of Rs 3 less every minute than the preceding minute. Find the time taken by him to count the entire amount.

#### Answer:

In the given problem, the total amount = Rs 10710.

For the first half and hour (30 minutes) he counts at a rate of Rs 180 per minute. So,

The amount counted in 30 minutes

So, amount left after half an hour

After 30 minutes he counts at a rate of Rs 3 less every minute. So,

At 31^{st} minute the rate of counting per minute = 177.

At 32^{nd} minute the rate of counting per minute = 174.

So, the rate of counting per minute for each minute will form an A.P. with the first term as 177 and common difference as −3.

So, the total time taken to count the amount left after half an hour can be calculated by using the formula for the sum of *n* terms of an A.P,

We get,

So, we get the following quadratic equation,

Solving the equation by splitting the middle term, we get,

So,

Or

$\mathrm{Now}\mathrm{let}\mathrm{n}=60\mathrm{then}\mathrm{finding}\mathrm{the}\mathrm{last}\mathrm{term},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[\mathrm{a}+\mathrm{l}\right]\phantom{\rule{0ex}{0ex}}5310=\frac{60}{2}\left[177+\mathrm{l}\right]\phantom{\rule{0ex}{0ex}}177=177+\mathrm{l}\phantom{\rule{0ex}{0ex}}\mathrm{l}=0\phantom{\rule{0ex}{0ex}}\mathrm{It}\mathrm{means}\mathrm{the}\mathrm{work}\mathrm{will}\mathrm{be}\mathrm{finesh}\mathrm{in}59\mathrm{th}\mathrm{minute}\mathrm{only}\mathrm{because}60\mathrm{th}\mathrm{term}\mathrm{is}0.\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{we}\mathrm{will}\mathrm{take}\mathrm{n}=59$

Therefore, the total time required for counting the entire amount

So, the total time required for counting the entire amount is.

#### Page No 5.54:

#### Question 68:

A piece of equipment cost a certain factory Rs 60,000. If it depreciates in value, 15% the first, 13.5% the next year, 12% the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost?

#### Answer:

In the given problem,

Cost of the equipment = Rs 600,000

It depreciates by 15% in the first year. So,

Depreciation in 1 year

It depreciates by 13.5% of the original cost in the 2 year. So,

Depreciation in 2 year

Further, it depreciates by 12% of the original cost in the 3 year. So,

Depreciation in 3 year

So, the depreciation in value of the equipment forms an A.P. with first term as 90000 and common difference as −9000.

So, the total depreciation in value in 10 years can be calculated by using the formula for the sum of *n* terms of an A.P,

We get,

So, the total depreciation in the value after 10 years is Rs 495000.

Therefore, the value of equipment

So, the value of the equipment after 10 years is.

#### Page No 5.55:

#### Question 69:

A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each prize.

#### Answer:

In the given problem,

Total amount of money (*S*_{n}) = Rs 700

There are a total of 7 prizes and each prize is Rs 20 less than the previous prize. So let us take the first prize as Rs *a.*

So, the second prize will be Rs, third prize will be Rs.

Therefore, the prize money will form an A.P. with first term *a* and common difference −20.

So, using the formula for the sum of *n* terms,

We get,

On further simplification, we get,

Therefore, the value of first prize is Rs 160.

Second prize = Rs 140

Third prize = Rs 120

Fourth prize = Rs 100

Fifth prize = Rs 80

Sixth prize = Rs 60

Seventh prize= Rs 40

So the values of prizes are

#### Page No 5.55:

#### Question 70:

If *S _{n}* denotes the sum of the first

*n*terms of an A.P., prove that

*S*

_{30}= 3(

*S*

_{20}−

*S*

_{10})

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, sum of first* n*^{ }terms = *S*_{n }= $\frac{n}{2}$[2*a* + (*n* − 1)*d*]

Now,

*S*_{10 }= $\frac{10}{2}$[2*a* + (10 − 1)*d*]

= 5(2*a* + 9*d*)

= 10*a* + 45*d* ....(1)

*S*_{20 }= $\frac{20}{2}$[2*a* + (20 − 1)*d*]

= 10(2*a* + 19*d*)

= 20*a* + 190*d* ....(2)

*S*_{30 }= $\frac{30}{2}$[2*a* + (30 − 1)*d*]

= 15(2*a* + 29*d*)

= 30*a* + 435*d* ....(3)

On subtracting (1) from (2), we get

*S*_{20} − *S*_{10 }= 20*a* + 190*d − *(10*a* + 45*d*)

= 10*a* + 145*d*

On multiplying both sides by 3, we get

3(*S*_{20} − *S*_{10}) = 3(10*a* + 145*d*)

= 30*a* + 435*d
= S*

_{30 }[From (3)]

Hence,

*S*

_{30}= 3(

*S*

_{20}−

*S*

_{10})

#### Page No 5.55:

#### Question 71:

Solve the question $\left(-4\right)+\left(-1\right)+2+5+....+x=437.$

#### Answer:

Suppose *x* is *n*th term of the given A.P.

${a}_{n}=x\phantom{\rule{0ex}{0ex}}Here,a=-4,d=3.\phantom{\rule{0ex}{0ex}}Itisgiventhat,{S}_{n}=437.\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{n}{2}\left[2\left(-4\right)+\left(n-1\right)3\right]=437\phantom{\rule{0ex}{0ex}}\Rightarrow 3{n}^{2}-11n-874=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3{n}^{2}-57n+46n-874=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3n\left(n-19\right)+46\left(n-19\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow n=-\frac{46}{3},19\phantom{\rule{0ex}{0ex}}Since,ncannotbeinfractionson=19.\phantom{\rule{0ex}{0ex}}Now,{a}_{n}=x\phantom{\rule{0ex}{0ex}}\Rightarrow \left(-4\right)+\left(19-1\right)3=x\phantom{\rule{0ex}{0ex}}\Rightarrow -4+54=x\phantom{\rule{0ex}{0ex}}\Rightarrow x=50$

#### Page No 5.55:

#### Question 72:

Which term of the A.P. $-2,-7,-12,....$will be $-77$ ? Find the sum of this A.P. upto the term $-77$ .

#### Answer:

Here, *a* = −2 and *d* = −5.

$If{a}_{n}=-77.\phantom{\rule{0ex}{0ex}}\Rightarrow \left(-2\right)+\left(n-1\right)\left(-5\right)=-77\phantom{\rule{0ex}{0ex}}\Rightarrow -2-5n+5=-77\phantom{\rule{0ex}{0ex}}\Rightarrow n=16\phantom{\rule{0ex}{0ex}}Therefore,\phantom{\rule{0ex}{0ex}}{S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}\Rightarrow {S}_{16}=\frac{16}{2}\left[2\left(-2\right)+\left(16-1\right)\left(-5\right)\right]\phantom{\rule{0ex}{0ex}}=8\left[-4-75\right]\phantom{\rule{0ex}{0ex}}=-632$

#### Page No 5.55:

#### Question 73:

The sum of first n terms of an A.P. whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another A.P. whose first term is $-30$ and common difference is 8 . Find n.

#### Answer:

According to the question, we have

$\frac{n}{2}\left[2\left(8\right)+\left(n-1\right)20\right]=\frac{{\displaystyle 2n}}{{\displaystyle 2}}\left[2\left(-30\right)+\left(2n-1\right)8\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \left[16+20n-20\right]=2\left[-60+16n-8\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 20n-4=-136+32n\phantom{\rule{0ex}{0ex}}\Rightarrow 12n=132\phantom{\rule{0ex}{0ex}}\Rightarrow n=11$

#### Page No 5.55:

#### Question 74:

The students of a school decided to beautify the school on the annual day by fixing colourful on the straight passage of the school . They have 27 flags to be fixed at intervals of every 2 metre . The flags are stored at the position of the middle most flag . Ruchi was given the responsibility of placing the flags . Ruchi kept her books where the flags were stored . She could carry only one flag at a time . How much distance did she cover in completing this job and returning back to collect her books ? What is the maximum distance she travelled carrying a flag ?

#### Answer:

Distance covered to place the first flag to the left of the middle flag = 2 × 2 m = 4 m.

Distance covered to place the second flag to the left of the middle flag = 2 × 4 m = 8 m.

Similarly,

Distance covered to place the thirteenth flag to the left of the middle flag = 2 × 26 m = 52 m.

Now,

The total distance covered = 2 ( 4 + 8 + 12 +....+ 52)

The sum is as follows:

$S=2\times \frac{13}{2}\left(4+52\right)\phantom{\rule{0ex}{0ex}}=13\times 56\phantom{\rule{0ex}{0ex}}=728m$

The total distance trvelled is 728 m and maximum distance travelled in carrying a flag is 26 m.

#### Page No 5.56:

#### Question 1:

*Mark the correct alternative in each of the following:*

If 7th and 13th terms of an A.P. be 34 and 64 respectively, then its 18th term is

(a) 87

(b) 88

(c) 89

(d) 90

#### Answer:

In the given problem, we are given 7^{th} and 13^{th}^{ }term of an A.P.

We need to find the 26^{th} term

Here,

Now, we will find and using the formula

So,

Also,

Further, to solve for *a* and *d*

On subtracting (1) from (2), we get

Substituting (3) in (1), we get

Thus,

So, for 18^{th} term (*n = *18),

Substituting the above values in the formula,

Therefore,

Hence, the correct option is **(c).**

#### Page No 5.56:

#### Question 2:

If the sum of *P* terms of an A.P. is *q* and the sum of *q* terms is *p*, then the sum of *p* + *q* terms will be

(a) 0

(b) *p* − *q*

(c) *p *+ *q*

(d) −(*p* + *q*)

#### Answer:

In the given problem, we are given and

We need to find

Now, as we know,

So,

Similarly,

Subtracting (2) from (1), we get

Now,

Thus,

Hence, the correct option is (d).

#### Page No 5.56:

#### Question 3:

If the sum of n terms of an A.P. be 3*n*^{2} + *n* and its common difference is 6, then its first term is

(a) 2

(b) 3

(c) 1

(d) 4

#### Answer:

In the given problem, the sum of *n* terms of an A.P. is given by the expression,

Here, we can find the first term by substitutingas sum of first term of the A.P. will be the same as the first term. So we get,

Therefore, the first term of this A.P is. So, the correct option is **(d)**.

#### Page No 5.56:

#### Question 4:

The first and last terms of an A.P. are 1 and 11. If the sum of its terms is 36, then the number of terms will be

(a) 5

(b) 6

(c) 7

(d) 8

#### Answer:

In the given problem, we need to find the number of terms in an A.P. We are given,

First term (*a*) = 1

Last term (*a*_{n}) = 11

Sum of its terms

Now, as we know,

Where, *a* = the first term

*l* = the last term

So, we get,

Therefore, the total number of terms in the given A.P. is

Hence the correct option is **(b)**.

#### Page No 5.56:

#### Question 5:

If the sum of *n* terms of an A.P. is 3*n*^{2} + 5*n** *then which of its terms is 164?

(a) 26th

(b) 27th

(c) 28th

(d) none of these.

#### Answer:

Here, the sum of first *n* terms is given by the expression,

We need to find which term of the A.P. is 164.

Let us take 164 as the *n*^{th} term

So we know that the *n*^{th}term of an A.P. is given by,

So,

Using the property,

We get,

Further solving for *n*, we get

Therefore,

Hence the correct option is **(b).**

#### Page No 5.56:

#### Question 6:

If the sum of *n* terms of an A.P. is 2*n*^{2} + 5n, then its *n*th term is

(a) 4*n* − 3

(b) 3*n* − 4

(c) 4*n** *+ 3

(d) 3*n* + 4

#### Answer:

Here, the sum of first *n* terms is given by the expression,

We need to find the *n*^{th} term.

So we know that the *n*^{th}term of an A.P. is given by,

So,

Using the property,

We get,

Therefore,

Hence the correct option is **(c).**

#### Page No 5.56:

#### Question 7:

If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is

(a) 13

(b) 9

(c) 21

(d) 17

#### Answer:

In the given problem, the sum of three consecutive terms of an A.P is 51 and the product of the first and the third terms is 273.

We need to find the third term.

Here,

Let the three terms be where, *a* is the first term and *d* is the common difference of the A.P

So,

Also,

Further solving for *d*,

Now, it is given that this is an increasing A.P. so *d *cannot be negative.

So, *d* = 4

Substituting the values of *a* and *d* in the expression for the third term, we get,

Third term =

So,

Therefore, the third term is

Hence, the correct option is **(c)**.

#### Page No 5.56:

#### Question 8:

If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times, the least, then the numbers are

(a) 5, 10, 15, 20

(b) 4, 101, 16, 22

(c) 3, 7, 11, 15

(d) none of these

#### Answer:

Here, we are given that four numbers are in A.P., such that their sum is 50 and the greatest number is 4 times the smallest.

So, let us take the four terms as.

Now, we are given that sum of these numbers is 50, so we get,

Also, the greatest number is 4 times the smallest, so we get,

Now, using (2) in (1), we get,

Now, using the value of *a* in (2), we get,

So, first term is given by,

Second term is given by,

Third term is given by,

Fourth term is given by,

Therefore, the four terms are

Hence, the correct option is **(a)**.

#### Page No 5.56:

#### Question 9:

Let *S _{n}* denote the sum of

*n*terms of an A.P. whose first term is

*a*. If the common difference

*d*is given by

*d*=

*S*

_{n}_{ }− kS_{n}_{−1}+

*S*

_{n−2}, then

*k*=

(a) 1

(b) 2

(c) 3

(d) none of these.

#### Answer:

In the given problem, we are given

We need to find the value of *k*

So here,

First term = *a*

Common difference = *d*

Sum of *n *terms = *S*_{n}

Now, as we know,

Also, for *n*-1* *terms,

Further, for *n*-2 terms,

Now, we are given,

Using (1), (2) and (3) in the given equation, we get

Taking common, we get,

Taking 2 common from the numerator, we get,

Therefore,

Hence, the correct option is **(b)**.

#### Page No 5.56:

#### Question 10:

The first and last term of an A.P. are *a* and *l* respectively. If *S* is the sum of all the terms of the A.P. and the common difference is given by $\frac{{l}^{2}-{a}^{2}}{k-(l+a)}$, then *k* =

#### Answer:

In the given problem, we are given the first, last term, sum and the common difference of an A.P.

We need to find the value of *k*

Here,

First term = *a*

Last term = *l*

Sum of all the terms = *S*

Common difference (*d*) =

Now, as we know,

Further, substituting (1) in the given equation, we get

Now, taking *d* in common, we get,

Taking (*n*-1) as common, we get,

Further, multiplying and dividing the right hand side by 2, we get,

Now, as we know,

Thus,

Therefore, the correct option is **(b).**

#### Page No 5.56:

#### Question 11:

If the sum of first *n* even natural numbers is equal to *k *times the sum of first *n* odd natural numbers, then *k* =

(a) $\frac{1}{n}$

(b) $\frac{n-1}{n}$

(c) $\frac{n+1}{2n}$

(d) $\frac{n+1}{n}$

#### Answer:

In the given problem, we are given that the sum of the first *n* even natural numbers is equal to *k* times the sum of first *n* odd natural numbers.

We need to find the value of *k*

Now, we know that the first odd natural number is 1. Also, all the odd terms will form an A.P. with the common difference of 2.

So here,

First term (*a*) = 1

Common difference (*d*) = 2

So, let us take the number of terms as *n*

Now, as we know,

So, for *n *terms,

Also, we know that the first even natural number is 2. Also, all the odd terms will form an A.P. with the common difference of 2.

So here,

First term (*a*) = 2

Common difference (*d*) = 2

So, let us take the number of terms as *n*

So, for *n *terms,

Solving further, we get

Now, as the sum of the first *n* even natural numbers is equal to *k* times the sum of first *n* odd natural numbers

Using (1) and (2), we get

Therefore,

Hence, the correct option is **(d)**.

#### Page No 5.57:

#### Question 12:

If the first, second and last term of an A.P. are *a*, *b* and 2*a* respectively, its sum is

(a) $\frac{ab}{2(b-a)}$

(b) $\frac{ab}{(b-a)}$

(c) $\frac{3ab}{2(b-a)}$

(d) none of these

#### Answer:

In the given problem, we are given first, second and last term of an A.P. We need to find its sum.

So, here

First term = *a*

Second term (*a*_{2}) = b

Last term (*l*__)__ = 2*a*

Now, using the formula

Also,

Further as we know,

Substituting (2) in the above equation, we get

Using (1), we get

Thus,

Therefore, the correct option is **(c).**

#### Page No 5.57:

#### Question 13:

If S_{1} is the sum of an arithmetic progression of '*n*' odd number of terms and S_{2} the sum of the terms of the series in odd places, then $\frac{{S}_{1}}{{S}_{2}}=$

(a) $\frac{2n}{n+1}$

(b) $\frac{n}{n+1}$

(c) $\frac{n+1}{2n}$

(d) $\frac{n+1}{n}$

#### Answer:

In the given problem, we are given as the sum of an A.P of ‘*n*’ odd number of terms and the sum of the terms of the series in odd places.

We need to find

Now, let *a*_{1},* **a*_{2}….* a*_{n}* *be the *n* terms of A.P

Where *n* is odd

Let *d* be the common difference of the A.P

Then,

And be the sum of the terms of the places in odd places,

Where, number of terms =

Common difference = 2*d*

So,

Now,

Thus,

Therefore, the correct option is **(a)**.

#### Page No 5.57:

#### Question 14:

If in an A.P. *S*_{n} = *n*^{2}p and *S*_{m} = m^{2}p, where *S*_{r} denotes the sum of *r* terms of the A.P., then *S*_{p} is equal to

(a) $\frac{1}{2}{p}^{3}$

(b) *m* *n* *p*

(c) *p*^{3}

(d) (*m* + *n*) *p*^{2}

#### Answer:

In the given problem, we are given an A.P whose and

We need to find

Now, as we know,

Where, first term = *a*

Common difference = *d*

Number of terms = *n*

So,

Similarly,

Equating (1) and (2), we get,

$\frac{1}{2n}\left(2a+nd-d\right)=\frac{1}{2m}\left(2a+md-d\right)$

$\Rightarrow m\left(2a+nd-d\right)=n\left(2a+md-d\right)$

$\Rightarrow 2am+mnd-md=2an+mnd-nd$

Solving further, we get,

Further, substituting (3) in (1), we get,

Now,

Thus,

Therefore, the correct option is **(C).**

#### Page No 5.57:

#### Question 15:

If S_{n} denote the sum of the first *n *terms of an A.P. If S_{2n} = 3S_{n}, then S_{3n} : S_{n} is equal to

(a) 4

(b) 6

(c) 8

(d) 10

#### Answer:

Here, we are given an A.P. whose sum of *n* terms is *S*_{n} and.

We need to find.

Here we use the following formula for the sum of *n* terms of an A.P.,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

So, first we find *S*_{3n},

Similarly,

Also,

Now,

So, using (2) and (3), we get,

On further solving, we get,

So,

Taking common, we get,

Therefore,

Hence, the correct option is **(b)**.

#### Page No 5.57:

#### Question 16:

In an AP. *S*_{p} = q, *S*_{q} = *p* and *S*_{r} denotes the sum of first *r* terms. Then, *S*_{p}_{+q} is equal to

(a) 0

(b) −(*p* + *q*)

(c) *p* + *q*

(d) *pq*

#### Answer:

In the given problem, we are given and

We need to find

Now, as we know,

So,

Similarly,

Subtracting (2) from (1), we get

Now,

Thus,

Hence, the correct option is **(b).**

#### Page No 5.57:

#### Question 17:

If S_{r} denotes the sum of the first *r* terms of an A.P. Then , *S*_{3n}: (*S*_{2n} − *S*_{n}) is

(a) *n*

(b) 3*n*

(c) 3

(d) none of these

#### Answer:

Here, we are given an A.P. whose sum of *r* terms is *S*_{r}. We need to find.

Here we use the following formula for the sum of *n* terms of an A.P.,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

So, first we find *S*_{3n},

Similarly,

Also,

So, using (1), (2) and (3), we get,

Taking common, we get,

Therefore,

Hence, the correct option is **(c)**.

#### Page No 5.57:

#### Question 18:

If the first term of an A.P. is 2 and common difference is 4, then the sum of its 40 terms is

(a) 3200

(b) 1600

(c) 200

(d) 2800

#### Answer:

In the given problem, we need to find the sum of 40 terms of an arithmetic progression, where we are given the first term and the common difference. So, here we use the following formula for the sum of *n* terms of an A.P.,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

Given,

First term (*a*) = 2

Common difference (*d*) = 4

Number of terms (*n*) = 40

So, using the formula we get,

Therefore, the sum of first 40 terms for the given A.P. is. So, the correct option is **(a)**.

#### Page No 5.57:

#### Question 19:

The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is

(a) 5

(b) 10

(c) 12

(d) 14

(e) 20

#### Answer:

In the given problem, we have an A.P.

Here, we need to find the number of terms *n* such that the sum of *n* terms is 406.

So here, we will use the formula,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

The first term (*a*) = 3

The sum of* n* terms (*S*_{n}) = 406

Common difference of the A.P. (*d*) =

So, on substituting the values in the formula for the sum of *n* terms of an A.P., we get,

So, we get the following quadratic equation,

On solving by splitting the middle term, we get,

Further,

Or,

Since, the number of terms cannot be a fraction, the number of terms (*n*) is

Hence, the correct option is **(d)**.

#### Page No 5.57:

#### Question 20:

Sum of n terms of the series $\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+..is$

(a) $\frac{n(n+1)}{2}$

(b) $2n(n+1)$

(c) $\frac{n(n+1)}{\sqrt{2}}$

(d) 1

#### Answer:

In the given problem, we need to find the sum of terms for a given arithmetic progression,

So, here we use the following formula for the sum of *n* terms of an A.P.,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

Here,

Common difference of the A.P. (*d*) =

Number of terms* *(*n*) = *n*

First term for the given A.P. (*a*) =

So, using the formula we get,

Now, taking common from both the terms inside the bracket we get,

Therefore, the sum of first *n* terms for the given A.P. is. So, the correct option is **(c).**

#### Page No 5.57:

#### Question 21:

The 9th term of an A.P. is 449 and 449th term is 9. The term which is equal to zero is

(a) 501^{th}

(b) 502^{th}

(c) 508^{th}

(d) none of these

#### Answer:

In the given problem, let us take the first term as *a* and the common difference as *d*.

Here, we are given that,

We need to find *n*

Also, we know,

For the 9^{th} term (*n = *9),

Similarly, for the 449^{th} term (*n = *449),

Subtracting (3) from (4), we get,

Now, to find *a*, we substitute the value of *d* in (3),

So, for the given A.P

So, let us take the term equal to zero as the *n*^{th} term. So,

So,

Therefore, the correct option is **(d).**

#### Page No 5.57:

#### Question 22:

If $\frac{1}{x+2},\frac{1}{x+3},\frac{1}{x+5}$ are in A.P. Then, *x* =

(a) 5

(b) 3

(c) 1

(d) 2

#### Answer:

Here, we are given three terms,

First term (*a*_{1}) =

Second term (*a*_{2}) =

Third term (*a*_{3}) =

We need to find the value of *x* for which these terms are in A.P. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,

Also,

Now, on equating (1) and (2), we get,

Therefore, for, these three terms will form an A.P.

So, the correct option is **(c)**.

#### Page No 5.57:

#### Question 23:

The *n*th term of an A.P., the sum of whose *n* terms is S_{n}, is

(a) S_{n} + S_{n−1}

(b) S_{n}_{ }− S_{n−1}

(c) S_{n} + S_{n+1}

(d) S_{n} − S_{n+1}

#### Answer:

A.P. we use following formula,

So, the* n*^{th} term of the A.P. is given by . Therefore, the correct option is **(b).**

#### Page No 5.57:

#### Question 24:

The common difference of an A.P., the sum of whose n terms is S_{n}, is

(a) S_{n} − 2S_{n−1}_{ }+ S_{n−2}

(b) S_{n} − 2S_{n−1}_{ }− S_{n−2}

(c) S_{n} − S_{n−2}

(d) S_{n} − S_{n−1}

#### Answer:

Here, we are given an A.P. the sum of whose *n* terms is *S*_{n}. So, to calculate the common difference of the A.P, we find two consecutive terms of the A.P.

Now, the *n*^{th} term of the A.P will be given by the following formula,

Next, we find the (*n *− 1)^{th} term using the same formula,

Now, the common difference of an A.P. (*d*) =

Therefore,

Hence the correct option is **(a).**

#### Page No 5.58:

#### Question 25:

If the sums of *n* terms of two arithmetic progressions are in the ratio $\frac{3n+5}{5n-7}$, then their *n*^{th} terms are in the ratio

(a) $\frac{3n-1}{5n-1}$

(b) $\frac{3n+1}{5n+1}$

(c) $\frac{5n+1}{3n+1}$

(d) $\frac{5n-1}{3n-1}$

#### Answer:

In the given problem, the ratio of the sum of *n* terms of two A.P’s is given by the expression,

We need to find the ratio of their *n*^{th }terms.

Here we use the following formula for the sum of *n* terms of an A.P.,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

So,

Where, *a* and *d* are the first term and the common difference of the first A.P.

Similarly,

Where, *a*^{’} and *d*^{’} are the first term and the common difference of the first A.P.

So,

Equating (1) and (2), we get,

Now, to find the ratio of the *n*^{th}* *term, we replace *n* by. We get,

As we know,

Therefore, we get,

Hence the correct option is **(b)**.

#### Page No 5.58:

#### Question 26:

If S_{n} denote the sum of *n* terms of an A.P. with first term *a *and common difference *d* such that $\frac{Sx}{Skx}$ is independent of *x*, then

(a) *d*= *a*

(b) *d* = 2*a*

(c) *a* = 2*d*

(d) *d *= −*a*

#### Answer:

Here, we are given an A.P. with *a* as the first term and *d* as the common difference. The sum of *n* terms of the A.P. is given by *S*_{n}*.*

We need to find the relation between *a* and *d* such thatis independent of

So, let us first find the values of *S*_{x}* *and *S*_{kx} using the following formula for the sum of *n* terms of an A.P.,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

So, we get,

Similarly,

So,

Now, to get a term independent of *x* we have to eliminate the other terms, so we get

So, if we substitute, we get,

Therefore,

Hence, the correct option is **(b)**.

#### Page No 5.58:

#### Question 27:

If the first term of an A.P. is a and nth term is *b*, then its common difference is

(a) $\frac{b-a}{n+1}$

(b) $\frac{b-a}{n-1}$

(c) $\frac{b-a}{n}$

(d) $\frac{b+a}{n-1}$

#### Answer:

Here, we are given the first term of the A.P. as *a* and the *n*^{th} term (*a*_{n}) as *b. *So, let us take the common difference of the A.P. as *d.*

Now, as we know,

On substituting the values given in the question, we get.

Therefore,

Hence the correct option is **(b).**

#### Page No 5.58:

#### Question 28:

The sum of first* n* odd natural numbers is

(a) 2*n** *− 1

(b) 2*n** *+ 1

(c) *n*^{2}

(d) *n*^{2} − 1

#### Answer:

In this problem, we need to find the sum of first *n* odd natural numbers.

So, we know that the first odd natural number is 1. Also, all the odd terms will form an A.P. with the common difference of 2.

So here,

First term (*a*) = 1

Common difference (*d*) = 2

So, let us take the number of terms as *n*

Now, as we know,

So, for *n *terms,

Therefore, the sum of first *n* odd natural numbers is.

Hence the correct option is **(c).**

#### Page No 5.58:

#### Question 29:

Two A.P.'s have the same common difference. The first term of one of these is 8 and that of the other is 3. The difference between their 30th term is

(a) 11

(b) 3

(c) 8

(d) 5

#### Answer:

Here, we are given two A.P.’s with same common difference. Let us take the common difference as *d.*

Given,

First term of first A.P. (*a*) = 8

First term of second A.P. (*a*_{’}) = 3

We need to find the difference between their 30^{th} terms.

So, let us first find the 30^{th} term of first A.P.

Similarly, we find the 30^{th} term of second A.P.

Now, the difference between the 30^{th} terms is,

Therefore,

Hence, the correct option is **(d).**

#### Page No 5.58:

#### Question 30:

If 18, *a*, *b*, −3 are in A.P., the *a* + *b* =

(a) 19

(b) 7

(c) 11

(d) 15

#### Answer:

Here, we are given four terms which are in A.P.,

First term (*a*_{1}) =

Second term (*a*_{2}) =

Third term (*a*_{3}) =

Fourth term (*a*_{4})=

So, in an A.P. the difference of two adjacent terms is always constant. So, we get,

Also,

Now, on equating (1) and (2), we get,

Therefore,

Hence the correct option is **(d)**.

#### Page No 5.58:

#### Question 31:

The sum of* n* terms of two A.P.'s are in the ratio 5*n* + 9 : 9*n* + 6. Then, the ratio of their 18th term is

(a) $\frac{179}{321}$

(b) $\frac{178}{321}$

(c) $\frac{175}{321}$

(d) $\frac{176}{321}$

#### Answer:

In the given problem, the ratio of the sum of *n* terms of two A.P’s is given by the expression,

We need to find the ratio of their *18*^{th }terms.

Here we use the following formula for the sum of *n* terms of an A.P.,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

So,

Where, *a* and *d* are the first term and the common difference of the first A.P.

Similarly,

Where, *a*’ and *d*^{’} are the first term and the common difference of the first A.P.

So,

Equating (1) and (2), we get,

Now, to find the ratio of the *n*^{th}* *term, we replace *n* by. We get,

As we know,

Therefore, for the 18^{th} terms, we get,

Hence

Hence no option is correct.

#### Page No 5.58:

#### Question 32:

If $\frac{5+9+13+...tonterms}{7+9+11+...to(n+1)terms}=\frac{17}{16},$then n =

(a) 8

(b) 7

(c) 10

(d) 11

#### Answer:

Here, we are given,

We need to find *n.*

So, first let us find out the sum of *n* terms of the A.P. given in the numerator. Here we use the following formula for the sum of *n* terms of an A.P.,

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

Here,

Common difference of the A.P. (*d*) =

Number of terms* *(*n*) = *n*

First term for the given A.P. (*a*) =

So, using the formula we get,

Similarly, we find out the sum of terms of the A.P. given in the denominator.

Here,

Common difference of the A.P. (*d*) =

Number of terms* *(*n*) = *n*

First term for the given A.P. (*a*) =

So, using the formula we get,

Now substituting the values of (2) and (3) in equation (1), we get,

Further solving the quadratic equation for *n* by splitting the middle term, we get,

So, we get

Or

Since *n *is a whole number, it cannot be a fraction. So,

Therefore, the correct option is **(b).**

#### Page No 5.58:

#### Question 33:

The sum of n terms of an A.P. is 3n^{2} + 5*n*, then 164 is its

(a) 24^{th} term

(b) 27^{th}^{ }term

(c) 26^{th} term

(d) 25^{th} term

#### Answer:

Here, the sum of first *n* terms is given by the expression,

We need to find which term of the A.P. is 164.

Let us take 164 as the *n*^{th} term.

So we know that the *n*^{th}term of an A.P. is given by,

So,

Using the property,

We get,

Further solving for *n*, we get

Therefore,

Hence the correct option is **(b).**

#### Page No 5.58:

#### Question 34:

If the nth term of an A.P. is 2*n* + 1, then the sum of first *n* terms of the A.P. is

(a) *n*(*n* − 2)

(b) *n*(*n* + 2)

(c) *n*(*n* + 1)

(d) *n*(*n* − 1)

#### Answer:

Here, we are given an A.P. whose *n*^{th} term is given by the following expression, . We need to find the sum of first *n* terms.

So, here we can find the sum of the *n* terms of the given A.P., using the formula,

Where, *a* = the first term

*l* = the last term

So, for the given A.P,

The first term (*a*) will be calculated using in the given equation for *n*^{th} term of A.P.

Now, the last term (*l*) or the *n*^{th} term is given

So, on substituting the values in the formula for the sum of *n* terms of an A.P., we get,

Therefore, the sum of the* n *terms of the given A.P. is. So the correct option is **(b).**

#### Page No 5.58:

#### Question 35:

If 18^{th} and 11^{th} term of an A.P. are in the ratio 3 : 2, then its 21^{st}^{ }and 5^{th} terms are in the ratio

(a) 3 : 2

(b) 3 : 1

(c) 1 : 3

(d) 2 : 3

#### Answer:

In the given problem, we are given an A.P whose 18^{th}^{ }and 11^{th}^{ }term are in the ratio 3:2

We need to find the ratio of its 21^{st} and 5^{th} terms

Now, using the formula

,

Where,

*a* = first tem of the A.P

*n *= number of terms

*d* = common difference of the A.P

So,

Also,

Thus,

Further solving for *a*, we get

Now,

Also,

So,

Using (1) in the above equation, we get

Thus, the ratio of the 21^{st} and 5^{th} term is

Therefore the correct option is **(b)**.

#### Page No 5.58:

#### Question 36:

The sum of first 20 odd natural numbers is

(a) 100

(b) 210

(c) 400

(d) 420

#### Answer:

Let *a* be the first term and *d* be the common difference.

We know that, sum of first* n*^{ }terms = *S*_{n }= $\frac{n}{2}$[2*a* + (*n* − 1)*d*]

The given series is 1 + 3 + 5 + ......

First term = *a *= 1.

Common difference = *d* = 3 − 1 = 2

∴ *S*_{20 }= $\frac{20}{2}$[2 × 1 + (20 − 1)2]

= 10(2 + 19 × 2)

= 10(40)

= 400

Hence, the correct option is (c).

#### Page No 5.58:

#### Question 37:

The common difference of the A.P. is $\frac{1}{2q},\frac{1-2q}{2q},\frac{1-4q}{2q},...$is

(a) −1

(b) 1

(c) *q*

(d) 2*q*

#### Answer:

Let *a* be the first term and *d* be the common difference.

The given A.P. is $\frac{1}{2q},\frac{1-2q}{2q},\frac{1-4q}{2q},...$

Common difference = *d* = Second term − First term

= $\frac{1-2q}{2q}-\frac{1}{2q}$

= $\frac{-2q}{2q}=-1$

Hence, the correct option is (a).

#### Page No 5.58:

#### Question 38:

The common difference of the A.P. $\frac{1}{3},\frac{1-3b}{3},\frac{1-6b}{3},...$ is

(a) $\frac{1}{3}$

(b) $-\frac{1}{3}$

(c) −*b*

(d) *b*

#### Answer:

Let *a* be the first term and *d* be the common difference.

The given A.P. is $\frac{1}{3},\frac{1-3b}{3},\frac{1-6b}{3},...$

Common difference = *d* = Second term − First term

= $\frac{1-3b}{3}-\frac{1}{3}$

= $\frac{-3b}{3}=-b$

Hence, the correct option is (c).

#### Page No 5.59:

#### Question 39:

The common difference of the A.P. $\frac{1}{2b},\frac{1-6b}{2b},\frac{1-12b}{2b},...$ is

(a) 2*b*

(b) −2*b*

(c) 3

(d) −3

#### Answer:

Let *a* be the first term and *d* be the common difference.

The given A.P. is $\frac{1}{2b},\frac{1-6b}{2b},\frac{1-12b}{2b},...$

Common difference = *d* = Second term − First term

= $\frac{1-6b}{2b}-\frac{1}{2b}$

= $\frac{-6b}{2b}=-3$

Hence, the correct option is (d).

#### Page No 5.59:

#### Question 40:

If *k*, 2*k* − 1 and 2*k* + 1 are three consecutive terms of an A.P., the value of *k* is

(a) −2

(b) 3

(c) −3

(d) 6

#### Answer:

Since, *k*, 2*k* − 1 and 2*k* + 1 are three consecutive terms of an A.P.

Then, Second term − First term = Third term − Second term = *d* (common difference)

⇒ 2*k* − 1 − *k* = 2*k* + 1 − (2*k* − 1)

⇒ *k* − 1 = 2*k* + 1 − 2*k* + 1

⇒ *k* − 1 = 2

⇒ *k* = 2 + 1

⇒ *k* = 3

Hence, the correct option is (b).

#### Page No 5.59:

#### Question 41:

The next term of the A.P. $\sqrt{7},\sqrt{28},\sqrt{63},....$

(a) $\sqrt{70}$

(b) $\sqrt{84}$

(c) $\sqrt{97}$

(d) $\sqrt{112}$

#### Answer:

Let *a* be the first term and *d* be the common difference.

The given A.P. is $\sqrt{7},\sqrt{28},\sqrt{63},....$

i.e., $\sqrt{7},\sqrt{4\times 7},\sqrt{9\times 7},....$

i.e., $\sqrt{7},2\sqrt{7},3\sqrt{7},....$

Common difference = *d* = Second term − First term

= $2\sqrt{7}-\sqrt{7}$

= $\sqrt{7}$

∴ Next term of the A.P. = $3\sqrt{7}+\sqrt{7}$

= $4\sqrt{7}$

= $\sqrt{16\times 7}$

= $\sqrt{112}$

Hence, the correct option is (d).

#### Page No 5.59:

#### Question 42:

The first three terms of an A.P. respectively are 3*y* − 1, 3*y* + 5 and 5*y* + 1. Then, *y* equals

(a) −3

(b) 4

(c) 5

(d) 2

#### Answer:

Since, 3*y* − 1, 3*y* + 5 and 5*y* + 1 are first three terms of an A.P.

Then, Second term − First term = Third term − Second term = *d* (common difference)

⇒ 3*y* + 5 − (3*y* − 1) = 5*y* + 1 − (3*y* + 5)

⇒ 3*y* + 5 − 3*y* + 1 = 5*y* + 1 − 3*y* − 5

⇒ 6 = 2*y* − 4

⇒ 2*y* = 6 + 4

⇒ 2*y* = 10

⇒ *y* = 5

Hence, the correct option is (c).

#### Page No 5.59:

#### Question 43:

The sum of first five multiples of 3 is

(a) 45

(b) 55

(c) 65

(d) 75

#### Answer:

First five multiples of 3 are:

3, 6, 9, 12, 15

They are in A.P. with

First term = *a* = 3

Common difference =* d* = 6 − 3 = 3

${S}_{5}=\frac{5}{2}\left[2a+\left(5-1\right)d\right]\phantom{\rule{0ex}{0ex}}=\frac{5}{2}\left[2a+4d\right]\phantom{\rule{0ex}{0ex}}=\frac{5}{2}\left[2\left(3\right)+4\left(3\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{5}{2}\left[6+12\right]\phantom{\rule{0ex}{0ex}}=\frac{5}{2}\times 18\phantom{\rule{0ex}{0ex}}=45$

Hence, the correct option is (a).

#### Page No 5.59:

#### Question 44:

The sum of first 16 terms of the A.P.: 10, 6, 2, ..., is

(a) –320

(b) 320

(c) –352

(d) –400

#### Answer:

The given A.P. is:

10, 6, 2, ...,

First term = *a* = 10

Common difference = *d* = 6 − 10 = −4

${S}_{16}=\frac{16}{2}\left[2a+\left(16-1\right)d\right]\phantom{\rule{0ex}{0ex}}=8\left[2a+15d\right]\phantom{\rule{0ex}{0ex}}=8\left[2\left(10\right)+15\left(-4\right)\right]\phantom{\rule{0ex}{0ex}}=8\left[20-60\right]\phantom{\rule{0ex}{0ex}}=8\times \left(-40\right)\phantom{\rule{0ex}{0ex}}=-320$

Hence, the correct option is (a).

#### Page No 5.59:

#### Question 45:

If the first term of an A.P. is –5 and the common difference is 2, then the sum of first 6 terms is

(a) 0

(b) 5

(c) 6

(d) 15

#### Answer:

Given:

First term = *a* = −5

Common difference = *d* = 2

${S}_{6}=\frac{6}{2}\left[2a+\left(6-1\right)d\right]\phantom{\rule{0ex}{0ex}}=3\left[2a+5d\right]\phantom{\rule{0ex}{0ex}}=3\left[2\left(-5\right)+5\left(2\right)\right]\phantom{\rule{0ex}{0ex}}=3\left[-10+10\right]\phantom{\rule{0ex}{0ex}}=3\times 0\phantom{\rule{0ex}{0ex}}=0$

Hence, the correct option is (a).

#### Page No 5.59:

#### Question 46:

The 4^{th} term from the end of the AP: –11, –8, –5, ..., 49 is

(a) 37

(b) 40

(c) 43

(d) 58

#### Answer:

The given A.P. is: –11, –8, –5, ..., 49

Rearranging the terms from last to first, we get

49, 46, 43, ..... –11

First term = *a* = 49

Common difference = *d* = 46 – 49

= –3

The fourth term is:

${a}_{4}=a+\left(4-1\right)d\phantom{\rule{0ex}{0ex}}=49+3\left(-3\right)\phantom{\rule{0ex}{0ex}}=49-9\phantom{\rule{0ex}{0ex}}=40$

Therefore, the 4^{th} term from the end of the AP: –11, –8, –5, ..., 49 is 40.

Hence, the correct option is (b).

#### Page No 5.59:

#### Question 47:

Which term of the A.P. 21, 42, 63, 84, ... is 210?

(a) 9^{th}

(b) 10^{th}

(c) 11^{th}

(d) 12^{th}

#### Answer:

The given A.P. is: 21, 42, 63, 84, ...

First term = *a* = 21

Common difference = *d* = 42 – 21

= 21

*n*^{th} term = *a _{n}* = 210

Now,

${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}\Rightarrow 210=21+\left(n-1\right)\left(21\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 210=21+21n-21\phantom{\rule{0ex}{0ex}}\Rightarrow 210=21n\phantom{\rule{0ex}{0ex}}\Rightarrow n=10$

Therefore, the 10

^{th}term of the A.P. 21, 42, 63, 84, ... is 210.

Hence, the correct option is (b).

#### Page No 5.59:

#### Question 48:

If the 2^{nd} term of an A.P, is 13 and 5^{th} term is 25, what is its 7^{th} term?

(a) 30

(b) 33

(c) 37

(d) 38

#### Answer:

Given:

2^{nd} term = *a*_{2} = 13

5^{th} term = *a*_{5} = 25

Now,

${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{a}_{2}=a+\left(2-1\right)d\phantom{\rule{0ex}{0ex}}\Rightarrow 13=a+d\phantom{\rule{0ex}{0ex}}\Rightarrow a+d=13...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{a}_{5}=a+\left(5-1\right)d\phantom{\rule{0ex}{0ex}}\Rightarrow 25=a+4d\phantom{\rule{0ex}{0ex}}\Rightarrow a+4d=25...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Subtracting}\left(1\right)\mathrm{from}\left(2\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}3d=12\phantom{\rule{0ex}{0ex}}\Rightarrow d=4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}d\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}a=9\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},a=9\mathrm{and}d=4.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\phantom{\rule{0ex}{0ex}}{a}_{7}=a+\left(7-1\right)d\phantom{\rule{0ex}{0ex}}=9+6\left(4\right)\phantom{\rule{0ex}{0ex}}=33$

Therefore, the 7^{th} term is 33.

Hence, the correct option is (b).

#### Page No 5.59:

#### Question 49:

The value of *x* for which 2*x*, *x* + 10 and 3*x* + 2 are the three consecutive terms of an A.P; is

(a) 6

(b) –6

(c) 18

(d) –18

#### Answer:

Given $2x,x+10,3x+2$ are the consecutive terms of an AP.

Therefore, the common difference will be same.

$\Rightarrow \left(x+10\right)-2x=\left(3x+2\right)-\left(x+10\right)\phantom{\rule{0ex}{0ex}}\Rightarrow x+10-2x=3x+2-x-10\phantom{\rule{0ex}{0ex}}\Rightarrow 10-x=2x-8\phantom{\rule{0ex}{0ex}}\Rightarrow 3x=18\phantom{\rule{0ex}{0ex}}\Rightarrow x=6$

Hence, the correct answer is option (a).

#### Page No 5.59:

#### Question 50:

The first term of an A.P. is *p* and the common difference is *q*, then its 10th term is

(a) *q* + 9*p*

(b) *p* – 9*q*

(c) *p* + 9*q*

(d) 2*p* + 9*q*

#### Answer:

The ${n}^{th}$ term of an AP $=a+\left(n-1\right)d,$where $a$ and $d$ are the first term and common difference respectively.

Therefore, ${10}^{th}$ term $=$$p+\left(10-1\right)q=p+9q$.

Hence, the correct answer is option (c).

#### Page No 5.59:

#### Question 1:

The sequence $\frac{2a-6b}{3b},\frac{2a-3b}{3b},\frac{2a}{3b},\frac{2a+3b}{3b},\frac{2a+6b}{3b}$.... is an A.P. with common difference ________

#### Answer:

The given A.P. is $\frac{2a-6b}{3b},\frac{2a-3b}{3b},\frac{2a}{3b},\frac{2a+3b}{3b},\frac{2a+6b}{3b}$....

Common difference (*d*) = *a*_{2} − *a*_{1}

= $\frac{2a-3b}{3b}-\frac{2a-6b}{3b}$

= $\frac{2a-3b-2a+6b}{3b}$

= $\frac{3b}{3b}$

= *b*

Hence, the sequence $\frac{2a-6b}{3b},\frac{2a-3b}{3b},\frac{2a}{3b},\frac{2a+3b}{3b},\frac{2a+6b}{3b}$.... is an A.P. with common difference __ b__.

#### Page No 5.59:

#### Question 2:

If the sequence $\frac{a-3b}{b},\frac{3a-3b}{b},\frac{5a-3b}{b},\frac{7a-3b}{b},...$is an A.P with common difference 3, then *a/b* = ________.

#### Answer:

The given A.P. is: $\frac{a-3b}{b},\frac{3a-3b}{b},\frac{5a-3b}{b},\frac{7a-3b}{b},...$

Common difference (*d*) = *a*_{2} − *a*_{1}

$\Rightarrow 3=\frac{3a-3b}{b}-\frac{a-3b}{b}\phantom{\rule{0ex}{0ex}}\Rightarrow 3=\frac{3a-3b-a+3b}{b}\phantom{\rule{0ex}{0ex}}\Rightarrow 3=\frac{2a}{b}\phantom{\rule{0ex}{0ex}}\Rightarrow 3b=2a\phantom{\rule{0ex}{0ex}}\Rightarrow 2a=3b\phantom{\rule{0ex}{0ex}}\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}2b,\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2a}{2b}=\frac{3b}{2b}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a}{b}=\frac{3}{2}$

Hence, if the sequence $\frac{a-3b}{b},\frac{3a-3b}{b},\frac{5a-3b}{b},\frac{7a-3b}{b},...$is an A.P with common difference 3, then *a/b* = $\overline{)\frac{3}{2}}$.

#### Page No 5.60:

#### Question 3:

The 11th term of the A.P.$-5,-\frac{5}{2},0,\frac{5}{2}$,..., is

#### Answer:

The given A.P. is: $-5,-\frac{5}{2},0,\frac{5}{2}$,...,

First term (*a*) = –5

Common difference (*d*) = *a*_{2} – *a*_{1}

= $-\frac{5}{2}-\left(-5\right)$

= $-\frac{5}{2}+5$

= $\frac{-5+10}{2}$

= $\frac{5}{2}$

The 11th term is:

${a}_{11}=a+\left(11-1\right)d\phantom{\rule{0ex}{0ex}}=-5+10\left(\frac{5}{2}\right)\phantom{\rule{0ex}{0ex}}=-5+5\left(5\right)\phantom{\rule{0ex}{0ex}}=-5+25\phantom{\rule{0ex}{0ex}}=20$

Hence, the 11th term of the A.P.$-5,-\frac{5}{2},0,\frac{5}{2}$,..., is __20__.

#### Page No 5.60:

#### Question 4:

The 21^{st} term of the A.P. whose first two terms are – 3 and 4 is ____________.

#### Answer:

Given:

First term (*a*) = –3

Second term (*a*_{2}) = 4

Common difference (*d*) = *a*_{2} – *a*_{1}

= $4-\left(-3\right)$

= 4 + 3

= 7

The 21^{st} term is:

${a}_{21}=a+\left(21-1\right)d\phantom{\rule{0ex}{0ex}}=-3+20\left(7\right)\phantom{\rule{0ex}{0ex}}=-3+140\phantom{\rule{0ex}{0ex}}=137$

Hence, the 21^{st} term of the A.P. whose first two terms are –3 and 4 is __137__.

#### Page No 5.60:

#### Question 5:

The famous Mathematician associated with finding the sum of the first 100 natural numbers was _________.

#### Answer:

Gauss is the famous Mathematician associated with finding the sum of the first 100 natural numbers.

Hence, the famous Mathematician associated with finding the sum of the first 100 natural numbers was __Gauss__.

#### Page No 5.60:

#### Question 6:

If the common difference of an A.P. is 5, then *a*_{18} – *a*_{13} = ______________.

#### Answer:

Given:

Common difference (*d*) = 5

Now,

${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{a}_{18}=a+\left(18-1\right)d=a+17d\phantom{\rule{0ex}{0ex}}{a}_{13}=a+\left(13-1\right)d=a+12d\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore {a}_{18}-{a}_{13}=\left(a+17d\right)-\left(a+12d\right)\phantom{\rule{0ex}{0ex}}=17d-12d\phantom{\rule{0ex}{0ex}}=5d\phantom{\rule{0ex}{0ex}}=5\times 5\left(\because d=5\right)\phantom{\rule{0ex}{0ex}}=25$

Hence, *a*_{18} – *a*_{13} = __25__.

#### Page No 5.60:

#### Question 7:

If *a*_{1}, *a*_{2}, *a*_{3}, ....., *a _{n}*, ..... is an A.P. such that

*a*

_{18}–

*a*

_{14}= 32, then its common difference is __________.

#### Answer:

Given:

*a*_{18} – *a*_{14} = 32

Now,

${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{a}_{18}=a+\left(18-1\right)d=a+17d\phantom{\rule{0ex}{0ex}}{a}_{14}=a+\left(14-1\right)d=a+13d\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore {a}_{18}-{a}_{14}=\left(a+17d\right)-\left(a+13d\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 32=17d-13d\phantom{\rule{0ex}{0ex}}\Rightarrow 32=4d\phantom{\rule{0ex}{0ex}}\Rightarrow d=8$

Hence, if *a*_{1}, *a*_{2}, *a*_{3}, ....., *a _{n}*, ..... is an A.P. such that

*a*

_{18}–

*a*

_{14}= 32, then its common difference is

__8__.

#### Page No 5.60:

#### Question 8:

If 5, *a*_{2}, *a*_{3}, ...., *a*_{20}, 145 is an A.P., then *a*_{2} + *a*_{20 }=__________.

#### Answer:

If *a*_{1}, *a*_{2}, *a*_{3}, ...., *a _{n}* are in A.P., then

*a*

_{1 }+

*a*=

_{n }*a*

_{2 }+

*a*

_{n−}_{1}

*=*

_{ }*a*

_{3 }+

*a*

_{n−}_{2 }= ....

Here,

*a*

_{1 }+

*a*

_{21}

*=*

_{ }*a*

_{2 }+

*a*

_{20}

⇒

*a*

_{2 }+

*a*

_{20}= 5 + 145 = 150

Hence,

*a*

_{2}+

*a*

_{20 }=

__150__.

#### Page No 5.60:

#### Question 9:

If *n *– 2, 4*n* – 1, and 5*n *+ 2 are in A.P., then *n *= _______.

#### Answer:

If *n *– 2, 4*n* – 1, and 5*n *+ 2 are in A.P.,

then their common difference must be same.

$\therefore {a}_{2}-{a}_{1}={a}_{3}-{a}_{2}=d\phantom{\rule{0ex}{0ex}}\Rightarrow \left(4n-1\right)-\left(n-2\right)=\left(5n+2\right)-\left(4n-1\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 4n-1-n+2=5n+2-4n+1\phantom{\rule{0ex}{0ex}}\Rightarrow 3n+1=n+3\phantom{\rule{0ex}{0ex}}\Rightarrow 3n-n=3-1\phantom{\rule{0ex}{0ex}}\Rightarrow 2n=2\phantom{\rule{0ex}{0ex}}\Rightarrow n=1$

Hence, *n *= __1__.

#### Page No 5.60:

#### Question 10:

The value of the middle term of the A.P. –11, –7, –3, ...., 49, 53 is ________.

#### Answer:

The given A.P. is: –11, –7, –3, ...., 49, 53

First term (*a*) = –11

Common difference (*d*) = *a*_{2} – *a*_{1}

= –7 – (–11)

= –7 + 11

= 4

Last term (*a _{n}*) = 53

Now,

${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}\Rightarrow 53=-11+\left(n-1\right)\left(4\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 53=-11+4n-4\phantom{\rule{0ex}{0ex}}\Rightarrow 53=-15+4n\phantom{\rule{0ex}{0ex}}\Rightarrow 4n=53+15\phantom{\rule{0ex}{0ex}}\Rightarrow 4n=68\phantom{\rule{0ex}{0ex}}\Rightarrow n=17$

∴ Total number of terms = 17

$\mathrm{Middle}\mathrm{term}={\left(\frac{n+1}{2}\right)}^{\mathrm{th}}\mathrm{term}\phantom{\rule{0ex}{0ex}}={\left(\frac{17+!}{2}\right)}^{\mathrm{th}}\mathrm{term}\phantom{\rule{0ex}{0ex}}={9}^{\mathrm{th}}\mathrm{term}$

Therefore,

${a}_{9}=a+\left(9-1\right)d\phantom{\rule{0ex}{0ex}}=-11+8\left(4\right)\phantom{\rule{0ex}{0ex}}=-11+32\phantom{\rule{0ex}{0ex}}=21$

Hence, the value of the middle term of the A.P. –11, –7, –3, ...., 49, 53 is

__21__.

#### Page No 5.60:

#### Question 11:

If 9^{th} term of an A.P. is zero, then its 29^{th} and 19^{th} terms are in the ratio __________.

#### Answer:

Given:

9^{th} term (*a*_{9}) = 0

Now,

${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{a}_{9}=a+\left(9-1\right)d\phantom{\rule{0ex}{0ex}}\Rightarrow 0=a+8d\phantom{\rule{0ex}{0ex}}\Rightarrow a=-8d$

Therefore,

$\frac{{a}_{29}}{{a}_{19}}=\frac{a+\left(29-1\right)d}{a+\left(19-1\right)d}\phantom{\rule{0ex}{0ex}}=\frac{a+28d}{a+18d}\phantom{\rule{0ex}{0ex}}=\frac{-8d+28d}{-8d+18d}\left(\because a=-8d\right)\phantom{\rule{0ex}{0ex}}=\frac{20d}{10d}\phantom{\rule{0ex}{0ex}}=\frac{2}{1}$

Hence, if 9^{th} term of an A.P. is zero, then its 29^{th} and 19^{th} terms are in the ratio __2 : 1__.

#### Page No 5.60:

#### Question 12:

If *a _{n}* denotes the

*n*

^{th}term of the A.P. 3, 8, 13, 18, ..., then the value of

*a*

_{30}–

*a*

_{20}is __________.

#### Answer:

The given A.P. is: 3, 8, 13, 18, ...

First term (*a*) = 3

Common difference (*d*) = 8 – 3 = 5

Now,

${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{a}_{30}=a+\left(30-1\right)d=a+29d\phantom{\rule{0ex}{0ex}}{a}_{20}=a+\left(20-1\right)d=a+19d\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore {a}_{30}-{a}_{20}=\left(a+29d\right)-\left(a+19d\right)\phantom{\rule{0ex}{0ex}}=29d-19d\phantom{\rule{0ex}{0ex}}=10d\phantom{\rule{0ex}{0ex}}=10\times 5\left(\because d=5\right)\phantom{\rule{0ex}{0ex}}=50$

Hence, the value of *a*_{30} –* a*_{20} is __50__.

#### Page No 5.60:

#### Question 13:

The sum of first 50 odd natural numbers is __________.

#### Answer:

First 50 odd natural numbers are: 1, 3, 5, 7, ...

First term (*a*) = 1

Common difference (*d*) = 3 – 1 = 2

Now,

${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}{S}_{50}=\frac{50}{2}\left[2a+\left(50-1\right)d\right]\phantom{\rule{0ex}{0ex}}=25\left[2\left(1\right)+49\left(2\right)\right]\phantom{\rule{0ex}{0ex}}=25\left[2+98\right]\phantom{\rule{0ex}{0ex}}=25\times 100\phantom{\rule{0ex}{0ex}}=2500$

Hence, the sum of first 50 odd natural numbers is __2500__.

#### Page No 5.60:

#### Question 14:

The sum of first *n* odd natural numbers is ________.

#### Answer:

First *n *odd natural numbers are: 1, 3, 5, 7, ...

First term (*a*) = 1

Common difference (*d*) = 3 – 1 = 2

Now,

${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}=\frac{n}{2}\left[2\left(1\right)+\left(n-1\right)\left(2\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{n}{2}\left[2+2n-2\right]\phantom{\rule{0ex}{0ex}}=\frac{n}{2}\left[2n\right]\phantom{\rule{0ex}{0ex}}={n}^{2}$

Hence, the sum of first *n* odd natural numbers is __ n^{2}__.

#### Page No 5.60:

#### Question 15:

In an A.P. *a*_{1}, *a*_{2}, *a*_{3}, ..., *a _{n}*, ...., if

*a*

_{1}= 1,

*a*= 20 and S

_{n }*= 399, then the value of*

_{n }*n*is __________.

#### Answer:

Given:

First term (*a*) = *a*_{1} = 1

*n*^{th} term = *a _{n }*= 20

Sum of

*n*terms = S

*= 399*

_{n }Now,

${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}\Rightarrow 20=1+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}\Rightarrow \left(n-1\right)d=20-1\phantom{\rule{0ex}{0ex}}\Rightarrow \left(n-1\right)d=19...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 399=\frac{n}{2}\left[2\left(1\right)+19\right]\left(\mathrm{From}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 399=\frac{n}{2}\left[21\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 798=21n\phantom{\rule{0ex}{0ex}}\Rightarrow n=\frac{798}{21}\phantom{\rule{0ex}{0ex}}\Rightarrow n=38$

Hence, the value of

*n*is

__38__.

#### Page No 5.60:

#### Question 16:

If 7 times the 7^{th} term of an A.P. is equal to 11 times its 11^{th} term, then the value of its 18^{th} term is ________.

#### Answer:

Given:

$7{a}_{7}=11{a}_{11}$

Now,

${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\phantom{\rule{0ex}{0ex}}7{a}_{7}=11{a}_{11}\phantom{\rule{0ex}{0ex}}\Rightarrow 7\left[a+\left(7-1\right)d\right]=11\left[a+\left(11-1\right)d\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 7\left[a+6d\right]=11\left[a+10d\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 7a+42d=11a+110d\phantom{\rule{0ex}{0ex}}\Rightarrow 11a-7a=42d-110d\phantom{\rule{0ex}{0ex}}\Rightarrow 4a=-68d\phantom{\rule{0ex}{0ex}}\Rightarrow a=-17d...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{a}_{18}=a+\left(18-1\right)d\phantom{\rule{0ex}{0ex}}=-17d+17d\left(\mathrm{From}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=0$

Hence, the value of its 18^{th} term is __0__.

#### Page No 5.60:

#### Question 17:

Two arithmetic progressions have the same common difference. Their first terms are *A *and *B *respectively. The difference between their *n*^{th} terms is _________.

#### Answer:

Let the first term of first A.P. be *A* and common difference be *d*.

And the first term of second A.P. be *B* and common difference be *d*.

*n*^{th} term of first A.P. = *a _{n}*

*n*

^{th}term of second A.P. =

*b*

_{n}Now,

${a}_{n}=A+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}\mathrm{and}{b}_{n}=B+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\phantom{\rule{0ex}{0ex}}{a}_{n}-{b}_{n}=A+\left(n-1\right)d-B-\left(n-1\right)d\phantom{\rule{0ex}{0ex}}=A-B\phantom{\rule{0ex}{0ex}}$

Hence, the difference between their

*n*

^{th}terms is

*.*

__A − B__#### Page No 5.60:

#### Question 18:

If the *n*^{th} terms of two A.P.s: 9,7,5.... and 24, 21, 18, ... are the same, then the value of *n *is _________.

#### Answer:

The given A.P.s are:

9,7,5.... and 24, 21, 18, ...

9, 7, 5,...

First term (*a*) = 9

Common difference (*d*) = 7 − 9 = −2

${n}^{\mathrm{th}}\mathrm{term}\left({a}_{n}\right)=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}=9+\left(n-1\right)\left(-2\right)\phantom{\rule{0ex}{0ex}}=9-2n+2\phantom{\rule{0ex}{0ex}}=11-2n...\left(1\right)$

24, 21, 18, ...

First term (*a*) = 24

Common difference (*d*) = 21 − 24 = −3

${n}^{\mathrm{th}}\mathrm{term}\left({b}_{n}\right)=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}=24+\left(n-1\right)\left(-3\right)\phantom{\rule{0ex}{0ex}}=24-3n+3\phantom{\rule{0ex}{0ex}}=27-3n...\left(2\right)$

It is given that,* a _{n} = b_{n}*

$\Rightarrow 11-2n=27-3n\phantom{\rule{0ex}{0ex}}\Rightarrow 3n-2n=27-11\phantom{\rule{0ex}{0ex}}\Rightarrow n=16$

Hence, the value of

*n*is

__16__.

#### Page No 5.60:

#### Question 19:

If* S*_{n} = *n*(4*n* + 1) is the sum of* n* terms of an A.P., then its common difference is __________.

#### Answer:

Given:

${S}_{n}=n\left(4n+1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{Sum}\mathrm{of}\mathrm{one}\mathrm{term}={S}_{1}=1\left(4+1\right)=5\phantom{\rule{0ex}{0ex}}\Rightarrow {a}_{1}=a=5...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Sum}\mathrm{of}\mathrm{two}\mathrm{term}s={S}_{2}=2\left(8+1\right)=18\phantom{\rule{0ex}{0ex}}\Rightarrow {a}_{1}+{a}_{2}=18...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Subtracting}\left(1\right)\mathrm{from}\left(2\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}{a}_{2}=13\phantom{\rule{0ex}{0ex}}\Rightarrow a+d=13\phantom{\rule{0ex}{0ex}}\Rightarrow d=8\left(\because a=5\right)$

Hence, if* S*_{n} = *n*(4*n* + 1) is the sum of* n* terms of an A.P., then its common difference is __8__.

#### Page No 5.60:

#### Question 20:

If the ratio of the sums of first *n* terms of two A.P.s is $\frac{5n+13}{7n+27}$, then the ratio of their 4^{th} terms is ________.

#### Answer:

Given:

$\frac{{S}_{n}}{{S}_{n}^{\text{'}}}=\frac{5n+13}{7n+27}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Let}\mathrm{the}\mathrm{first}\mathrm{term}\mathrm{of}{1}^{\mathrm{st}}\mathrm{A}.\mathrm{P}.\mathrm{be}a\mathrm{and}\mathrm{of}{2}^{\mathrm{nd}}\mathrm{A}.\mathrm{P}.\mathrm{be}{a}^{\text{'}}.\phantom{\rule{0ex}{0ex}}\mathrm{Let}\mathrm{the}\mathrm{common}\mathrm{difference}\mathrm{of}{1}^{\mathrm{st}}\mathrm{A}.\mathrm{P}.\mathrm{be}d\mathrm{and}\mathrm{of}{2}^{\mathrm{nd}}\mathrm{A}.\mathrm{P}.\mathrm{be}{d}^{\text{'}}.\phantom{\rule{0ex}{0ex}}\mathrm{Let}{\mathrm{S}}_{\mathrm{n}}\mathrm{be}\mathrm{the}\mathrm{sum}\mathrm{of}n\mathrm{terms}\mathrm{of}{1}^{\mathrm{st}}\mathrm{A}.\mathrm{P}.\mathrm{and}{\mathrm{S}}_{\mathrm{n}}^{\text{'}}\mathrm{be}\mathrm{the}\mathrm{sum}\mathrm{of}n\mathrm{terms}\mathrm{of}{2}^{\mathrm{nd}}\mathrm{A}.\mathrm{P}..\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\frac{{S}_{n}}{{S}_{n}^{\text{'}}}=\frac{{\displaystyle \frac{n}{2}}\left[2a+\left(n-1\right)d\right]}{{\displaystyle \frac{n}{2}}\left[2{a}^{\text{'}}+\left(n-1\right){d}^{\text{'}}\right]}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{5n+13}{7n+27}=\frac{2a+\left(n-1\right)d}{2{a}^{\text{'}}+\left(n-1\right){d}^{\text{'}}}....\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Let}{a}_{4}\mathrm{be}\mathrm{the}{4}^{\mathrm{th}}\mathrm{term}\mathrm{of}{1}^{\mathrm{st}}\mathrm{A}.\mathrm{P}.\mathrm{and}{a}_{4}^{\text{'}}\mathrm{be}\mathrm{the}{4}^{\mathrm{th}}\mathrm{term}\mathrm{of}{2}^{\mathrm{nd}}\mathrm{A}.\mathrm{P}..\phantom{\rule{0ex}{0ex}}\frac{{a}_{4}}{{a}_{4}^{\text{'}}}=\frac{a+\left(4-1\right)d}{{a}^{\text{'}}+\left(4-1\right){d}^{\text{'}}}\phantom{\rule{0ex}{0ex}}=\frac{a+3d}{{a}^{\text{'}}+3{d}^{\text{'}}}\phantom{\rule{0ex}{0ex}}\mathrm{Multiplying}\mathrm{and}\mathrm{dividing}\mathrm{the}\mathrm{RHS}\mathrm{by}2,\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\frac{{a}_{4}}{{a}_{4}^{\text{'}}}=\frac{2a+6d}{2{a}^{\text{'}}+6{d}^{\text{'}}}....\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Putting}\mathrm{n}=7\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\frac{5\left(7\right)+13}{7\left(7\right)+27}=\frac{2a+\left(7-1\right)d}{2{a}^{\text{'}}+\left(7-1\right){d}^{\text{'}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{35+13}{49+27}=\frac{2a+6d}{2{a}^{\text{'}}+6{d}^{\text{'}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2a+6d}{2{a}^{\text{'}}+6{d}^{\text{'}}}=\frac{48}{76}=\frac{12}{19}....\left(3\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{From}\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\frac{{a}_{4}}{{a}_{4}^{\text{'}}}=\frac{12}{19}$

Hence, the ratio of their 4^{th} terms is __12 : 19__.

#### Page No 5.60:

#### Question 1:

Define an arithmetic progression.

#### Answer:

An arithmetic progression is a sequence of terms such that the difference between any two consecutive terms of the sequence is always same.

Suppose we have a sequence

So, if these terms are in A.P., then,

And so on…

Here, *d* is the common difference of the A.P.

Example: 1, 3, 5, 7, 9 … is an A.P. with common difference (*d*) as 2.

#### Page No 5.60:

#### Question 2:

Write the common difference of an A.P. whose *n*th term is a_{n} = 3n + 7.

#### Answer:

In the given problem, *n*th term is given by,

.

To find the common difference of the A.P., we need two consecutive terms of the A.P.

So, let us find the first and the second term of the given A.P.

First term,

Second term (),

Now, the common difference of the A.P. (*d*) =

Therefore, the common difference is.

#### Page No 5.60:

#### Question 3:

Which term of the sequence 114, 109, 104, ... is the first negative term?

#### Answer:

Here, A.P is

So, first term,

Now,

Common difference (*d*) =

Now, we need to find the first negative term,

Further simplifying, we get,

Thus,

Therefore, the first negative term is the of the given A.P.

#### Page No 5.60:

#### Question 4:

Write the value of a_{30} − a_{10} for the A.P. 4, 9, 14, 19, ....

#### Answer:

In this problem, we are given an A.P. and we need to find.

A.P. is

Here,

First term (*a*) = 4

Common difference of the A.P. (*d*)

Now, as we know,

Here, we find *a*_{30} and *a*_{20}*.*

So, for 30^{th} term,

Also, for 10^{th} term,

So,

Therefore, for the given A.P.

#### Page No 5.61:

#### Question 5:

Write 5th term from the end of the A.P. 3, 5, 7, 9, ..., 201.

#### Answer:

In the given problem, we need to find the 5^{th} term from the end for the given A.P.

3, 5, 7, 9 …201

Here, to find the 5^{th} term from the end let us first find the common difference of the A.P. So,

First term (*a*) = 3

Last term (*a*_{n}) = 201

Common difference (*d*) =

Now, as we know, the *n*^{th} term from the end can be given by the formula,

So, the 5^{th} term from the end,

Therefore, the 5^{th} term from the end of the given A.P. is.

#### Page No 5.61:

#### Question 6:

Write the value of x for which 2*x*, *x* + 10 and 3*x* + 2 are in A.P.

#### Answer:

Here, we are given three terms,

First term (*a*_{1}) =

Second term (*a*_{2}) =

Third term (*a*_{3}) =

We need to find the value of *x* for which these terms are in A.P. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,

Also,

Now, on equating (1) and (2), we get,

Therefore, for, these three terms will form an A.P.

#### Page No 5.61:

#### Question 7:

Write the *n*th term of an A.P. the sum of whose* n* terms is S_{n}.

#### Answer:

We are given an A.P. the sum of whose *n* terms is *S*_{n}. So, to calculate the *n*^{th} term of the A.P. we use following formula,

So, the* n*^{th} term of the A.P. is given by .

#### Page No 5.61:

#### Question 8:

Write the sum of first *n* odd natural numbers.

#### Answer:

In this problem, we need to find the sum of first *n* odd natural numbers.

So, we know that the first odd natural number is 1. Also, all the odd terms will form an A.P. with the common difference of 2.

So here,

First term (*a*) = 1

Common difference (*d*) = 2

So, let us take the number of terms as *n*

Now, as we know,

So, for *n *terms,

Therefore, the sum of first *n* odd natural numbers is.

#### Page No 5.61:

#### Question 9:

Write the sum of first *n* even natural numbers.

#### Answer:

In this problem, we need to find the sum of first *n* even natural numbers.

So, we know that the first odd natural number is 2. Also, all the odd terms will form an A.P. with the common difference of 2.

So here,

First term (*a*) = 2

Common difference (*d*) = 2

So, let us take the number of terms as *n*

Now, as we know,

So, for *n *terms,

Therefore, the sum of first *n* even natural numbers is.

#### Page No 5.61:

#### Question 10:

If the sum of *n* terms of an A.P. is S_{n} = 3n^{2} + 5n. Write its common difference.

#### Answer:

Here, we are given,

Let us take the first term as *a* and the common difference as *d*.

Now, as we know,

So, we get,

Also,

On comparing the terms containing *n *in (1) and (2), we get,

Therefore, the common difference is