## RD Sharma Solutions Class 10 Maths Chapter 9 – Free PDF Download

**RD Sharma Solutions for Class 10 Maths Chapter 9 â€“ Arithmetic Progressions** is provided here for students to study and excel in their board exams. In order to succeed in the Class 10 Mathematics examinations, a structured way of understanding the concepts and solving problems is a must. The RD Sharma Solutions is on par with these needs for a student to secure good marks in the qualification exam. Furthermore, the solutions have been created by our expert team at BYJUâ€™S, which is in simple language with appropriate explanations to give clarity on the concepts in the chapter.

Arithmetic Progressions of** RD Sharma Solutions Class 10** is an exciting chapter where students can score good marks easily. But due to some misconceptions, not all students can grab those marks. In order to help students, the RD Sharma Solutions for Class 10 provides descriptive answers to all six exercises. This chapterâ€™s key aspects are:

- Understanding sequences
- Arithmetic Progressions
- Describing the sequence by writing the algebraic formula for its terms
- Find the sum of terms in an A.P.
- Solving various word problems related to Arithmetic Progressions

## Download the PDF of RD Sharma Solutions For Class 10 Maths Chapter 9 Arithmetic Progressions here

### Access the RD Sharma Solutions For Class 10 Maths Chapter 9 – Arithmetic Progression

### RD Sharma Solutions for Class 10 Maths Chapter 9 Exercise 9.1 Page No: 9.5

**1. Write the first terms of each of the following sequences whose n ^{th}Â term are:**

**(i) a _{n}Â = 3n + 2**

**(ii) a _{n}Â = (n â€“ 2)/3**

**(iii) a _{n}Â = 3^{n}**

**(iv) a _{n}Â = (3n â€“ 2)/ 5**

**(v) a _{n}Â = (-1)^{n }. 2^{n}**

**(vi) a _{n}Â = n(n â€“ 2)/2**

**(vii) a _{n}Â = n^{2}Â – n + 1**

**(viii) a _{n}Â = n^{2}Â – n + 1**

**(ix) a _{n}Â = (2nÂ â€“ 3)/ 6**

**Solutions:**

(i) a_{n}Â = 3n + 2

Given sequence whose a_{n}Â = 3nÂ + 2

To get the first five terms of given sequence, put n = 1, 2, 3, 4, 5 and we get

a_{1}Â = (3 Ã— 1) + 2 = 3 + 2 = 5

a_{2}Â = (3 Ã— 2) + 2 = 6 + 2 = 8

a_{3}Â = (3 Ã— 3) + 2 = 9 + 2 = 11

a_{4}Â = (3 Ã— 4) + 2 = 12 + 2 = 14

a_{5}Â = (3 Ã— 5) + 2 = 15 + 2 = 17

*âˆ´ the required first five terms of the sequence whose n ^{th} term, a_{n}Â = 3n + 2 are 5, 8, 11, 14, 17.*

(ii) a_{n}Â = (n â€“ 2)/3

Given sequence whose** **

On putting n = 1, 2, 3, 4, 5 then can get the first five terms

âˆ´ the required first five terms of the sequence whose n^{th} term**,**

(iii) a_{n}Â = 3^{n}

Given sequence whose a_{n}Â = 3^{n}

To get the first five terms of given sequence, put n = 1, 2, 3, 4, 5 in the above

a_{1}Â = 3^{1}Â = 3;

a_{2}Â = 3^{2}Â = 9;

a_{3}Â = 27;

a_{4}Â = 3^{4}Â = 81;

a_{5}Â = 3^{5}Â = 243.

*âˆ´ the required first five terms of the sequence whose n ^{th} term, a_{n}Â = 3^{n} are 3, 9, 27, 81, 243.*

(iv) a_{n}Â = (3n â€“ 2)/ 5

Given sequence whose

To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5 in the above

And, we get

*âˆ´ the required first five terms of the sequence are 1/5, 4/5, 7/5, 10/5, 13/5*

(v) a_{n}Â = (-1)^{n}2^{n}

Given sequence whose a_{n}Â = (-1)^{n}2^{n}

To get first five terms of the sequence, put n = 1, 2, 3, 4, 5 in the above.

a_{1}Â = (-1)^{1}.2^{1}Â = (-1).2 = -2

a_{2}Â = (-1)^{2}.2^{2}Â = (-1).4 = 4

a_{3}Â = (-1)^{3}.2^{3}Â = (-1).8 = -8

a_{4}Â = (-1)^{4}.2^{4}Â = (-1).16 = 16

a_{5}Â = (-1)^{5}.2^{5}Â = (-1).32 = -32

*âˆ´ the first five terms of the sequence are â€“ 2, 4, – 8, 16, – 32.*

(vi) a_{n}Â = n(n â€“ 2)/2

The given sequence is,

To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5.

And, we get

*âˆ´ the required first five terms are -1/2, 0, 3/2, 4, 15/2*

(vii) a_{n}Â = n^{2}Â – n + 1

The given sequence whose, a_{n}Â = n^{2}Â – n + 1

To get the first five terms of given sequence, put n = 1, 2, 3, 4, 5.

And, we get

a_{1}Â = 1^{2}Â – 1 + 1 = 1

a_{2}Â = 2^{2}Â – 2 + 1 = 3

a_{3}Â = 3^{2}Â – 3 + 1 = 7

a_{4}Â = 4^{2}Â – 4 + 1 = 13

a_{5}Â = 5^{2}Â – 5 + 1 = 21

*âˆ´ the required first five terms of the sequence are 1, 3, 7, 13, 21.*

(viii) a_{n}Â = 2n^{2}Â – 3n + 1

The given sequence whose a_{n}Â = 2n^{2}Â – 3n + 1

To get the first five terms of the sequence, put n = 1, 2, 3, 4, 5.

And, we get

a_{1}Â = 2.1^{2}Â – 3.1 + 1 = 2 – 3 + 1 = 0

a_{2}Â = 2.2^{2}Â – 3.2 + 1 = 8 – 6 + 1 = 3

a_{3}Â = 2.3^{2}Â – 3.3 + 1 = 18 – 9 + 1 = 10

a_{4}Â = 2.4^{2}Â – 3.4 + 1 = 32 – 12 + 1 = 21

a_{5}Â = 2.5^{2}Â – 3.5 + 1 = 50 – 15 + 1 = 36

*âˆ´ the required first five terms of the sequence are 0, 3, 10, 21, 36.*

(ix) a_{n}Â = (2nÂ â€“ 3)/ 6

Given sequence whose,

To get the first five terms of the sequence we put n = 1, 2, 3, 4, 5.

And, we get

*âˆ´ the required first five terms of the sequence are -1/6, 1/6, 1/2, 5/6 and 7/6*

### RD Sharma Solutions for Class 10 Maths Chapter 9 Exercise 9.2 Page No: 9.8

**1. Show that the sequence defined by a _{n} = 5n â€“ 7 is an A.P., find its common difference. **

**Solution: **

Given, a_{n} = 5n â€“ 7

Now putting n = 1, 2, 3, 4 we get,

a_{1} = 5(1) â€“ 7 = 5 â€“ 7 = -2

a_{2} = 5(2) â€“ 7 = 10 â€“ 7 = 3

a_{3} = 5(3) â€“ 7 = 15 â€“ 7 = 8

a_{4} = 5(4) â€“ 7 = 20 â€“ 7 = 13

We can see that,

a_{2} â€“ a_{1 }= 3 â€“ (-2) = 5

a_{3} â€“ a_{2 }= 8 â€“ (3) = 5

a_{4} â€“ a_{3 }= 13 â€“ (8) = 5

*Since the difference between the terms is common, we can conclude that the given sequence defined by a _{n} = 5n â€“ 7 is an A.P with common difference 5.*

**2. Show that the sequence defined by a _{n} = 3n^{2} â€“ 5 is not an A.P.**

**Solution: **

Given, a_{n} = 3n^{2} â€“ 5

Now putting n = 1, 2, 3, 4 we get,

a_{1} = 3(1)^{2} â€“ 5= 3 â€“ 5 = -2

a_{2} = 3(2)^{2} â€“ 5 = 12 â€“ 5 = 7

a_{3} = 3(3)^{2} â€“ 5 = 27 â€“ 5 = 22

a_{4} = 3(4)^{2} â€“ 5 = 48 â€“ 5 = 43

We can see that,

a_{2} â€“ a_{1 }= 7 â€“ (-2) = 9

a_{3} â€“ a_{2 }= 22 â€“ 7 = 15

a_{4} â€“ a_{3 }= 43 â€“ 22 = 21

*Since the difference between the terms is not common and varying, we can conclude that the given sequence defined by a _{n} = 3n^{2} â€“ 5 is not an A.P.*

**3. The general term of a sequence is given by a _{n} = -4n + 15. Is the sequence an A.P.? If so, find its 15^{th} term and the common difference. **

**Solution: **

Given, a_{n} = -4n + 15

Now putting n = 1, 2, 3, 4 we get,

a_{1} = -4(1) + 15 = -4 + 15 = 11

a_{2} = -4(2) + 15 = -8 + 15 = 7

a_{3} = -4(3) + 15 = -12 + 15 = 3

a_{4} = -4(4) + 15 = -16 + 15 = -1

We can see that,

a_{2} â€“ a_{1 }= 7 â€“ (11) = -4

a_{3} â€“ a_{2 }= 3 â€“ 7 = -4

a_{4} â€“ a_{3 }= -1 â€“ 3 = -4

Since the difference between the terms is common, we can conclude that the given sequence defined by a_{n} = -4n + 15 is an A.P with common difference of -4.

Hence, the 15^{th} term will be

*a _{15} = -4(15) + 15 = -60 + 15 = -45*

### RD Sharma Solutions for Class 10 Maths Chapter 9 Exercise 9.3 Page No: 9.11

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

**(i) – 5, -1, 3, 7,â€¦ **

**(ii) 1/5, 3/5, 5/5, 7/5,â€¦ **

**(iii) 0.3, 0.55, 0.80, 1.05,â€¦ **

**(iv) -1.1, â€“ 3.1, â€“ 5.1, â€“ 7.1,â€¦ **

**Solution:**

We know that if a is the first term and d is the common difference, the arithmetic progression is a, a + d, a + 2d + a + 3d,â€¦.

(i) â€“ 5, â€“1, 3, 7,â€¦

Given arithmetic series is â€“ 5, â€“1, 3, 7,…

c a, a + d, a + 2d + a + 3d,â€¦.

Thus, by comparing these two we get, a = â€“ 5, a + d = 1, a + 2d = 3, a + 3d = 7

First term (a) = â€“ 5

By subtracting second and first term, we get

(a + d) – (a) = d

-1 – (- 5) = d

4 = d

*â‡’ Common difference (d) = 4.*

(ii) 1/5, 3/5, 5/5, 7/5, ………….

Given arithmetic series is 1/5, 3/5, 5/5, 7/5, ……………

It is seen that, itâ€™s of the form of 1/5, 2/5, 5/5, 7/5, ……….. a, a + d, a + 2d, a + 3d,

Thus, by comparing these two, we get

a = 1/5, a + d = 3/5, a + 2d = 5/5, a + 3d = 7/5

First term (a) = 1/5

By subtracting first term from second term, we get

d = (a + d)-(a)

d = 3/5 – 1/5

d = 2/5

*â‡’ common difference (d) = 2/5*

(iii) 0.3, 0.55, 0.80, 1.05, …………

Given arithmetic series 0.3, 0.55, 0.80, 1.05, ……….

It is seen that, itâ€™s of the form of a, a + d, a + 2d, a + 3d,

Thus, by comparing we get,

a = 0.3, a + d = 0.55, a + 2d = 0.80, a + 3d = 1.05

First term (a) = 0.3.

By subtracting first term from second term. We get

d = (a + d) – (a)

d = 0.55 – 0.3

d = 0.25

*â‡’ Common difference (d) = 0.25*

(iv) â€“1.1, â€“ 3.1, â€“ 5.1, â€“7.1, ……..

General series is â€“1.1, â€“ 3.1, â€“ 5.1, â€“7.1, ……..

It is seen that, itâ€™s of the form of a, a + d, a + 2d, a + 3d, ………..

Thus, by comparing these two, we get

a = â€“1.1, a + d = â€“3.1, a + 2d = â€“5.1, a + 3d = â€“7.1

First term (a) = â€“1.1

Common difference (d) = (a + d) – (a)

= -3.1 â€“Â ( â€“ 1.1)

*â‡’ Common difference (d) = â€“ 2*

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

**(i) a = 4, d = â€“ 3**

**(ii) a = â€“1, d = 1/2**

**(iii) a = â€“1.5, d = â€“ 0.5**

**Solution:**

We know that, if first term (a) = a and common difference = d, then the arithmetic series is: a, a + d, a + 2d, a + 3d,

(i) a = 4, d = -3

Given, first term (a) = 4

Common difference (d) = -3

Then arithmetic progression is: a, a + d, a + 2d, a + 3d, ……

â‡’ 4, 4 – 3, 4 + 2(-3), 4 + 3(-3), ……

*â‡’ 4, 1, â€“ 2, â€“ 5, â€“ 8 ……..*

(ii) a = -1, d = 1/2

Given, first term (a) = -1

Common difference (d) = 1/2

Then arithmetic progression is: a, a + d, a + 2d, a + 3d,

â‡’ -1, -1 + 1/2, -1 + 2Â½, -1 + 3Â½, …

*â‡’ -1, -1/2, 0, 1/2*

(iii) a = â€“1.5, d = â€“ 0.5

Given First term (a) = â€“1.5

Common difference (d) = â€“ 0.5

Then arithmetic progression is; a, a + d, a + 2d, a + 3d, ……

â‡’ -1.5, -1.5 + (-0.5), â€“1.5 + 2(â€“Â 0.5), â€“1.5 + 3(â€“ 0.5)

*â‡’ â€“ 1.5, â€“ 2, â€“ 2.5, â€“ 3, …….*

**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 1/4 of their remaining in the cylinder.Â **

**(iii) Divya deposited Rs 1000 at compound interest at the rate of 10% per annum. The amount at the end of first year, second year, third year, â€¦, and so on. **

**Solution:**

(i) Given,

Cost of digging a well for the first meter (c_{1}) = Rs.150.

And, the cost rises by Rs.20 for each succeeding meter

Then,

Cost of digging for the second meter (c_{2}) = Rs.150 + Rs 20 = Rs 170

Cost of digging for the third meter (c_{3}) = Rs.170 + Rs 20 = Rs 210

Hence, its clearly seen that the costs of digging a well for different lengths are 150, 170, 190, 210, ….

*Evidently, this series is in Aâˆ™P.*

With first term (a) = 150, common difference (d) = 20

(ii) Given,

Let the initial volume of air in a cylinder be V liters each time 3^{th}/4 of air in a remaining i.e

1 -1/4

First time, the air in cylinder is V.

Second time, the air in cylinder is 3/4 V.

Third time, the air in cylinder is (3/4)^{2}Â V.

Thus, series is V, 3/4 V, (3/4)^{2Â }V,(3/4)^{3Â }V, ….

*Hence, the above series is not a A.P.*

(iii) Given,

Divya deposited Rs 1000 at compound interest of 10% p.a

So, the amount at the end of first year is = 1000 + 0.1(1000) = Rs 1100

And, the amount at the end of second year is = 1100 + 0.1(1100) = Rs 1210

And, the amount at the end of third year is = 1210 + 0.1(1210) = Rs 1331

*Cleary, these amounts 1100, 1210 and 1331 are not in an A.P since the difference between them is not the same.*

### RD Sharma Solutions for Class 10 Maths Chapter 9 Exercise 9.4 Page No: 9.24

**1. Find:**

**(i) 10 ^{th}Â tent of the AP 1, 4, 7, 10….**

**(ii) 18 ^{th}Â term of the AP âˆš2, 3âˆš2, 5âˆš2, â€¦â€¦.Â **

**(iii) n ^{th}Â term of the AP 13, 8, 3, -2, ……….Â **

**(iv) 10 ^{th}Â term of the AP -40, -15, 10, 35, ………….Â **

**(v) 8 ^{th}Â term of the AP 11, 104, 91, 78, ……………Â **

**(vi) 11 ^{th}Â tenor of the AP 10.0, 10.5, 11.0, 11.2, …………..Â **

**(vii) 9 ^{th}Â term of the AP 3/4, 5/4, 7/4 + 9/4, ………..**

**Solution:**

**(i**) Given A.P. is 1, 4, 7, 10, ……….

First term (a) = 1

Common difference (d) = Second term – First term

= 4 – 1 = 3.

We know that, n^{th}Â term in an A.P = a + (n – 1)d

Then, 10^{th}Â term in the A.P is 1 + (10 – 1)3

= 1 + 9×3

= 1 + 27

= 28

*âˆ´ 10 ^{th}Â term of A. P. is 28*

(ii) Given A.P. is âˆš2, 3âˆš2, 5âˆš2, â€¦â€¦.

First term (a) = âˆš2

Common difference = Second term â€“ First term

= 3âˆš2 – âˆš2

â‡’ d = 2âˆš2

We know that, n^{th}Â term in an A. P. = a + (n – 1)d

Then, 18^{th}Â term of A. P. = âˆš2 + (18 – 1)2âˆš2

= âˆš2 + 17.2âˆš2

= âˆš2 (1+34)

= 35âˆš2

*âˆ´ 18 ^{th}Â term of A. P. is 35âˆš2*

(iii) Given A. P. is 13, 8, 3, – 2, Â …………

First term (a) = 13

Common difference (d) = Second term first term

= 8 – 13 = â€“ 5

We know that, n^{th}Â term of an A.P. a_{n}Â = a +(n – 1)d

= 13 + (n – 1) – 5

= 13 – 5n + 5

*âˆ´ n ^{th} term of the A.P is a_{n}Â = 18 – 5n*

(iv) Given A. P. is – 40, -15, 10, 35, ……….

First term (a) = -40

Common difference (d) = Second term – fast term

= -15 – (- 40)

= 40 – 15

= 25

We know that, n^{th}Â term of an A.P. a_{n}Â = a + (n – 1)d

Then, 10^{th}Â term of A. P. a_{10}Â = -40 + (10 – 1)25

= â€“ 40 + 9.25

= â€“ 40 + 225

= 185

*âˆ´ 10 ^{th}Â term of the A. P. is 185*

(v) Given sequence is 117, 104, 91, 78, ………….

First term (a) = 117

Common difference (d) = Second term – first term

= 104 – 117

= â€“ 13

We know that, n^{th}Â term = a + (n – 1)d

Then, 8^{th}Â term = a + (8 – 1)d

= 117 + 7(-13)

= 117 – 91

= 26

*âˆ´ 8 ^{th}Â term of the A. P. is 26*

(vi) Given A. P is 10.0, 10.5, 11.0, 11.5,

First term (a) = 10.0

Common difference (d) = Second term – first term

= 10.5 – 10.0 = 0.5

We know that, n^{th}Â term a_{n}Â = a + (n – 1)d

Then, 11^{th}Â term a_{11}Â = 10.0 + (11 – 1)0.5

= 10.0 + 10 x 0.5

= 10.0 + 5

=15.0

*âˆ´ 11 ^{th}Â term of the A. P. is 15.0*

(vii) Given A. P is 3/4, Â 5/4, 7/4, 9/4, …………

First term (a) = 3/4

Common difference (d) = Second term – first term

= 5/4 – 3/4

= 2/4

We know that, n^{th}Â term a_{n}Â = a + (n – 1)d

Then, 9^{th}Â term a_{9}Â = a + (9 – 1)d

*âˆ´ 9 ^{th}Â term of the A. P. is 19/4.*

**Â **

**2.(i) Which term of the AP 3, 8, 13, …. is 248?**

**(ii) Which term of the AP 84, 80, 76, … is 0?**

**(iii) Which term of the AP 4. 9, 14, …. is 254?**

**(iv) Which term of the AP 21. 42, 63, 84, … is 420?**

**(v) Which term of the AP 121, 117. 113, … is its first negative term?**

**Solution:**

(i) Given A.P. is 3, 8, 13, ………..

First term (a) = 3

Common difference (d) = Second term – first term

= 8 – 3

= 5

We know that, n^{th}Â term (a_{n}) = a + (n – 1)d

And, given n^{th}Â term a_{n}Â = 248

248 = 3+(n – 1)5

248 = -2 + 5n

5n = 250

n =250/5 = 50

*âˆ´ 50 ^{th}Â term in the A.P is 248.*

(ii) Given A. P is 84, 80, 76, …………

First term (a) = 84

Common difference (d) = a_{2}Â – a

= 80 – 84

= â€“ 4

We know that, n^{th}Â term (a_{n}) = a +(n – 1)d

And, given nth term is 0

0 = 84 + (n – 1) – 4

84 = +4(n – Â 1)

n – 1 = 84/4 = 21

n = 21 + 1 = 22

*âˆ´ 22 ^{nd}Â term in the A.P is 0.*

(iii) Given A. P 4, 9, 14, …………

First term (a) = 4

Common difference (d) = a_{2}Â â€“ a_{1}

= 9 – 4

= 5

We know that, n^{th}Â term (a_{n}) = a + (n – 1)d

And, given n^{th}Â term is 254

4 + (n – 1)5 = 254

(n – 1)âˆ™5 = 250

n – 1 = 250/5 = 50

n = 51

*âˆ´Â 51 ^{th}Â term in the A.P is 254.*

(iv) Given A. P 21, 42, 63, 84, ………

a = 21, d = a_{2}Â â€“ a_{1}

= 42 – 21

= 21

We know that, n^{th}Â term (a_{n}) = a +(n – 1)d

And, given n^{th} term = 420

21 + (n – 1)21 = 420

(n – 1)21 = 399

n – 1 = 399/21 = 19

n = 20

*âˆ´ 20 ^{th}Â term is 420.*

(v) Given A.P is 121, 117, 113, ………..

Fiat term (a) = 121

Common difference (d) = 117 – 121

= – 4

We know that, n^{th}Â term a_{n} = a + (n – 1)d

And, for some n^{th}Â term is negative i.e., a_{n}Â < 0

121 + (n – 1) – 4 < 0

121 + 4 – 4n < 0

125 – 4n < 0

4n > 125

n > 125/4

n > 31.25

The integer which comes after 31.25 is 32.

*âˆ´ 32 ^{nd}Â term in the A.P will be the first negative term.*

**3.(i) Is 68 a term of the A.P. 7, 10, 13,â€¦ ? **

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

**(iii) Is -150 a term of the A.P. 11, 8, 5, 2, â€¦ ?**

**Solutions: **

(i) Given, A.P. 7, 10, 13,â€¦

Here, a = 7 and d = a_{2} â€“ a_{1} = 10 â€“ 7 = 3

We know that, n^{th}Â term a_{n} = a + (n – 1)d

Required to check n^{th}Â term a_{n}Â = 68

a + (n – 1)d = 68

7 + (n – 1)3 = 68

7 + 3n â€“ 3 = 68

3n + 4 = 68

3n = 64

â‡’ n = 64/3, which is not a whole number.

*Therefore, 68 is not a term in the A.P.*

(ii) Given, A.P. 3, 8, 13,â€¦

Here, a = 3 and d = a_{2} â€“ a_{1} = 8 â€“ 3 = 5

We know that, n^{th}Â term a_{n} = a + (n – 1)d

Required to check n^{th}Â term a_{n}Â = 302

a + (n – 1)d = 302

3 + (n – 1)5 = 302

3 + 5n â€“ 5 = 302

5n – 2 = 302

5n = 304

â‡’ n = 304/5, which is not a whole number.

*Therefore, 302 is not a term in the A.P.*

(iii) Given, A.P. 11, 8, 5, 2, â€¦

Here, a = 11 and d = a_{2} â€“ a_{1} = 8 â€“ 11 = -3

We know that, n^{th}Â term a_{n} = a + (n – 1)d

Required to check n^{th}Â term a_{n}Â = -150

a + (n – 1)d = -150

11 + (n – 1)(-3) = -150

11 â€“ 3n + 3 = -150

3n = 150 + 14

3n = 164

â‡’ n = 164/3, which is not a whole number.

*Therefore, -150 is not a term in the A.P.*

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

**(i) 7, 10, 13, â€¦.., 43**

**(ii) -1, -5/6, -2/3, -1/2, â€¦ , 10/3**

**(iii) 7, 13, 19, â€¦, 205 **

**(iv) 18, 15Â½, 13, â€¦., -47**

**Solution: **

(i) Given, A.P. 7, 10, 13, â€¦.., 43

Here, a = 7 and d = a_{2} â€“ a_{1} = 10 â€“ 7 = 3

We know that, n^{th}Â term a_{n} = a + (n – 1)d

And, given n^{th}Â term a_{n}Â = 43

a + (n – 1)d = 43

7 + (n – 1)(3) = 43

7 + 3n â€“ 3 = 43

3n = 43 â€“ 4

3n = 39

â‡’ *n = 13*

*Therefore, there are 13 terms in the given A.P.*

(ii) Given, A.P. -1, -5/6, -2/3, -1/2, â€¦ , 10/3

Here, a = -1 and d = a_{2} â€“ a_{1} = -5/6 â€“ (-1) = 1/6

We know that, n^{th}Â term a_{n} = a + (n – 1)d

And, given n^{th}Â term a_{n}Â = 10/3

a + (n – 1)d = 10/3

-1 + (n – 1)(1/6) = 10/3

-1 + n/6 â€“ 1/6 = 10/3

n/6 = 10/3 + 1 + 1/6

n/6 = (20 + 6 + 1)/6

n = (20 + 6 + 1)

â‡’ *n = 27*

*Therefore, there are 27 terms in the given A.P.*

(iii) Given, A.P. 7, 13, 19, â€¦, 205

Here, a = 7 and d = a_{2} â€“ a_{1} = 13 â€“ 7 = 6

We know that, n^{th}Â term a_{n} = a + (n – 1)d

And, given n^{th}Â term a_{n}Â = 205

a + (n – 1)d = 205

7 + (n – 1)(6) = 205

7 + 6n â€“ 6 = 205

6n = 205 â€“ 1

n = 204/6

â‡’ *n = 34*

*Therefore, there are 34 terms in the given A.P.*

(iv) Given, A.P. 18, 15Â½, 13, â€¦., -47

Here, a = 7 and d = a_{2} â€“ a_{1} = 15Â½ â€“ 18 = 5/2

We know that, n^{th}Â term a_{n} = a + (n – 1)d

And, given n^{th}Â term a_{n}Â = -47

a + (n – 1)d = 43

18 + (n – 1)(-5/2) = -47

18 â€“ 5n/2 + 5/2 = -47

36 â€“ 5n + 5 = -94

5n = 94 + 36 + 5

5n = 135

â‡’ *n = 27*

*Therefore, there are 27 terms in the given A.P.*

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

**Solution: **

Given,

a = 5 and d = 3

We know that, n^{th}Â term a_{n} = a + (n – 1)d

So, for the given A.P. a_{n} = 5 + (n – 1)3 = 3n + 2

Also given, last term = 80

â‡’ 3n + 2 = 80

3n = 78

n = 78/3 = *26*

*Therefore, there are 26 terms in the A.P.*

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

**Solution:**

Given,

a_{6} = 19 and a_{17} = 41

We know that, n^{th}Â term a_{n} = a + (n – 1)d

So,

a_{6} = a + (6-1)d

â‡’ a + 5d = 19 â€¦â€¦ (i)

Similarity,

a_{17} = a + (17 – 1)d

â‡’ a + 16d = 41 â€¦â€¦ (ii)

Solving (i) and (ii),

(ii) â€“ (i) â‡’

a + 16d â€“ (a + 5d) = 41 â€“ 19

11d = 22

â‡’ d = 2

Using d in (i), we get

a + 5(2) = 19

a = 19 â€“ 10 = 9

Now, the 40^{th} term is given by a_{40} = 9 + (40 – 1)2 = 9 + 78 = *87*

*Therefore the 40 ^{th} term is 87.*

**7. If 9 ^{th} term of an A.P. is zero, prove its 29^{th} term is double the 19^{th} term. **

** Solution: **

Given,

a_{9} = 0

We know that, n^{th}Â term a_{n} = a + (n – 1)d

So, a + (9 – 1)d = 0 â‡’ a + 8d = 0 â€¦â€¦(i)

Now,

29^{th} term is given by a_{29} = a + (29 – 1)d

â‡’ a_{29} = a + 28d

And, a_{29} = (a + 8d) + 20d [using (i)]

â‡’ a_{29} = 20d â€¦.. (ii)

Similarly, 19^{th} term is given by a_{19} = a + (19 – 1)d

â‡’ a_{19} = a + 18d

And, a_{19} = (a + 8d) + 10d [using (i)]

â‡’ a_{19} = 10d â€¦..(iii)

On comparing (ii) and (iii), itâ€™s clearly seen that

a_{29} = 2(a_{19})

*Therefore, 29 ^{th} term is double the 19^{th} term.*

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

**Solution: **

Given,

10 times the 10^{th} term of an A.P. is equal to 15 times the 15^{th} term.

We know that, n^{th}Â term a_{n} = a + (n – 1)d

â‡’ 10(a_{10}) = 15(a_{15})

10(a + (10 – 1)d) = 15(a + (15 – 1)d)

10(a + 9d) = 15(a + 14d)

10a + 90d = 15a + 210d

5a + 120d = 0

5(a + 24d) = 0

a + 24d = 0

a + (25 â€“ 1)d = 0

â‡’ a_{25} = 0

*Therefore, the 25 ^{th} term of the A.P. is zero.*

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

**Solution: **

Given,

A_{10} = 41 and a_{18} = 73

We know that, n^{th}Â term a_{n} = a + (n – 1)d

So,

a_{10} = a + (10 – 1)d

â‡’ a + 9d = 41 â€¦â€¦ (i)

Similarity,

a_{18} = a + (18 – 1)d

â‡’ a + 17d = 73 â€¦â€¦ (ii)

Solving (i) and (ii),

(ii) â€“ (i) â‡’

a + 17d â€“ (a + 9d) = 73 â€“ 41

8d = 32

â‡’ d = 4

Using d in (i), we get

a + 9(4) = 41

a = 41 â€“ 36 = 5

Now, the 26^{th} term is given by a_{26} = 5 + (26 – 1)4 = 5 + 100 = 105

*Therefore the 26 ^{th} term is 105.*

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

**Solution: **

Given,

24^{th} term is twice the 10^{th} term.

We know that, n^{th}Â term a_{n} = a + (n – 1)d

â‡’ a_{24} = 2(a_{10})

a + (24 – 1)d = 2(a + (10 – 1)d)

a + 23d = 2(a + 9d)

a + 23d = 2a + 18d

a = 5d â€¦. (1)

Now, the 72^{nd} term can be expressed as

a_{72} = a + (72 – 1)d

= a + 71d

= a + 5d + 66d

= a + a + 66d [using (1)]

= 2(a + 33d)

= 2(a + (34 – 1)d)

= 2(a_{34})

â‡’ a_{72 }= 2(a_{34})

*Hence, the 72 ^{nd} term is twice the 34^{th} term of the given A.P.*

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

**Solution: **

Given,

a_{26 }= 0 , a_{11 }= 3 and a_{n }(last term) = -1/5 of an A.P.

We know that, n^{th}Â term a_{n} = a + (n – 1)d

Then,

a_{26} = a + (26 – 1)d

â‡’ a + 25d = 0 â€¦..(1)

And,

a_{11} = a + (11 – 1)d

â‡’ a + 10d = 3 â€¦â€¦ (2)

Solving (1) and (2),

(1) â€“ (2) â‡’

a + 25d â€“ (a + 10d) = 0 â€“ 3

15d = -3

â‡’ d = -1/5

Using d in (1), we get

a + 25(-1/5) = 0

a = 5

Now, given that the last term a_{n} = -1/5

â‡’ 5 + (n – 1)(-1/5) = -1/5

5 + -n/5 + 1/5 = -1/5

25 â€“ n + 1 = -1

*n = 27*

*Therefore, the A.P has 27 terms and its common difference is -1/5.*

**12. If the n ^{th} term of the A.P. 9, 7, 5, â€¦. is same as the n^{th} term of the A.P. 15, 12, 9, â€¦ find n. **

**Solution: **

Given,

A.P_{1 }= 9, 7, 5, â€¦. and A.P_{2 }= 15, 12, 9, â€¦

And, we know that, n^{th}Â term a_{n} = a + (n – 1)d

For A.P_{1},

a = 9, d = Second term â€“ first term = 9 â€“ 7 = -2

And, its n^{th}Â term a_{n }= 9 + (n – 1)(-2) = 9 â€“ 2n + 2

a_{n }= 11 â€“ 2n â€¦..(i)

Similarly, for A.P_{2}

a = 15, d = Second term â€“ first term = 12 â€“ 15 = -3

And, its n^{th}Â term a_{n }= 15 + (n – 1)(-3) = 15 â€“ 3n + 3

a_{n }= 18 – 3n â€¦..(ii)

According to the question, its given that

n^{th} term of the A.P_{1} = n^{th} term of the A.P_{2}

â‡’ 11 â€“ 2n = 18 – 3n

*n = 7*

*Therefore, the 7 ^{th} term of the both the A.Ps are equal.*

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

**(i) 3, 5, 7, 9, â€¦. 201**

**(ii) 3,8,13, â€¦ ,253**

**(iii) 1, 4, 7, 10, â€¦ ,88**

**Solution: **

In order the find the 12^{th} term for the end of an A.P. which has n terms, its done by simply finding the ((n -12) + 1)^{th} of the A.P

And we know, n^{th}Â term a_{n} = a + (n – 1)d

(i) Given A.P = 3, 5, 7, 9, â€¦. 201

Here, a = 3 and d = (5 – 3) = 2

Now, find the number of terms when the last term is known i.e, 201

a_{n} = 3 + (n – 1)2 = 201

3 + 2n â€“ 2 = 201

2n = 200

n = 100

Hence, the A.P has 100 terms.

So, the 12^{th} term from the end is same as (100 â€“ 12 + 1)^{th} of the A.P which is the 89^{th} term.

â‡’ a_{89 }= 3 + (89 – 1)2

= 3 + 88(2)

= 3 + 176

*= 179*

*Therefore, the 12 ^{th} term from the end of the A.P is 179.*

(ii) Given A.P = 3,8,13, â€¦ ,253

Here, a = 3 and d = (8 – 3) = 5

Now, find the number of terms when the last term is known i.e, 253

a_{n} = 3 + (n – 1)5 = 253

3 + 5n â€“ 5 = 253

5n = 253 + 2 = 255

n = 255/5

*n = 51*

*Hence, the A.P has 51 terms.*

So, the 12^{th} term from the end is same as (51 â€“ 12 + 1)^{th} of the A.P which is the 40^{th} term.

â‡’ a_{40 }= 3 + (40 – 1)5

= 3 + 39(5)

= 3 + 195

*= 198*

*Therefore, the 12 ^{th} term from the end of the A.P is 198.*

(iii) Given A.P = 1, 4, 7, 10, â€¦ ,88

Here, a = 1 and d = (4 – 1) = 3

Now, find the number of terms when the last term is known i.e, 88

a_{n} = 1 + (n – 1)3 = 88

1 + 3n â€“ 3 = 88

3n = 90

*n = 30*

*Hence, the A.P has 30 terms.*

So, the 12^{th} term from the end is same as (30 â€“ 12 + 1)^{th} of the A.P which is the 19^{th} term.

â‡’ a_{89 }= 1 + (19 – 1)3

= 1 + 18(3)

= 1 + 54

*= 55*

*Therefore, the 12 ^{th} term from the end of the A.P is 55.*

**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.**

**Solution: **

Letâ€™s consider the first term and the common difference of the A.P to be a and d respectively.

Then, we know that *a _{n} = a + (n – 1)d*

Given conditions,

4^{th} term of an A.P. is three times the first

Expressing this by equation we have,

â‡’ a_{4} = 3(a)

a + (4 – 1)d = 3a

3d = 2a â‡’ a = 3d/2â€¦â€¦.(i)

And,

7^{th} term exceeds twice the third term by 1

â‡’ a_{7} = 2(a_{3}) + 1

a + (7 â€“ 1)d = 2(a + (3â€“1)d) + 1

a + 6d = 2a + 4d + 1

a â€“ 2d +1 = 0 â€¦.. (ii)

Using (i) in (ii), we have

3d/2 â€“ 2d + 1 = 0

3d â€“ 4d + 2 = 0

d = 2

So, putting d = 2 in (i), we get a

â‡’ *a = 3*

*Therefore, the first term is 3 and the common difference is 2.*

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

**Solution: **

Given, in an A.P

a_{6} = 12 and a_{8} = 22

We know that *a _{n} = a + (n – 1)d*

So,

a_{6 }= a + (6-1)d = a + 5d = 12 â€¦. (i)

And,

a_{8} = a + (8-1)d = a + 7d = 22 â€¦â€¦. (ii)

Solving (i) and (ii), we have

(ii) – (i) â‡’

a + 7d â€“ (a + 5d) = 22 â€“ 12

2d = 10

d = 5

Putting d in (i) we get,

a + 5(5) = 12

a = 12 â€“ 25

a = -13

Thus, for the A.P: a = -13 and d = 5

So, the n^{th} term is given by a_{n} = a + (n-1)d

a_{n }= -13 + (n-1)5 = -13 + 5n â€“ 5

â‡’ *a _{n }= 5n â€“ 18*

*Hence, the second term is given by a _{2 }= 5(2) â€“ 18 = 10 â€“ 18 = -8*

**16. How many numbers of two digit are divisible by 3? **

**Solution: **

The first 2 digit number divisible by 3 is 12. And, the last 2 digit number divisible by 3 is 99.

So, this forms an A.P.

12, 15, 18, 21, â€¦. , 99

Where, a = 12 and d = 3

Finding the number of terms in this A.P

â‡’ 99 = 12 + (n-1)3

99 = 12 + 3n â€“ 3

90 = 3n

n = 90/3 = *30*

*Therefore, there are 30 two digit numbers divisible by 3.*

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

**Solution: **

Given, an A.P of 60 terms

And, a = 7 and a_{60} = 125

We know that a_{n} = a + (n – 1)d

â‡’ a_{60} = 7 + (60 – 1)d = 125

7 + 59d = 125

59d = 118

d = 2

So, the 32^{nd} term is given by

a_{32} = 7 + (32 -1)2 = 7 + 62 = 69

â‡’ *a _{32} = 69*

**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.**

**Solution: **

Given, in an A.P

The sum of 4^{th} and 8^{th} terms of an A.P. is 24

â‡’ a_{4} + a_{8} = 24

And, we know that *a _{n} = a + (n – 1)d*

2a + 10d = 24

a + 5d = 12 â€¦. (i)

Also given that,

the sum of the 6^{th} and 10^{th} terms is 34

â‡’ a_{6} + a_{10} = 34

2a + 14d = 34

a + 7d = 17 â€¦â€¦ (ii)

Subtracting (i) form (ii), we have

a + 7d â€“ (a + 5d) = 17 â€“ 12

2d = 5

d = 5/2

Using d in (i) we get,

a + 5(5/2) = 12

a = 12 â€“ 25/2

*a = -1/2*

*Therefore, the first term is -1/2 and the common difference is 5/2.*

**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. **

**Solution: **

Given, an A.P whose

a = 5 and a_{100} = -292

We know that a_{n} = a + (n – 1)d

a100 = 5 + 99d = -292

99d = -297

d = -3

Hence, the 50^{th} term is

*a _{50} = a + 49d = 5 + 49(-3) = 5 â€“ 147 = -142*

**20. Find a _{30} â€“ a_{20} for the A.P.**

**(i) -9, -14, -19, -24 (ii) a, a+d, a+2d, a+3d, â€¦â€¦**

**Solution: **

We know that a_{n} = a + (n – 1)d

So, a_{30} â€“ a_{20} = (a + 29d) â€“ (a + 19) =10d

(i) Given A.P. -9, -14, -19, -24

Here, a = -9 and d = -14 â€“ (-9) = = -14 + 9 = -5

So, a_{30} â€“ a_{20} = 10d

= 10(-5)

*= -50*

(ii) Given A.P. a, a+d, a+2d, a+3d, â€¦â€¦

So, a_{30} â€“ a_{20} = (a + 29d) â€“ (a + 19d)

*=10d*

**21. Write the expression a _{n} â€“ a_{k} for the A.P. a, a+d, a+2d, â€¦..**

**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 }â€“ a_{5} = 200 **

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

**Solution: **

Given A.P. a, a+d, a+2d, â€¦..

So, a_{n }= a + (n-1)d = a + nd â€“d

And, a_{k }= a + (k-1)d = a + kd â€“ d

a_{n }– a_{k }= (a + nd â€“ d) â€“ (a + kd â€“ d)

*= (n â€“ k)d*

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

Here n = 13 and k = 11,

a_{13 }â€“ a_{11 }= (13 â€“ 11)d = 2d

â‡’ 79 â€“ 5 = 2d

*d = 74/2 = 37*

(ii) Given, a_{10 }â€“ a_{5} = 200

â‡’ (10 – 5)d = 200

5d = 200

*d = 40*

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

â‡’ a_{20} â€“ a_{18} = 10

(20 – 18)d = 10

2d = 10

*d = 5*

**22. Find n if the given value of x is the n ^{th} term of the given A.P. **

**(i) 25, 50, 75, 100, ; x = 1000 (ii) -1, -3, -5, -7, â€¦; x = -151**

**(iii) 5Â½, 11, 16Â½, 22, â€¦.; x = 550 (iv) 1, 21/11, 31/11, 41/11, â€¦; x = 171/11**

**Solution: **

(i) Given A.P. 25, 50, 75, 100, â€¦â€¦ ,1000

Here, a = 25 d = 50 â€“ 25 = 25

Last term (n^{th} term) = 1000

We know that *a _{n} = a + (n – 1)d*

â‡’ 1000 = 25 + (n-1)25

1000 = 25 + 25n â€“ 25

n = 1000/25

*n = 40*

(ii) Given A.P. -1, -3, -5, -7, â€¦., -151

Here, a = -1 d = -3 â€“ (-1) = -2

Last term (n^{th} term) = -151

We know that a_{n} = a + (n – 1)d

â‡’ -151 = -1 + (n-1)(-2)

-151 = -1 – 2n + 2

n = 152/2

*n = 76*

(iii) Given A.P. 5Â½, 11, 16Â½, 22, â€¦ , 550

Here, a = 5Â½ d = 11 â€“ (5Â½) = 5Â½ = 11/2

Last term (n^{th} term) = 550

We know that a_{n} = a + (n – 1)d

â‡’ 550 = 5Â½ + (n-1)(11/2)

550 x 2 = 11+ 11n â€“ 11

1100 = 11n

*n = 100*

(iv) Given A.P. 1, 21/11, 31/11, 41/11, 171/11

Here, a = 1 d = 21/11 â€“ 1 = 10/11

Last term (n^{th} term) = 171/11

We know that a_{n} = a + (n – 1)d

â‡’ 171/11 = 1 + (n-1)10/11

171 = 11 + 10n â€“ 10

n = 170/10

*n = 17*

**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. **

**Solution: **

Given, an A.P in which,

a_{8} = 1/2(a_{2})

a_{11} = 1/3(a_{4}) + 1

We know that *a _{n} = a + (n – 1)d*

â‡’ a_{8} = 1/2(a_{2})

a + 7d = 1/2(a + d)

2a + 14d = a + d

a + 13d = 0 â€¦â€¦ (i)

And, a_{11} = 1/3(a_{4}) + 1

a + 10d = 1/3(a + 3d) + 1

3a + 30d = a + 3d + 3

2a + 27d = 3 â€¦â€¦ (ii)

Solving (i) and (ii), by (ii) â€“ 2x(i) â‡’

2a + 27d â€“ 2(a + 13d) = 3 – 0

d = 3

Putting d in (i) we get,

a + 13(3) = 0

a = -39

*Thus, the 15 ^{th} term a_{15 }= -39 + 14(3) = -39 + 42 = 3*

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

**Solution: **

Given, in an A.P

a_{3} = 16 and a_{7} = a_{5} + 12

We know that a_{n} = a + (n – 1)d

â‡’ a + 2d = 16â€¦â€¦ (i)

And,

a + 6d = a + 4d + 12

2d = 12

â‡’ d = 6

Using d in (i), we have

a + 2(6) = 16

a = 16 â€“ 12 = 4

*Hence, the A.P is 4, 10, 16, 22, â€¦â€¦.*

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

**Solution: **

Given,

a_{7 }= 32 and a_{13 }= 62

From a_{n }– a_{k }= (a + nd â€“ d) â€“ (a + kd â€“ d)

= (n â€“ k)d

a_{13} â€“ a_{7} = (13 – 7)d = 62 â€“ 32 = 30

6d = 30

d = 5

Now,

a_{7} = a + (7 – 1)5 = 32

a + 30 = 32

a = 2

*Hence, the A.P is 2, 7, 12, 17, â€¦â€¦*

**26. Which term of the A.P. 3, 10, 17, â€¦. will be 84 more than its 13 ^{th} term ?**

**Solution: **

Given, A.P. 3, 10, 17, â€¦.

Here, a = 3 and d = 10 â€“ 3 = 7

According the question,

a_{n} = a_{13} + 84

Using a_{n} = a + (n – 1)d,

3 + (n – 1)7 = 3 + (13 – 1)7 + 84

3 + 7n â€“ 7 = 3 + 84 + 84

7n = 168 + 7

n = 175/7

*n = 25*

Therefore, it the 25^{th} term which is 84 more than its 13^{th} term.

**27. Two arithmetic progressions have the same common difference. The difference between their 100 ^{th} terms is 100, what is the difference between their 1000^{th} terms?**

**Solution: **

Let the two A.Ps be A.P_{1} and A.P_{2}

For A.P_{1} the first term = a and the common difference = d

And for A.P_{2} the first term = b and the common difference = d

So, from the question we have

a_{100} â€“ b_{100} = 100

(a + 99d) â€“ (b + 99d) = 100

a – b = 100

Now, the difference between their 1000^{th} terms is,

(a + 999d) â€“ (b + 999d) = a â€“ b = *100*

*Therefore, the difference between their 1000 ^{th} terms is also 100.*

### RD Sharma Solutions for Class 10 Maths Chapter 9 Exercise 9.5 Page No: 9.30

**1. Find the value of x for which (8x + 4), (6x â€“ 2) and (2x + 7) are in A.P.
Solution: **

Given,

(8x + 4), (6x â€“ 2) and (2x + 7) are in A.P.

So, the common difference between the consecutive terms should be the same.

(6x â€“ 2) â€“ (8x + 4) = (2x + 7) â€“ (6x â€“ 2)

â‡’ 6x â€“ 2 â€“ 8x â€“ 4 = 2x + 7 â€“ 6x + 2

â‡’ -2x â€“ 6 = -4x + 9

â‡’ -2x + 4x = 9 + 6

â‡’ 2x = 15

*Therefore, x = 15/2*

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

Given,

x + 1, 3x and 4x + 2 are in A.P.

So, the common difference between the consecutive terms should be the same.

3x â€“ x â€“ 1 = 4x + 2 â€“ 3x

â‡’ 2x â€“ 1 = x + 2

â‡’ 2x â€“ x = 2 + 1

â‡’ x = 3

*Therefore, x = 3*

**3. Show that (a â€“ b)Â², (aÂ² + bÂ²) and (a + b)Â² are in A.P.
Solution:**

If (a â€“ b)Â², (aÂ² + bÂ²) and (a + b)Â² have to be in A.P. then,

It should satisfy the condition,

2b = a + c [for a, b, c are in A.P]

Thus,

2 (aÂ² + bÂ²) = (a â€“ b)Â² + (a + b)Â²

2 (aÂ² + bÂ²) = aÂ² + bÂ² â€“ 2ab + aÂ² + bÂ² + 2ab

2 (aÂ² + bÂ²) = 2aÂ² + 2bÂ² = 2 (aÂ² + bÂ²)

*LHS = RHS*

Hence proved.

**4. 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.
Solution:**

Letâ€™s consider the three terms of the A.P. to be a â€“ d, a, a + d

so, the sum of three terms = 21

â‡’ a â€“ d + a + a + d = 21

â‡’ 3a = 21

â‡’ a = 7

And, product of the first and 3rd = 2nd term + 6

â‡’ (a â€“ d) (a + d) = a + 6

aÂ² â€“ dÂ² = a + 6

â‡’ (7 )Â² â€“ dÂ² = 7 + 6

â‡’ 49 â€“ dÂ² = 13

â‡’ dÂ² = 49 â€“ 13 = 36

â‡’ dÂ² = (6)Â²

â‡’ d = 6

*Hence, the terms are 7 â€“ 6, 7, 7 + 6 â‡’ 1, 7, 13*

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

Let the three numbers of the A.P. be a â€“ d, a, a + d

From the question,

Sum of these numbers = 27

a â€“ d + a + a + d = 27

â‡’ 3a = 27

a = 27/3 = 9

Now, product of these numbers = 648

(a – d)(a)(a + d) = 648

a(a^{2} â€“ d^{2}) = 648

a^{2} â€“ 648/a = d^{2}

9^{2} â€“ (648/9) = d^{2}

9^{3} â€“ 648 = 9d^{2}

729 â€“ 648 = 9d^{2}

81 = 9d^{2}

d^{2} = 9

d = 3 or -3

*Hence, the terms are 9-3, 9 and 9+3 â‡’ 6, 9, 12 or 12, 9, 6 (for d = -3)*

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

Letâ€™s consider the four terms of the A.P. to be (a â€“ 3d), (a â€“ d), (a + d) and (a + 3d).

From the question,

Sum of these terms = 50

â‡’ (a â€“ 3d) + (a â€“ d) + (a + d) + (a + 3d) = 50

â‡’ a â€“ 3d + a â€“ d + a + d + a â€“ 3d= 50

â‡’ 4a = 50

â‡’ a =Â 50/4 = 25/2

And, also given that the greatest number = 4 x least number

â‡’ a + 3d = 4 (a â€“ 3d)

â‡’ a + 3d = 4a â€“ 12d

â‡’ 4a â€“ a = 3d + 12d

â‡’3a = 15d

â‡’a = 5d

Using the value of a in the above equation, we have

â‡’25/2 = 5d

â‡’ d = 5/2

So, the terms will be:

(a â€“ 3d) = (25/2 â€“ 3(5/2)), (a â€“ d) = (25/2 â€“ 5/2), (25/2 + 5/2) and (25/2 + 3(5/2)).

*â‡’ 5, 10, 15, 20*

### RD Sharma Solutions for Class 10 Maths Chapter 9 Exercise 9.6 Page No: 9.50

**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}Â + y^{2}), (x + y)^{2}, to 22 tams**

**(viii) â€“ 26, â€“ 24, â€“ 22, …. to 36 terms**

**Solution:**

In an A.P if the first term = a, common difference = d, and if there are n terms.

Then, sum of n terms is given by:

(i) Given A.P.is 50, 46, 42 to 10 term.

First term (a) = 50

Common difference (d) = 46 – 50 = â€“ 4

n^{thÂ }term (n) = 10

= 5{100 – 9.4}

= 5{100 – 36}

= 5 Ã— 64

*âˆ´ S _{10}Â = 320*

(ii) Given A.P is, 1, 3, 5, 7, …..to 12 terms.

First term (a) = 1

Common difference (d) = 3 – 1 = 2

n^{th}Â term (n) = 12

= 6 Ã— {2 + 22} = 6.24

*âˆ´ S _{12}Â = 144*

(iii) Given A.P. isÂ 3, 9/2, 6, 15/2, … to 25 terms

First term (a) = 3

Common difference (d) = 9/2 – 3 = 3/2

Sum of n terms S_{n}, given n = 25

(iv) Given expression is 41, 36, 31, â€¦.. to 12 terms.

First term (a) = 41

Common difference (d) = 36 – 41 = -5

Sum of nÂ terms S_{n}, given n = 12

(v) a + b, a â€“ b, a – 3b, â€¦.. to 22 terms

First term (a) = a + b

Common difference (d) = a – b – a – b = -2b

Sum of nÂ terms S_{n}Â = n/2{2a(n – 1). d}

Here n = 22

S_{22}Â = 22/2{2.(a + b) + (22 – 1). -2b}

= 11{2(a + b) – 22b)

= 11{2a – 20b}

= 22a – 440b

*âˆ´S _{22}Â = 22a – 440b*

(vi) (x – y)^{2},(x^{2}Â + y^{2}), (x + y)^{2},… to n terms

First term (a) = (x – y)^{2}

Common difference (d) = x^{2}Â + y^{2}Â – (x – y)^{2}

= x^{2}Â + y^{2}Â – (x^{2}Â + y^{2}Â – 2xy)

= x^{2}Â + y^{2}Â – x^{2}Â + y^{2}Â + 2xy

= 2xy

Sum of n^{th}Â terms S_{n}Â = n/2{2a + (n – 1). d}

= n/2{2(x – y)^{2}Â + (n – 1). 2xy}

= n{(x – y)^{2}Â + (n – 1)xy}

*âˆ´ S _{n}Â = n{(x â€” y)^{2}Â + (n â€” 1). xy)*

(viii) Given expression -26, – 24. -22, to 36 terms

First term (a) = -26

Common difference (d) = -24 – (-26)

= -24 + 26 = 2

Sum of n terms, S_{n}Â = n/2{2a + (n – 1)d) for n = 36

S_{n}Â = 36/2{2(-26) + (36 – 1)2}

= 18[-52 + 70]

= 18×18

= 324

*âˆ´ S _{n}Â = 324*

**2. Find the sum to n terms of the A.P. 5, 2, â€“1, â€“ 4, â€“7, …**

**Solution:**

Given AP is 5, 2, -1, -4, -7, …..

Here, a = 5, d = 2 – 5 = -3

We know that,

S_{n}Â = n/2{2a + (n – 1)d}

= n/2{2.5 + (n – 1) – 3}

= n/2{10 – 3(n – 1)}

= n/2{13 – 3n)

*âˆ´ S _{n}Â = n/2(13 – 3n)*

**3. Find the sum of n terms of an A.P. whose the terms is given by a _{n} = 5 – 6n.**

**Solution:**

Given nth term of the A.P as a_{n}Â = 5 – 6n

Put n = 1, we get

a_{1}Â = 5 – 6.1 = -1

So, first term (a) = -1

Last term (a_{n}) = 5 – 6n = 1

Then, S_{n}Â = n/2(-1 + 5 – 6n)

*= n/2(4 – 6n) = n(2 – 3n)*

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

**Solution:**

Given A.P. 8, 10, 12, 14, .. , 126

Here, a = 8 , d = 10 â€“ 8 = 2

We know that, a_{n} = a + (n – 1)d

So, to find the number of terms

126 = 8 + (n – 1)2

126 = 8 + 2n – 2

2n = 120

n = 60

Next, letâ€™s find the 51^{st} term

a_{51} = 8 + 50(2) = 108

So, the sum of last ten terms is the sum of a_{51} + a_{52} + a_{53} + â€¦â€¦. + a_{60}

Here, n = 10, a = 108 and l = 126

S = 10/2 [108 + 126]

= 5(234)

*= 1170*

*Hence, the sum of last ten terms of the A.P is 1170.*

**5. Find the sum of first 15 terms of each of the following sequences having n ^{th}Â term as:**

**(i) a _{n}Â = 3 + 4nÂ **

**(ii) b _{n}Â = 5 + 2nÂ **

**(iii) x _{n}Â = 6 – nÂ **

**(iv) y _{n}Â = 9 â€“ 5n**

**Solution:**

(i) Given an A.P. whose n^{th} term is given by a_{n}Â = 3 + 4n

To find the sum of the n terms of the given A.P., using the formula,

S_{n}Â = n(a + l)/ 2

Where, a = the first term l = the last term.

Putting n = 1 in the given a_{n}, we get

a = 3 + 4(1) = 3 + 4 = 7

For the last term (l), here n = 15

a_{15} = 3 + 4(15) = 63

So,Â S_{n}Â = 15(7 + 63)/2

= 15 x 35

*= 525*

*Therefore, the sum of the 15 terms of the given A.P. is S _{15}Â = 525*

(ii) Given an A.P. whose n^{th} term is given by b_{n}Â = 5 + 2n

To find the sum of the n terms of the given A.P., using the formula,

S_{n}Â = n(a + l)/ 2

Where, a = the first term l = the last term.

Putting n = 1 in the given b_{n}, we get

a = 5 + 2(1) = 5 + 2 = 7

For the last term (l), here n = 15

a_{15} = 5 + 2(15) = 35

So,Â S_{n}Â = 15(7 + 35)/2

= 15 x 21

*= 315*

*Therefore, the sum of the 15 terms of the given A.P. is S _{15}Â = 315*

(iii) Given an A.P. whose n^{th} term is given by x_{n}Â = 6 – n

To find the sum of the n terms of the given A.P., using the formula

S_{n}Â = n(a + l)/ 2

Where, a = the first term l = the last term.

Putting n = 1 in the given x_{n}, we get

a = 6 â€“ 1 = 5

For the last term (l), here n = 15

a_{15} = 6 â€“ 15 = -9

So,Â S_{n}Â = 15(5 â€“ 9)/2

= 15 x (-2)

*= -30*

*Therefore, the sum of the 15 terms of the given A.P. is S _{15}Â = -30*

(iv) Given an A.P. whose n^{th} term is given by y_{n}Â = 9 â€“ 5n

To find the sum of the n terms of the given A.P., using the formula,

S_{n}Â = n(a + l)/ 2

Where, a = the first term l = the last term.

Putting n = 1 in the given y_{n}, we get

a = 9 – 5(1) = 9 â€“ 5 = 4

For the last term (l), here n = 15

a_{15} = 9 – 5(15) = -66

So,Â S_{n}Â = 15(4 – 66)/2

= 15 x (-31)

*= -465*

*Therefore, the sum of the 15 terms of the given A.P. is S _{15}Â = -465*

**6. Find the sum of first 20 terms the sequence whose n ^{th}Â term is a_{n}Â = An + B.**

**Solution:
**

Given an A.P. whose nth term is given by, a_{n}Â = An + B

We need to find the sum of first 20 terms.

To find the sum of the n terms of the given A.P., we use the formula,

S_{n}Â = n(a + l)/ 2

Where, a = the first term l = the last term,

Putting n = 1 in the given a_{n}, we get

a = A(1) + B = A + B

For the last term (l), here n = 20

A_{20} = A(20) + B = 20A + B

S_{20}Â = 20/2((A + B) + 20A + B)

= 10[21A + 2B]

*= 210A + 20B*

*Therefore, the sum of the first 20 terms of the given A.P. is 210 A + 20B*

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

**Solution:**

Given an A.P. whose n^{th} term is given by a_{n}Â = 2 â€“ 3n

To find the sum of the n terms of the given A.P., we use the formula,

S_{n}Â = n(a + l)/ 2

Where, a = the first term l = the last term.

Putting n = 1 in the given a_{n}, we get

a = 2 – 3(1) = -1

For the last term (l), here n = 25

a_{25} = 2 – 3(25) = -73

So,Â S_{n}Â = 25(-1 – 73)/2

= 25 x (-37)

*= -925*

*Therefore, the sum of the 25 terms of the given A.P. is S _{25}Â = -925*

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

**Solution:**

Given an A.P. whose n^{th} term is given by a_{n}Â = 7 â€“ 3n

To find the sum of the n terms of the given A.P., we use the formula,

S_{n}Â = n(a + l)/ 2

Where, a = the first term l = the last term.

Putting n = 1 in the given a_{n}, we get

a = 7 – 3(1) = 7 â€“ 3 = 4

For the last term (l), here n = 25

a_{15} = 7 – 3(25) = -68

So,Â S_{n}Â = 25(4 – 68)/2

= 25 x (-32)

*= -800*

*Therefore, the sum of the 15 terms of the given A.P. is S _{25}Â = -800*

**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.**

**Solution:**

Given the sum of the certain number of terms of an A.P. = 116

We know that, S_{n}Â = n/2[2a + (n âˆ’ 1)d]

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.(25, 22, 19,…)

Here we have, the first term (a) = 25

The sum of n terms S_{n}Â = 116

Common difference of the A.P. (d) = a_{2}Â – a_{1}Â = 22 â€“ 25 = -3

Now, substituting values in S_{n}

âŸ¹Â 116 = n/2[2(25) + (n âˆ’ 1)(âˆ’3)]

âŸ¹Â (n/2)[50 + (âˆ’3n + 3)]Â = 116

âŸ¹Â (n/2)[53 âˆ’ 3n]Â = 116

âŸ¹ 53n – 3n^{2} = 116 x 2

Thus, we get the following quadratic equation,

3n^{2}Â – 53n + 232 = 0

By factorization method of solving, we have

âŸ¹ 3n^{2}Â – 24n – 29n + 232 = 0

âŸ¹ 3n( n – 8 ) – 29 ( n – 8 ) = 0

âŸ¹ (3n – 29)( n – 8 ) = 0

So, 3n – 29 = 0

âŸ¹ n =Â 29/3

Also, n – 8 = 0

âŸ¹ n = 8

Since, n cannot be a fraction, so the number of terms is taken as 8.

So, the term is:

a_{8}Â = a_{1}Â + 7d = 25 + 7(-3) = 25 – 21 =Â *4*

Hence, the last term of the given A.P. such that the sum of the terms is 116 isÂ 4.

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

**(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?Â **

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

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

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

**Solution: **

(i) Given AP. is 18, 16, 14, …

We know that,

*S _{n}Â = n/2[2a + (n âˆ’ 1)d]*

Here,

The first term (a) = 18

The sum of n terms (S_{n}) = 0 (given)

Common difference of the A.P.

(d) = a_{2}Â Â – a_{1}Â = 16 – 18 = â€“ 2

So, on substituting the values in S_{n}

âŸ¹Â 0 = n/2[2(18) + (n âˆ’ 1)(âˆ’2)]

âŸ¹Â 0 = n/2[36 + (âˆ’2n + 2)]

âŸ¹Â 0 = n/2[38 âˆ’ 2n]Â Further,Â n/2

âŸ¹ n = 0 Or, 38 – 2n = 0

âŸ¹ 2n = 38

âŸ¹ *n = 19*

Since, the number of terms cannot be zer0, hence the number of terms (n) should be 19.

(ii) Given, the first term (a) = -14, Filth term (a_{5}) = 2, Sum of terms (S_{n}) = 40 of the A.P.

If the common difference is taken as d.

Then, a_{5}Â = a_{Â }+ 4d

âŸ¹ 2 = -14 + 4d

âŸ¹ 2 + 14 = 4d

âŸ¹ 4d = 16

âŸ¹ d = 4

Next, we know that S_{nÂ }= n/2[2a + (n âˆ’ 1)d]

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

d = common difference of the given A.P.

n = number of terms

Now, on substituting the values in S_{n}

âŸ¹Â 40 = n/2[2(âˆ’14) + (n âˆ’ 1)(4)]

âŸ¹Â 40 = n/2[âˆ’28 + (4n âˆ’ 4)]

âŸ¹Â 40 = n/2[âˆ’32 + 4n]

âŸ¹ 40(2) = – 32n + 4n^{2}

So, we get the following quadratic equation,

4n^{2}Â – 32n – 80 = 0

âŸ¹ n^{2}Â – 8n – 20 = 0

On solving by factorization method, we get

n^{2}Â – 10n + 2n – 20 = 0

âŸ¹ n(n – 10) + 2( n – 10 ) = 0

âŸ¹ (n + 2)(n – 10) = 0

Either, n + 2 = 0

âŸ¹ n = -2

Or, n – 10 = 0

âŸ¹ *n = 10*

Since the number of terms cannot be negative**.**

*Therefore, the number of terms (n) is 10.*

(iii) Given AP is 9, 17, 25,…

We know that,

S_{n}Â = n/2[2a + (n âˆ’ 1)d]

Here we have,

The first term (a) = 9 and the sum of n terms (S_{n}) = 636

Common difference of the A.P. (d) = a_{2}Â Â – a_{1}Â = 17 – 9 = 8

Substituting the values in S_{n}, we get

âŸ¹Â 636 = n/2[2(9) + (n âˆ’ 1)(8)]

âŸ¹Â 636 = n/2[18 + (8n âˆ’ 8)]

âŸ¹ 636(2) = (n)[10 + 8n]

âŸ¹ 1271 = 10n + 8n^{2}

Now, we get the following quadratic equation,

âŸ¹ 8n^{2}Â + 10n – 1272 = 0

âŸ¹ 4n^{2}+ 5n – 636 = 0

On solving by factorisation method, we have

âŸ¹ 4n^{2}Â – 48n + 53n – 636 = 0

âŸ¹ 4n(n – 12) + 53(n – 12) = 0

âŸ¹ (4n + 53)(n – 12) = 0

Either 4n + 53 = 0 âŸ¹Â n = -53/4

Or, n – 12 = 0 âŸ¹ *n = 12*

Since, the number of terms cannot be a fraction.

*Therefore, the number of terms (n) is 12.*

(iv) Given A.P. is 63, 60, 57,…

We know that,

S_{n}Â = n/2[2a + (n âˆ’ 1)d]

Here we have,

the first term (a) = 63

The sum of n terms (S_{n}) = 693

Common difference of the A.P. (d) = a_{2}Â – a_{1}Â = 60 – 63 = â€“3

On substituting the values in S_{n }we get

âŸ¹Â 693 = n/2[2(63) + (n âˆ’ 1)(âˆ’3)]

âŸ¹Â 693 = n/2[126+(âˆ’3n + 3)]

âŸ¹Â 693 = n/2[129 âˆ’ 3n]

âŸ¹ 693(2) = 129n – 3n^{2}

Now, we get the following quadratic equation.

âŸ¹ 3n^{2}Â – 129n + 1386 = 0

âŸ¹ n^{2}Â – 43n + 462

Solving by factorisation method, we have

âŸ¹ n^{2}Â – 22n – 21n + 462 = 0

âŸ¹ n(n – 22) -21(n – 22) = 0

âŸ¹ (n – 22) (n – 21) = 0

Either, n – 22 = 0 âŸ¹ n = 22

Or,Â n – 21 = 0 âŸ¹ n = 21

Now, the 22^{nd}Â term will be a_{22}Â = a_{1}Â + 21d = 63 + 21( -3 ) = 63 – 63 = 0

So, the sum of 22 as well as 21 terms is 693.

*Therefore, the number of terms (n) is 21 or 22.*

(v) Given A.P. is 27, 24, 21. . .

We know that,

S_{n}Â = n/2[2a + (n âˆ’ 1)d]

Here we have, the first term (a) = 27

The sum of n terms (S_{n}) = 0

Common difference of the A.P. (d) = a_{2}Â – a_{1}Â = 24 – 27 = -3

On substituting the values in S_{n}, we get

âŸ¹Â 0 = n/2[2(27) + (n âˆ’ 1)( âˆ’ 3)]

âŸ¹ 0 = (n)[54 + (n – 1)(-3)]

âŸ¹ 0 = (n)[54 – 3n + 3]

âŸ¹ 0 = n [57 – 3n] Further we have, n = 0 Or, 57 – 3n = 0

âŸ¹ 3n = 57

âŸ¹ *n = 19*

The number of terms cannot be zero,

*Hence, the numbers of terms (n) is 19.*

**11. Find the sum of the first**

**(i) 11 terms of the A.P. : 2, 6, 10, 14,Â . . .Â **

**(ii) 13 terms of the A.P. : -6, 0, 6, 12, . . .Â **

**(iii) 51 terms of the A.P. : whose second term is 2 and fourth term is 8.**

**Solution:**

We know that the sum of terms for different arithmetic progressions is given by

*S _{n}Â = n/2[2a + (n âˆ’ 1)d]*

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

(i) Given A.P 2, 6, 10, 14,… to 11 terms.

Common difference (d) = a_{2}Â – a_{1}Â = 10 – 6 = 4

Number of terms (n) = 11

First term for the given A.P. (a) = 2

So,

S_{11}Â = 11/2[2(2) + (11 âˆ’ 1)4]

=Â 11/2[2(2) + (10)4]

=Â 11/2[4 + 40]

= 11 Ã— 22

*= 242*

Hence, the sum of first 11 terms for the given A.P. is 242

(ii) Given A.P. â€“ 6, 0, 6, 12, … to 13 terms.

Common difference (d) = a_{2}Â – a_{1}Â = 6 – 0 = 6

Number of terms (n) = 13

First term (a) = -6

So,

S_{13}Â = 13/2[2(âˆ’ 6) + (13 â€“1)6]

=Â 13/2[(âˆ’12) + (12)6]

=Â 13/2[60]Â = *390*

*Hence, the sum of first 13 terms for the given A.P. is 390*

(iii) 51 terms of an AP whose a_{2}Â = 2 and a_{4}Â = 8

We know that, a_{2}Â = a + d

2 = a + dÂ …(2)

Also, a_{4}Â = a + 3d

8 = a + 3dÂ Â … (2)

Subtracting (1) from (2), we have

2d = 6

d = 3

Substituting d = 3 in (1), we get

2 = a + 3

âŸ¹ a = -1

Given that the number of terms (n) = 51

First term (a) = -1

So,

S_{n}Â = 51/2[2(âˆ’1) + (51 âˆ’ 1)(3)]

=Â 51/2[âˆ’2 + 150]

=Â 51/2[148]

*= 3774*

*Hence, the sum of first 51 terms for the A.P. is 3774.*

**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.**

**Solution:**

We know that the sum of terms for an A.P is given by

S_{n}Â = n/2[2a + (n âˆ’ 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

(i) Given, first 15 multiples of 8.

These multiples form an A.P: 8, 16, 24, â€¦â€¦ , 120

Here, a = 8 , d = 61 â€“ 8 = 8 and the number of terms(n) = 15

Now, finding the sum of 15 terms, we have

*Hence, the sum of the first 15 multiples of 8 is 960*

(ii)(a) First 40 positive integers divisible by 3.

Hence, the first multiple is 3 and the 40^{th} multiple is 120.

And, these terms will form an A.P. with the common difference of 3.

Here, First term (a) = 3

Number of terms (n) = 40

Common difference (d) = 3

So, the sum of 40 terms

S_{40}Â = 40/2[2(3) + (40 âˆ’ 1)3]

= 20[6 + (39)3]

= 20(6 + 117)

= 20(123) = 2460

*Thus, the sum of first 40 multiples of 3 is 2460.*

(b) First 40 positive integers divisible by 5

Hence, the first multiple is 5 and the 40^{th} multiple is 200.

And, these terms will form an A.P. with the common difference of 5.

Here, First term (a) = 5

Number of terms (n) = 40

Common difference (d) = 5

So, the sum of 40 terms

S_{40 Â }= 40/2[2(5) + (40 âˆ’ 1)5]

= 20[10 + (39)5]

= 20 (10 + 195)

= 20 (205) = 4100

*Hence, the sum of first 40 multiples of 5 is 4100.*

(c) First 40 positive integers divisible by 6

Hence, the first multiple is 6 and the 40^{th} multiple is 240.

And, these terms will form an A.P. with the common difference of 6.

Here, First term (a) = 6

Number of terms (n) = 40

Common difference (d) = 6

So, the sum of 40 terms

S_{40}Â = 40/2[2(6) + (40 âˆ’ 1)6]

= 20[12 + (39)6]

=20(12 + 234)

= 20(246) = 4920

*Hence, the sum of first 40 multiples of 6 is 4920.*

(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.

And, these terms form an A.P. with the common difference of 13.

Here, first term (a) = 104 and the last term (l) = 988

Common difference (d) = 13

Finding the number of terms in the A.P. by, a_{n}Â = a + (n âˆ’ 1)d

We have,

988 = 104 + (n – 1)13

âŸ¹ 988 = 104 + 13n -13

âŸ¹ 988 = 91 + 13n

âŸ¹ 13n = 897

âŸ¹ n = 69

Now, using the formula for the sum of n terms, we get

S_{69}Â = 69/2[2(104) + (69 âˆ’ 1)13]

=Â 69/2[208 + 884]

=Â 69/2[1092]

= 69(546)

*= 37674*

*Hence, the sum of all 3 digit multiples of 13 is 37674.*

(iv) All 3 digit natural number which are multiples of 11.

So, we know that the first 3 digit multiple of 11 is 110 and the last 3 digit multiple of 13 is 990.

And, these terms form an A.P. with the common difference of 11.

Here, first term (a) = 110 and the last term (l) = 990

Common difference (d) = 11

Finding the number of terms in the A.P. by, *a _{n}Â = a + (n âˆ’ 1)d*

We get,

990 = 110 + (n – 1)11

âŸ¹ 990 = 110 + 11n -11

âŸ¹ 990 = 99 + 11n

âŸ¹ 11n = 891

âŸ¹ n = 81

Now, using the formula for the sum of n terms, we get

S_{81}Â = 81/2[2(110) + (81 âˆ’ 1)11]

=Â 81/2[220 + 880]

=Â 81/2[1100]

= 81(550)

*= 44550*

*Hence, the sum of all 3 digit multiples of 11 is 44550.*

**13. Find the sum:**

**(i) 2 + 4 + 6 + . . . + 200Â **

**(ii) 3 + 11 + 19 + . . . + 803Â **

**(iii) (-5) + (-8) + (-11) + . . . + (- 230)Â **

**(iv) 1 + 3 + 5 + 7 + . . . + 199Â **

**(vi) 34 + 32 + 30 + . . . + 10Â **

**(vii) 25 + 28 + 31 + . . . + 100Â **

**Solution:**

We know that the sum of terms for an A.P is given by

S_{n}Â = n/2[2a + (n âˆ’ 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

Or S_{n}Â = n/2[a + l]

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

(i) Given series. 2 + 4 + 6 + . . . + 200Â which is an A.P

Where, a = 2 ,d = 4 â€“ 2 = 2 and last term (a_{n }= l) = 200

We know that, a_{n} = a + (n – 1)d

So,

200 = 2 + (n – 1)2

200 = 2 + 2n â€“ 2

n = 200/2 = 100

Now, for the sum of these 100 terms

S_{100 }= 100/2 [2 + 200]

= 50(202)

*= 10100*

*Hence, the sum of terms of the given series is 10100.*

(ii) Given series. 3 + 11 + 19 + . . . + 803Â which is an A.P

Where, a = 3 ,d = 11 â€“ 3 = 8 and last term (a_{n }= l) = 803

We know that, a_{n} = a + (n – 1)d

So,

803 = 3 + (n – 1)8

803 = 3 + 8n â€“ 8

n = 808/8 = 101

Now, for the sum of these 101 terms

S_{101 }= 101/2 [3 + 803]

= 101(806)/2

= 101 x 403

*= 40703*

*Hence, the sum of terms of the given series is 40703.*

(iii) Given series (-5) + (-8) + (-11) + . . . + (- 230)**Â **which is an A.P

Where, a = -5 ,d = -8 â€“ (-5) = -3 and last term (a_{n }= l) = -230

We know that, *a _{n} = a + (n – 1)d*

So,

-230 = -5 + (n – 1)(-3)

-230 = -5 – 3n + 3

3n = -2 + 230

n = 228/3 = 76

Now, for the sum of these 76 terms

S_{76 }= 76/2 [-5 + (-230)]

= 38 x (-235)

*= -8930*

*Hence, the sum of terms of the given series is -8930.*

(iv) Given series. 1 + 3 + 5 + 7 + . . . + 199**Â **which is an A.P

Where, a = 1 ,d = 3 â€“ 1 = 2 and last term (a_{n }= l) = 199

We know that, a_{n} = a + (n – 1)d

So,

199 = 1 + (n – 1)2

199 = 1 + 2n â€“ 2

n = 200/2 = 100

Now, for the sum of these 100 terms

S_{100 }= 100/2 [1 + 199]

= 50(200)

*= 10000*

*Hence, the sum of terms of the given series is 10000.*

(v) Given series which is an A.P

Where, a = 7, d = 10 Â½ – 7 = (21 â€“ 14)/2 = 7/2 and last term (a_{n }= l) = 84

We know that, *a _{n} = a + (n – 1)d*

So,

84 = 7 + (n – 1)(7/2)

168 = 14 + 7n â€“ 7

n = (168 â€“ 7)/7 = 161/7 = 23

Now, for the sum of these 23 terms

S_{23 }= 23/2 [7 + 84]

= 23(91)/2

*= 2093/2*

*Hence, the sum of terms of the given series is 2093/2.*

(vi) Given series, 34 + 32 + 30 + . . . + 10**Â **which is an A.P

Where, a = 34 ,d = 32 â€“ 34 = -2 and last term (a_{n }= l) = 10

We know that, *a _{n} = a + (n – 1)d*

So,

10 = 34 + (n – 1)(-2)

10 = 34 – 2n + 2

n = (36 â€“ 10)/2 = 13

Now, for the sum of these 13 terms

S_{13 }= 13/2 [34 + 10]

= 13(44)/2

= 13 x 22

*= 286*

*Hence, the sum of terms of the given series is 286.*

(vii) Given series, 25 + 28 + 31 + . . . + 100**Â **which is an A.P

Where, a = 25 ,d = 28 â€“ 25 = 3 and last term (a_{n }= l) = 100

We know that, *a _{n} = a + (n – 1)d*

So,

100 = 25 + (n – 1)(3)

100 = 25 + 3n – 3

n = (100 â€“ 22)/3 = 26

Now, for the sum of these 26 terms

S_{100 }= 26/2 [25 + 100]

= 13(125)

*= 1625*

*Hence, the sum of terms of the given series is 1625.*

**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?**

**Solution:**

Given, the first term of the A.P (a) = 17

The last term of the A.P (l) = 350

The common difference (d) of the A.P. = 9

Let the number of terms be n. And, we know that; l = a + (n – 1)d

So, 350 = 17 + (n- 1) 9

âŸ¹ 350 = 17 + 9n – 9

âŸ¹ 350 = 8 + 9n

âŸ¹ 350 – 8 = 9n

Thus we get, n = 38

Now, finding the sum of terms

*S _{n}Â = n/2[a + l]*

= 38/2(17 + 350)

= 19 Ã— 367

*= 6973*

*Hence, the number of terms is of the A.P is 38 and their sum is 6973.*

**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.**

**Solution:**

Letâ€™s consider the first term as a and the common difference as d.

Given,

a_{3}Â = 7Â …. (1) and,

a_{7}Â = 3a_{3}Â + 2Â Â Â …. (2)

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

a_{7Â }= 3(7) + 2 = 21 + 2 = 23 Â …. (3)

Also, we know that

*a _{n}Â = a +(n – 1)d*

So, the 3th term (for n = 3),

a_{3}Â = a + (3 – 1)d

âŸ¹ 7 = a + 2dÂ Â (Using 1)

âŸ¹ a = 7 – 2d Â Â Â Â …. (4)

Similarly, for the 7th term (n = 7),

a_{7}Â = a + (7 – 1) d 24 = a + 6dÂ = 23Â (Using 3)

a = 23 – 6d Â …. (5)

Subtracting (4) from (5), we get,

a – a = (23 – 6d) – (7 – 2d)

âŸ¹ 0 = 23 – 6d – 7 + 2d

âŸ¹ 0 = 16 – 4d

âŸ¹ 4d = 16

âŸ¹ d = 4

Now, to find a, we substitute the value of d in Â (4), a =7 – 2(4)

âŸ¹ a = 7 – 8

a = -1

Hence, for the A.P. a = -1 and d = 4

For finding the sum, we know that

S_{n}Â = n/2[2a + (n âˆ’ 1)d]Â and n = 20 (given)

S_{20}Â = 20/2[2(âˆ’1) + (20 âˆ’ 1)(4)]

= (10)[-2 + (19)(4)]

= (10)[-2 + 76]

= (10)[74]

*= 740*

*Hence, the sum of first 20 terms for the given A.P. is 740*

**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.**

**Solution:**

Given,

The first term of the A.P (a) = 2

The last term of the A.P (l) = 50

Sum of all the terms S_{n} = 442

So, let the common difference of the A.P. be taken as d.

The sum of all the terms is given as,

442 = (n/2)(2 + 50)

âŸ¹Â 442 = (n/2)(52)

âŸ¹ 26n = 442

âŸ¹ n = 17

Now, the last term is expressed as

50 = 2 + (17 – 1)d

âŸ¹ 50 = 2 + 16d

âŸ¹ 16d = 48

*âŸ¹ d = 3*

*Thus, the common difference of the A.P. is d = 3.*

**17. If 12 ^{th}Â 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?**

**Solution:**

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

Given,

a_{12}Â = -13 S_{4}Â = 24

Also, we know that *a _{nÂ }= a + (n – 1)d*

So, for the 12th term

a_{12}Â = a + (12 – 1)d = -13

âŸ¹ a + 11d = -13

a = -13 – 11d Â …. (1)

And, we that for sum of terms

*S _{n}Â = n/2[2a + (n âˆ’ 1)d]*

Here, n = 4

S_{4}Â = 4/2[2(a) + (4 âˆ’ 1)d]

âŸ¹ 24 = (2)[2a + (3)(d)]

âŸ¹ 24 = 4a + 6d

âŸ¹ 4a = 24 – 6d

Subtracting (1) from (2), we have

Further simplifying for d, we get,

âŸ¹ -19 Ã— 2 = 19d

âŸ¹ d = â€“ 2

On substituting the value of d in (1), we find a

a = -13 – 11(-2)

a = -13 + 22

a = 9

Next, the sum of 10 term is given by

S_{10}Â = 10/2[2(9) + (10 âˆ’ 1)(âˆ’2)]

= (5)[19 + (9)(-2)]

= (5)(18 – 18) = 0

*Thus, the sum of first 10 terms for the given A.P. is S _{10}Â = 0.*

**18.**

**19. In an A.P., if the first term is 22, the common difference is â€“ 4 and the sum to n terms is 64, find n.**

**Solution:**

Given that,

a = 22, d = â€“ 4 and S_{n} = 64

Let us consider the number of terms as n.

For sum of terms in an A.P, we know that

S_{n}Â = n/2[2a + (n âˆ’ 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

So,

âŸ¹Â S_{n}Â = n/2[2(22) + (n âˆ’ 1)(âˆ’4)]

âŸ¹Â 64 = n/2[2(22) + (n âˆ’ 1)(âˆ’4)]

âŸ¹ 64(2) = n(48 – 4n)

âŸ¹ 128 = 48n – 4n^{2}

After rearranging the terms, we have a quadratic equation

4n^{2}Â – 48n + 128 = 0,

n^{2}Â – 12n + 32 = 0 [dividing by 4 on both sides]

n^{2}Â – 12n + 32 = 0

Solving by factorisation method,

n^{2}Â – 8n – 4n + 32 = 0

n ( n – 8 ) – 4 ( n – 8 ) = 0

(n – 8) (n – 4) = 0

So, we get n – 8 = 0 âŸ¹ n = 8

Or, n – 4 = 0 âŸ¹ n = 4

*Hence, the number of terms can be either n = 4 or 8.*

**20. In an A.P., if the 5 ^{th}Â and 12^{th}Â terms are 30 and 65 respectively, what is the sum of first 20 terms?Â **

**Solution:**

Letâ€™s take the first term as a and the common difference to be d

Given that,

a_{5}Â = 30 Â and a_{12}Â = 65

And, we know that *a _{n}Â = a + (n – 1)d*

So,

a_{5}Â = a + (5 – 1)d

30 = a + 4d

a = 30 – 4dÂ Â …. (i)

Similarly, a_{12}Â = a + (12 – 1) d

65 = a + 11d

a = 65 – 11d …. (ii)

Subtracting (i) from (ii), we have

a – a = (65 – 11d) – (30 – 4d)

0 = 65 – 11d – 30 + 4d

0 = 35 – 7d

7d = 35

d = 5

Putting d in (i), we get

a = 30 – 4(5)

a = 30 – 20

a = 10

Thus for the A.P; d = 5 and a = 10

Next, 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.,

*S _{n}Â = n/2[2a + (n âˆ’ 1)d]*

Where;

a = first term of the given A.P.

d = common difference of the given A.P.

n = number of terms

Here n = 20, so we have

S_{20 }= 20/2[2(10) + (20 âˆ’ 1)(5)]

= (10)[20 + (19)(5)]

= (10)[20 + 95]

= (10)[115]

*= 1150*

*Hence, the sum of first 20 terms for the given A.P. is 1150*

**21. Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.**

**Solution:**

Letâ€™s take the first term as a and the common difference as d.

Given that,

a_{2}Â = 14 and a_{3}Â = 18

And, we know that *a _{n}Â = a + (n – 1)d*

So,

a_{2}Â = a + (2 – 1)d

âŸ¹ 14 = a + d

âŸ¹ a = 14 – d …. (i)

Similarly,

a_{3}Â = a + (3 – 1)d

âŸ¹ 18 = a + 2d

âŸ¹ a = 18 – 2d …. (ii)

Subtracting (i) from (ii), we have

a – a = (18 – 2d) – (14 – d)

0 = 18 – 2d – 14 + d

0 = 4 – d

d = 4

Putting d in (i), to find a

a = 14 – 4

a = 10

Thus, for the A.P. d = 4 and a = 10

Now, to find sum of terms

*S _{n}Â = n/2(2a + (n âˆ’ 1)d)*

Where,

a = the first term of the A.P.

d = common difference of the A.P.

n = number of terms So, using the formula for

n = 51,

âŸ¹ S_{51}Â = 51/2[2(10) + (51 – 1)(4)]

=Â 51/2[20 + (40)4]

=Â 51/2[220]

= 51(110)

*= 5610*

*Hence, the sum of the first 51 terms of the given A.P. is 5610*

**22. 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. **

**Solution: **

Given,

Sum of 7 terms of an A.P. is 49

âŸ¹ S_{7} = 49

And, sum of 17 terms of an A.P. is 289

âŸ¹ S_{17} = 289

Let the first term of the A.P be a and common difference as d.

And, we know that the sum of n terms of an A.P is

*S _{n}Â = n/2[2a + (n âˆ’ 1)d]*

So,

S_{7 }= 49 = 7/2[2a + (7 – 1)d]

= 7/2 [2a + 6d]

= 7[a + 3d]

âŸ¹ 7a + 21d = 49

a + 3d = 7 â€¦.. (i)

Similarly,

S_{17 }= 17/2[2a + (17 – 1)d]

= 17/2 [2a + 16d]

= 17[a + 8d]

âŸ¹ 17[a + 8d] = 289

a + 8d = 17 â€¦.. (ii)

Now, subtracting (i) from (ii), we have

a + 8d â€“ (a + 3d) = 17 â€“ 7

5d = 10

d = 2

Putting d in (i), we find a

a + 3(2) = 7

a = 7 â€“ 6 = 1

So, for the A.P: a = 1 and d = 2

For the sum of n terms is given by,

S_{n}Â = n/2[2(1) + (n âˆ’ 1)(2)]

= n/2[2 + 2n – 2]

= n/2[2n]

= n^{2}

*Therefore, the sum of n terms of the A.P is given by n ^{2}.*

**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.**

**Solution:**

Sum of first n terms of an A.P is given by *S _{n}Â = n/2(2a + (n âˆ’ 1)d)*

Given,

First term (a) = 5, last term (a_{n}) = 45 and sum of n terms (S_{n}) = 400

Now, we know that

a_{n}Â = a + (n – 1)d

âŸ¹ 45 = 5 + (n – 1)d

âŸ¹ 40 = nd – d

âŸ¹ nd – d = 40 …. (1)

Also,

S_{nÂ }= n/2(2(a) + (n âˆ’ 1)d)

400 = n/2(2(5) + (n âˆ’ 1)d)

800 = n (10 + nd – d)

800 = n (10 + 40) [using (1)]

âŸ¹ nÂ = 16

Putting n in (1), we find d

nd – d = 40

16d – d = 40

15d = 40

d =Â 8/3

*Therefore, the common difference of the given A.P. is 8/3.*

**24. In an A.P. the first term is 8, n ^{th}Â term is 33 and the sum of first n term is 123. Find n and the d, the common difference.Â **

**Solution:**

Given,

The first term of the A.P (a) = 8

The nth term of the A.P (l) = 33

And, the sum of all the terms S_{n}Â = 123

Let the common difference of the A.P. be d.

So, find the number of terms by

123 = (n/2)(8 + 33)

123 = (n/2)(41)

n = (123 x 2)/ 41

n = 246/41

n = 6

Next, to find the common difference of the A.P. we know that

*l = a + (n – 1)d*

33 = 8 + (6 – 1)d

33 = 8 + 5d

5d = 25

*d = 5*

*Thus, the number of terms is n = 6 and the common difference of the A.P. is d = 5.*

**25. In an A.P. the first term is 22, n ^{th}Â term is -11 and the sum of first n term is 66. Find n and the d, the common difference.Â **

**Solution:**

Given,

The first term of the A.P (a) = 22

The nth term of the A.P (l) = -11

And, sum of all the terms S_{n} = 66

Let the common difference of the A.P. be d.

So, finding the number of terms by

66 = (n/2)[22 + (âˆ’11)]

66 = (n/2)[22 âˆ’ 11]

(66)(2) = n(11)

6 Ã— 2 = n

n = 12

Now, for finding d

We know that, l = a + (n – 1)d

– 11 = 22 + (12 – 1)d

-11 = 22 + 11d

11d = â€“ 33

*d = â€“ 3*

*Hence, the number of terms is n = 12 and the common difference d = -3*

**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 the common difference.**

**Solution:**

Given,

First term (a) = 7, last term (a_{n}) = 49 and sum of n terms (S_{n}) = 420

Now, we know that

*a _{n}Â = a + (n – 1)d*

âŸ¹ 49 = 7 + (n – 1)d

âŸ¹ 43 = nd – d

âŸ¹ nd – d = 42Â Â ….. (1)

Next,

S_{n}Â = n/2(2(7) + (n âˆ’ 1)d)

âŸ¹ 840 = n[14 + nd – d]

âŸ¹ 840 = n[14 + 42] [using (1)]

âŸ¹ 840 = 54n

âŸ¹ n = 15Â …. (2)

So, by substituting (2) in (1), we have

nd – d = 42

âŸ¹ 15d – d = 42

âŸ¹ 14d = 42

*âŸ¹ d = 3*

*Therefore, the common difference of the given A.P. is 3.*

**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. **

**Solution: **

Given,

First term (a) = 5 and the last term (l) = 45

Also, S_{n} = 400

We know that,

*a _{n}Â = a + (n – 1)d*

âŸ¹ 45 = 5 + (n – 1)d

âŸ¹ 40 = nd – d

âŸ¹ nd – d = 40Â Â ….. (1)

Next,

S_{n}Â = n/2(2(5) + (n âˆ’ 1)d)

âŸ¹ 400 = n[10 + nd – d]

âŸ¹ 800 = n[10 + 40] [using (1)]

âŸ¹ 800 = 50n

âŸ¹ n = 16Â …. (2)

So, by substituting (2) in (1), we have

nd – d = 40

âŸ¹ 16d – d = 40

âŸ¹ 15d = 40

*âŸ¹ d = 8/3*

*Therefore, the common difference of the given A.P. is 8/3.*

**28. The sum of first q terms of an A.P. is 162. The ratio of its 6 ^{th}Â term to its 13^{th}Â term is 1: 2. Find the first and 15^{th}Â term of the A.P.**

**Solution:**

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

And we know that, sum of first n terms is:

*S _{n}Â = n/2(2a + (n âˆ’ 1)d)*

Also, nth term is given by *a _{n}Â = a + (n – 1)d*

From the question, we have

S_{q}Â = 162 andÂ a_{6Â }: a_{13}Â = 1 : 2

So,

2a_{6Â }= a_{13}

âŸ¹ 2 [a + (6 – 1d)] = a + (13 – 1)d

âŸ¹ 2a + 10d = a + 12d

âŸ¹ a = 2dÂ Â …. (1)

And, S_{9}Â = 162

âŸ¹Â S_{9}Â = 9/2(2a + (9 âˆ’ 1)d)

âŸ¹ 162 =Â 9/2(2a + 8d)

âŸ¹ 162 Ã— 2 = 9[4d + 8d]Â [from (1)]

âŸ¹ 324 = 9 Ã— 12d

âŸ¹ d = 3

âŸ¹ a = 2(3) [from (1)]

âŸ¹ a = 6

Hence, the first term of the A.P. is 6

For the 15^{th} term, a_{15}Â = a + 14d = 6 + 14 Ã— 3 = 6 + 42

*Therefore, a _{15}Â = 48*

**29. If the 10 ^{th}Â term of an A.P. is 21 and the sum of its first 10 terms is 120, find its n^{th}Â term.**

**Solution:**

Letâ€™s consider a to be the first term and d be the common difference.

And we know that, sum of first n terms is:

*S _{n}Â = n/2(2a + (n âˆ’ 1)d)Â and n^{th}Â term is given by: a_{n}Â = a + (n – 1)d*

Now, from the question we have

S_{10}Â = 120

âŸ¹Â 120 = 10/2(2a + (10 âˆ’ 1)d)

âŸ¹Â 120 = 5(2a + 9d)

âŸ¹ 24 = 2a + 9dÂ …. (1)

Also given that, a_{10}Â = 21

âŸ¹Â 21 = a + (10 – 1)d

âŸ¹Â 21 = a + 9dÂ …. (2)

Subtracting (2) from (1), we get

24 – 21 = 2a + 9d – a – 9d

âŸ¹a = 3

Now, on putting a = 3 in equation (2), we get

3 + 9d = 21

9d = 18

*d = 2*

*Thus, we have the first term(a) = 3 and the common difference(d) = 2*

Therefore, the n^{th}Â term is given by

a_{n}Â = a + (n – 1)d = 3 + (n – 1)2

= 3 + 2n -2

*= 2n + 1*

*Hence, the n ^{th}Â term of the A.P is (a_{n}) = 2n + 1.*

**30. The sum of first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28 ^{th}Â term of this A.P.**

**Solution:**

Letâ€™s take a to be the first term and d to be the common difference.

And we know that, sum of first n terms

*S _{n}Â = n/2(2a + (n âˆ’ 1)d)*

Given that sum of the first 7 terms of an A.P. is 63.

S_{7}Â = 63

And sum of next 7 terms is 161.

So, the sum of first 14 terms = Sum of first 7 terms + sum of next 7 terms

S_{14 }= 63 + 161 = 224

Now, having

S_{7}Â = 7/2(2a + (7 âˆ’ 1)d)

âŸ¹ 63(2) = 7(2a + 6d)

âŸ¹ 9 Ã— 2 = 2a + 6d

âŸ¹ 2a + 6d = 18Â . . . . (1)

And,

S_{14}Â = 14/2(2a + (14 âˆ’ 1)d)

âŸ¹ 224 = 7(2a + 13d)

âŸ¹ 32 = 2a + 13dÂ …. (2)

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

âŸ¹ 13d – 6d = 32 – 18

âŸ¹ 7d = 14

âŸ¹ d = 2

Using d in (1), we have

2a + 6(2) = 18

2a = 18 – 12

a = 3

Thus, from n^{th}Â term

âŸ¹ a_{28Â }= a + (28 – 1)d

= 3 + 27 (2)

= 3 + 54 = 57

*Therefore, the 28 ^{th}Â term is 57.*

**31. The sum of first seven terms of an A.P. is 182. If its 4 ^{th}Â and 17^{th}Â terms are in ratio 1: 5, find the A.P.**

**Solution:**

Given that,

S_{17}Â = 182

And, we know that the sum of first n term is:

*S _{nÂ }= n/2(2a + (n âˆ’ 1)d)*

So,

S_{7}Â = 7/2(2a + (7 âˆ’ 1)d)

182 Ã— 2 = 7(2a + 6d)

364 = 14a + 42d

26 = a + 3d

a = 26 – 3dÂ … (1)

Also, itâ€™s given that 4^{th}Â term and 17^{th}Â term are in a ratio of 1: 5. So, we have

âŸ¹ 5(a_{4}) = 1(a_{17})

âŸ¹ 5 (a + 3d) = 1 (a + 16d)

âŸ¹ 5a + 15d = a + 16d

âŸ¹ 4a = dÂ …. (2)

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

âŸ¹ 4 ( 26 – 3d ) = d

âŸ¹ 104 – 12d = d

âŸ¹ 104 = 13d

âŸ¹ d = 8

Putting d in (2), we get

âŸ¹ 4a = d

âŸ¹ 4a = 8

âŸ¹ a = 2

*Therefore, the first term is 2 and the common difference is 8. So, the A.P. is 2, 10, 18, 26, . ..*

**32. The n ^{th} term of an A.P is given by (-4n + 15). Find the sum of first 20 terms of this A.P.**

**Solution: **

Given,

The n^{th} term of the A.P = (-4n + 15)

So, by putting n = 1 and n = 20 we can find the first ans 20^{th} term of the A.P

a = (-4(1) + 15) = 11

And,

a_{20} = (-4(20) + 15) = -65

Now, for find the sum of 20 terms of this A.P we have the first and last term.

So, using the formula

S_{n}Â = n/2(a + l)

S_{20}Â = 20/2(11 + (-65))

= 10(-54)

= -540

*Therefore, the sum of first 20 terms of this A.P. is -540.*

**33. In an A.P. the sum of first ten terms is -150 and the sum of its next 10 term is -550. Find the A.P.**

**Solution:**

Letâ€™s take a to be the first term and d to be the common difference.

And we know that, sum of first n terms

S_{n}Â = n/2(2a + (n âˆ’ 1)d)

Given that sum of the first 10 terms of an A.P. is -150.

S_{10}Â = -150

And the sum of next 10 terms is -550.

So, the sum of first 20 terms = Sum of first 10 terms + sum of next 10 terms

S_{20 }= -150 + -550 = -700

Now, having

S_{10}Â = 10/2(2a + (10 âˆ’ 1)d)

âŸ¹ -150 = 5(2a + 9d)

âŸ¹ -30 = 2a + 9d

âŸ¹ 2a + 9d = -30Â . . . . (1)

And,

S_{20}Â = 20/2(2a + (20 âˆ’ 1)d)

âŸ¹ -700 = 10(2a + 19d)

âŸ¹ -70 = 2a + 19dÂ …. (2)

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

âŸ¹ 19d – 9d = -70 â€“ (-30)

âŸ¹ 10d = -40

âŸ¹ d = -4

Using d in (1), we have

2a + 9(-4) = -30

2a = -30 + 36

a = 6/2 = 3

Hence, we have a = 3 and d = -4

*So, the A.P is 3, -1, -5, -9, -13,â€¦..*

**34. Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25 ^{th} term. **

**Solution **

Given,

First term of the A.P is 1505 and

S_{14} = 1505

We know that, the sum of first n terms is

S_{n}Â = n/2(2a + (n âˆ’ 1)d)

So,

S_{14} = 14/2(2(10) + (14 âˆ’ 1)d)Â = 1505

7(20 + 13d) = 1505

20 + 13d = 215

13d = 215 â€“ 20

d = 195/13

d =15

Thus, the 25^{th} term is given by

a_{25} = 10 + (25 -1)15

= 10 + (24)15

= 10 + 360

= 370

*Therefore, the 25 ^{th} term of the A.P is 370*

**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.**

**Solution:**

Given,

The first term of the A.P. (a) = 2

The last term of the A.P. (l) = 29

And, sum of all the terms (S_{n}) = 155

Let the common difference of the A.P. be d.

So, find the number of terms by sum of terms formula

S_{n} = n/2 (a + l)

155 = n/2(2 + 29)

155(2) = n(31)

31n = 310

n = 10

Using n for the last term, we have

l = a + (n – 1)d

29 = 2 + (10 – 1)d

29 = 2 + (9)d

29 – 2 = 9d

9d = 27

d = 3

*Hence, the common difference of the A.P. is d = 3*

**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?**

**Solution: **

Given,

In an A.P first term (a) = 17 and the last term (l) = 350

And, the common difference (d) = 9

We know that,

a_{n} = a + (n – 1)d

so,

a_{n} = l = 17 + (n – 1)9 = 350

17 + 9n â€“ 9 = 350

9n = 350 â€“ 8

n = 342/9

n = 38

So, the sum of all the term of the A.P is given by

S_{n} = n/2 (a + l)

= 38/2(17 + 350)

= 19(367)

= 6973

*Therefore, the sum of terms of the A.P is 6973.*

**37. Find the number of terms of the A.P. â€“12, â€“9, â€“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.**

**Solution:**

Given,

First term, aÂ = -12

Common difference, d = a_{2}Â – a_{1}Â = â€“ 9 â€“Â (- 12)

d = – 9 + 12 = 3

And, we know that n^{th}Â term =Â a_{n}Â = a + (n – 1)d

âŸ¹ 21 = -12 + (n – 1)3

âŸ¹ 21 = -12 + 3n – 3

âŸ¹ 21 = 3n – 15

âŸ¹ 36 = 3n

âŸ¹Â n = 12

Thus, the number of terms is 12.

Now, if 1 is added to each of the 12 terms, the sum will increase by 12.

Hence, the sum of all the terms of the A.P. so obtained is

âŸ¹ S_{12Â }+ 12Â = 12/2[a + l] + 12

= 6[-12 + 21] + 12

= 6 Ã— 9 + 12

= 66

*Therefore, the sum after adding 1 to each of the terms in the A.P is 66.*

## Frequently Asked Questions on RD Sharma Solutions for Class 10 Maths Chapter 9

### Is RD Sharma Solutions for Class 10 Maths Chapter 9 important from an exam point of view?

### Give a short summary on RD Sharma Solutions for Class 10 Maths Chapter 9?

1. Understanding sequences

2. Arithmetic Progressions

3. Describing the sequence by writing the algebraic formula for its terms

4. Find the sum of terms in an A.P.

5. Solving various word problems related to Arithmetic Progressions