# ML Aggarwal Solutions for Chapter 9 Arithmetic and Geometric Progression Class 10 Maths ICSE

**Exercise 9.1 **

**1. For the following A.P.s, write the first term a and the common difference d:**

**(i) 3, 1, -1, - 3,…….**

**(ii) 1/3, 5/3, 9/3, 13/3,……**

**(iii) – 3.2, - 3, - 2.8, - 2.6,……**

**Answer**

**(i)** 3, 1, -1, - 3, ……..

Here first term (a) = 3

And the common difference (d)

= 1 – 3 = - 2,

- 1 – 1 = - 2,...

⇒ d = - 2

**(ii) **1/3, 5/3, 9/3, 13/3, ….

Here first common term (a) = 1/3

And common difference (d) = 5/3 – 1/3 = 4/3,

9/3 – 5/3 = 4/3,…

⇒ d = 4/3

**(iii)** – 3.2, - 3, - 2.8, - 2.6,……

Here first term (a) = 3.2

And common difference (d) = -3 – (- 3.2)

= - 3 + 3.2

= 0.2

⇒ d = 0.2

**2.Write first terms of the A.P., when the first term a and the common difference d are given as follows: **

**(i) a = 10, d = 10**

**(ii) a = - 2, d = 10 **

**(iii) a = 4, d = - 3 **

**(iv) a = 1/2, d = - 1/2 **

**Answer**

**(i)** a = 10, d = 10

∴ A.P. = 10, 20, 30, 40....

**(ii)** a = - 2, d = 0

∴ A.P. = -2, - 2, - 2, - 2,…

**(iii)** a = 4, d = - 3

∴ A.P. = 4, 1, - 2, - 5, …

**(iv)** a = 1/2, d = - 1/6

A.P. is 1/2,

(1/2 – 1/6) = 2/6

2/6 – 1/6

1/6, …..

A.P. = 1/2, 2/6, 1/6, 0, …

= 1/2, 1/3, 1/6, 0,…..

**3. Which of the following lists of numbers form an A.P. ? If they form an A.P., find the common difference d and write the next three terms:**

**(i) 4, 10, 16, 22…..**

**(ii) – 2, 2, -2, 2,…**

**(iii) 2, 4, 8, 16,……**

**(iv) 2, 5/2, 3, 7/2, …..**

**(v) – 10, - 6, - 2, 2……**

**(vi) 1 ^{2}, 3^{2}, 5^{2}, 7^{2},…**

**(vii) 1, 3, 9, 27,…..**

**(ix) a, 2a, 3a, 4a, …….**

**(x) a, 2a + 1, 3a + 2, 4a + 3, ……**

**Answer**

**(i)** 4, 10, 16, 22,…

Here a = 4,

d = 10 – 4,

16 – 10 = 6 and,

22 – 16 = 6

∵ common difference is same

∵ It is in AP

And next three terms are 28, 34, 40

**(ii) **– 2, 2, - 2, 2, …..

Here, a = - 2

d = 2 – (-2) = 2 + 2 = 4 and

- 2 – 2 = - 4

∵ Common difference is not same.

∴ It is not an A.P.

**(iii)** 2, 4, 8, 16,…..

Here a = 2

d = 4 – 2 = 2,

8 – 4 = 4 and,

16 – 8 = 8

∴ Common differences is not same.

∴ It is not an A.P.

**(iv)** 2, 5/2, 3, 7/2, ……

Here a = 2,

d = 5/2 – 2 = 1/2

3 – 5/2 = 1/2

7/2 – 3 = 1/2

∵ Common difference is same.

∴ It is not an A.P.

**(iv)** 2, 5/2, 3, 7/2, …..

Here a = 2,

d = 5/2 – 2 = 1/2

3 – 5/2 = 1/2

7/2 – 3 = 1/2

∵ Common difference is same.

∴ It is an A.P.

And next three terms are 4, 9/2, 5

**(v) **– 10, - 6, -2, 2, …..

Here, first term (a) = - 10

d = - 6 – (-10)

= - 6 + 10 = 4,

- 2 – (-6) = - 2 + 6 = 4 and,

2 – (-2) = 2 + 2 = 4

∴ Common difference is same.

∴ It is an A.P.

And next three terms are 6, 10, 14, 20

**(vi)** 1^{2}, 3^{2}, 5^{2}, 7^{2}, ….

= 1, 9, 25, 49, …..

Here, first term (a) = 1^{2} = 1

d = 9 – 1 = 8,

25 – 9 = 16 and,

49 – 25 = 24

∵ Common difference is not same.

∴ It is not an A.P.

**(vii) **1, 3, 9, 27, …..

Here, first term (a) = 1

d = 3 – 1 = 2,

9 – 3 = 6 and,

27 – 9 = 18

∵ Common difference is not same.

∴ It is not an A.P.

⇒ √2, 2√2, 3√2, 4√2, …

Here, first term (a) = √2

And common difference (d) = 2√2 - √2 = √2

3√2 - 2√2 = √2

4√2 - 3√2 = √2

∵ The common difference is same.

**(ix)** a, 2a, 3a, 4a,…..

Here first term (a) = a

Common difference (d) = 2a – a = a

3a – 2a = a

4a – 3a = a

The common difference is same.

It is an A.P.

**(x)** And next three terms are 5a, 6a, 7a

a, 2a + 1, 3a + 2, 4a + 3,…..

Here first term (a) = a

And common difference (d) = 2a + 1 – a = a + 1

3a + 2 – 2a – 1 = a + 1

4a + 3 – 3a- 2 = a + 1

Common difference is same.

∴ It is an A.P.

And three next terms are 5a + 4, 6a + 5, 7a + 6, ….

**Exercise 9.2 **

**1. Find the A.P. whose nth term is 7 – 3K. Also find the 20 ^{th} term. **

**Answer**

T_{n} = 7 – 3n

Giving values 1, 2, 3, 4, ….. to n, we get

T_{1 }= 7 – 3 × 1 = 7 – 3 = 4

T_{2} = 7 – 3 × 2 = 7 – 6 = 1

T_{3} = 7 – 3 × 3 = 7 – 9 = - 2

T_{4} = 7 – 3 × 4 = 7 – 12 = - 5

T_{20} = 7 – 3 × 20 = 7 – 60 = - 53

A.P. is 4, 1, -2, -5, …..

20^{th} term = - 53

**2. ****Find the indicated terms in each of following A.P.s: **

**(i) 1, 6, 11, 16, ….; a _{20} **

**(ii) -4, -7, - 10, - 13, …, a _{25}, a_{n} **

**Answer**

**(i)** 1, 6, 11, 6, …..

Here, a = 1, d = 6 – 1 – 5

a_{20} = a + (n – 1)d

= 1 + (20 – 1) × 5

= 1 + 19 × 5

= 96

**(ii)** -4, -7, - 10, - 13, …., a25, an

Here, a = - 4,

d = - 7 – (-4) = - 7 + 4 - 3

a_{25} = a + (n – 1)d

= - 4 + (25 – 1) × - 3

= - 4 + 24 × (-3)

= - 4 – 72

= - 76

And a_{n} = a + (n – 1)d

= - 4 + (n – 1)(- 3)

= - 4 – 3n + 3

= - 1 – 3n

= - 3n – 1

**3. Find the nth term and the 12 ^{th} term of the list of numbers; 5, 2, -1, -4, **

**Answer**

5, -2, -1, - 4, …..

Here, a = 5

d = 2 – 5 = - 3

**(i)** T_{n }= a + (n – 1)d

= 5 + (n – 1)(-3)

= 5 – 3n + 3

= 8 – 3n

**(ii) **T_{12} = a + 11d

= 5 + 11(-3)

= 5 – 33

= - 28

**4. Find the 8 ^{th} term of the A.P. whose first term is 7 and common difference is 3. **

**Answer:**

First term (a) = 7

And common difference (d) = 3

A.P. = 7, 10, 13, 16, 19,….

T_{8} = a + (n – 1)d

= 7 + (8 – 1) × 3

= 7 + 7 × 3

= 7 + 21

= 28

**5. (i) If the common difference of an A.P. is – 3 and the 18 ^{th} term is – 5, then find its first term. **

**(ii) If the first term of an A.P. is – 18 and its 10 ^{th} term is zero, then find its common difference. A**

**Answer**

**(i) **Common difference (d) = 3

T_{18} = - 5

a + (n – 1)d = T_{n}

a + (18 – 1) (-3) = - 5

⇒ a + 17(-3) = - 5

⇒ a – 51 = - 5

⇒ a = - 5 + 51 = 46

∴ First term = 46

**(ii) **First term (a) = - 18

T_{10} = 0

a + (n – 1)d = T_{n}

- 18 + (10 – 1)d = 0

- 18 + 9d = 0

⇒ 9d = 18

⇒ d = 18/9 = 2

∴ Common difference = 2

**6. Which term of the A.P. **

**(i) 3, 8, 13, 18, … is 78 ?**

**(ii) 7, 13, 19, … is 205 ?**

**(iii) 18, 15.1/2, 13, … is – 47? **

**Answer**

**(i)** 3, 8, 13, 18, ….. is 78

Let 78 is nth term

Here, a = 3, d = 8 – 3 = 5

∴ 78 = a + (n – 1)d

⇒ 78 = 3 + (n – 1)5

⇒ 78 = 3 + 5n – 5

⇒ 78 + 5 – 3 = 5n

⇒ 5n = 80

⇒ n = 80/5 = 16

∴ 78 is 16^{th} term

**(ii)** 7, 13, 19, …. is 205

Let nth term is 205

Here, a = 7, d = 13 – 7 = 6

205 = a + (n – 1)d

⇒ 205 = 7 + (n – 1) × 6

⇒ 205 = 7 + 6n – 6

⇒ 6n = 205 – 7 + 6 = 204

n = 204/6 = 34

∴ 205 is 34^{th} term.

**(iii) **18, 15.1/2, 13, ……. Is – 47

Let nth term is – 47

a = 18, d = 15.1/2 – 18

= -2.1/2

= - 5/2

∴ - 47 = a + (n – 1)d

⇒ - 47 = 18 + (n – 1)(-5/2)

⇒ - 47 – 18 = -5/2.n + 5/2

⇒ - 65 – 5/2 = -5/2n

⇒ (- 135)/2 = (-5/2).n

∴ n = (-135)/2 × 2/(-5)

= 27

∴ - 47 is 27^{th} term.

**7. (Check whether – 150 is a term of the A.P. 11, 8, 5, 2, ……)**

**(ii) Find whether 55 is a term of the A.P. 7, 10, 10, 13, ….. or not. If yes, find which term is it. **

**(iii) Is 0 a term of the A.P. 31, 28, 25,……? Justify your answer.**

**Answer**

**(i) **A.P. is 11, 8, 5, 2, ….

Here, a = 11, d = 8 – 11 = - 3

Let – 150 = n, then

Tn = a + (n – 1)d

⇒ - 150 = 11 + (n – 1)(-3)

⇒ - 150 = 3 – 3n + 11

⇒ 3n = 3 + 150

= 153 + 11

= 164

n = 164/3 = 54.2

No, -150 is not any terms of the A.P.

**(ii)** A.P. 7, 10, 13, ……

Here, a = 7, d = 10 – 7 = 3

Let 55 is the nth term, then

T_{n} = a + (n – 1)d

⇒ 55 = 7 + (n – 1)× 3

⇒ 55 = 7 + 3n – 3

⇒ 3n = 55 – 7 + 3

= 51

∴ n = 51/3 = 17

∴ 55 is a term of the given A.P. and it is 17^{th} term.

**(iii) **A.P. 31, 28, 25, …..

Here, a = 31, d = 28 - 31 = - 3

Let 0 be the nth term, then

Tn = a + (n – 1)d

0 = 31 + (n – 1)(-3)

0 = 31 – 3n + 3

⇒ 3n = 34

n = 34/3 = 11.1

Hence, 0 is not any term of the A.P.

**8. (i) Find the 20 ^{th} term from the last term of the A.P. 3, 8, 13, …… 253. **

**(ii) Find the 12 ^{th} from the end of the A.P. -2, -4, -6, ……; - 100.**

**Answer**

**(i)** A.P. is 3, 8, 13, ……253

12^{th} term from the end

Last term = 253

Here, a = 3, d = 8 – 3 = 5

∴ last term (n) = a + (n – 1)d

253 = 3 + (n – 1) × 5

⇒ 253 = 3 + 5n – 5

⇒ 253 – 3 + 5 = 5n

⇒ 5n = 255

⇒ n = 255/5 = 51

∴ 253 is 51th term

Let m be the 20^{th} term from the last term

Then m be the 20^{th} term from the last term

Then m = l – (n – 1)d

= 253 – (20 – 1) × 5

= 253 – 19 × 5

= 253 – 95

= 158

∴ 20^{th} term from the end = 158

**(ii)** A.P. = -2, -4, -6, ……, - 100

a = -2, d = -4 – (-2)

= - 4 + 2

= - 2

l = - 100

∴ T_{n} = a + (n – 1)d

⇒ - 100 = - 2 + (n – 1) ×(-2)

-100 = -2 – 2n + 2

⇒ + 2n = 100

⇒ n = 100/2

= 50

Let m^{th }term is the 12^{th} term from the end

Then m = l – (n – 1)d

= - 100 – (12 – 1) × (-2)

= - 100 + 22

= - 78

**9. Find the sum of the two middle most terms of the A.P.. – 4/3, -1, - 2/3, …..4.1/3 **

**Answer**

Given,

A.P. is – 4/3, -1, - 2/3,….. 4.1/3

Here, a = - 4/3, d = 1 – (-4)/3 – 1 + 4/3

= 1/3

L = 4.1/3 4

∴ T_{n} = l = 4.1/3 = a + (n – 1)d

⇒ 4.1/3 = -4/3 + (n – 1) ×1/3

∴ 13/3 + 4/3 = 1/3(n – 1)

⇒ 17/3 × 3/1 = (n - 1)

(n - 1) = 17

⇒ n = 17 + 1 = 18

∴ Two middle term are 18/2 and 18/2 + 1

= 9^{th} and 10^{th} term

∴ a_{9} + a_{10}

= a + 8d + a + 9d

= 2a + 17d

= 2 × (-4)/3 + 17 × 1/3

= -8/3 + 17/3

= 9/3

= 3

**10. Which term of A.P. 53, 48, 43, …. Is the first negative term ? **

**Answer**

Let nth term is the first negative term of the A.P. 53, 48, 43, ……

Here, a = 53, d = 48 – 53 = - 5

∴ T_{n} = a + (n – 1)d

= 53 + (n – 1) × (-5)

= 53 – 5n + 5

= 58 – 5n

5n = 58

n = 58/5

= 11.3

∴ 12^{th} term will be negative.

**11. Determine the A.P. whose fifth term is 19 and the difference of the eighth term from the thirteenth term is 20. **

**Answer: **

In an AP.,

T_{5} = 19

T_{13} – T_{8} = 20

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

∴ T_{5} = a + 4d = 19

T_{13} – T_{8} = (a + 12d) – (a + 7d)

⇒ 20 = a + 12d – a – 7d

⇒ 20 = 5d

⇒ d = 20/5 = 4

Substitute the value of d in eq. (i), we get

∴ a + 4 × 4 = 19

⇒ a + 16 = 19

⇒ a = 19 – 16 = 3

∴ A.P. is 3, 7, 11, 15,…..

**12. Determine the A.P. whose third term is 16 and the 7 ^{th} term exceeds the 5^{th} term by 12.**

**Answer**

T_{3} = 16

T_{7 }– T_{5 }= 12

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

T_{3} = a + 2d = 16

T_{7} – T_{5} = (a + 6d) – (a + 4d) = 12

⇒ a + 6d – a - 4d = 12

⇒ 2d = 12

⇒ d = 12/2

= 6

Substitute the value of d in eq. (i), we get

∴ a + 2 × 6 = 16

⇒ a + 12 = 16

⇒ a = 16 – 12 = 4

∴ A.P. is 4, 10, 16, 22, 28, ….

**13. Find the 20 ^{th} term of the A.P. whose 7^{th} term is 24 less than the 11^{th} term, first term being 12. **

**Answer**

T_{1} – T_{7} = 24

a = 12

Let a be the first term and d be the common difference, then (a + 10d) – (a + 6d) = 24

a + 10d – a – 6d = 24 ⇒ 4d = 24

⇒ d = 24/4 = 6

a = 12

∴ T_{20} = a + 19d = 12 + 19 × 6

= 12 + 114

= 126

**14. Find the 31 ^{st} term of an A.P. whose 11^{th} term is 38 and 6^{th} term is 73. **

**Answer**

T_{11} = 38, T6 = 73

Let a be the first term and d be the common difference, then

a + 10d = 38 **…(i)**

a + 5d = 73 **…(iii)**

Subtracting, 5d = 35

d = 35/5

Substituting, 5d = 35

d = 35/3 = 7

Substitute the value of d in eq. (i), we get

a + 10d = 38

a + 70 = 38

⇒ a = 38 – 70 = - 32

∴ T_{31} = a + 30d

= - 32 + 30 × 7

= - 32 + 210

= 178

∴ 31st term = 178

**15. If the seventh term of an A.P. is 1/9 and its ninth term is 1/7, find its 63 ^{rd} term. **

**Answer**

a_{7} = 1/9

⇒ a + 6d = 1/9 **…(i)**

a_{9 }= 1/7

⇒ a + 8d = 1/7 **…(ii)**

On subtracting, - 2d = 1/9 – 1/7

⇒ -2d = (7 – 9)/63

⇒ -2d = -2/63

∴ d= 1/63

Now, substitute the value of d in eq. (i),

We get,

a + 6(1/63) = 1/9

⇒ a = 1/9 – 1/63

= (7 – 6)/63

= 1/63

∴ a_{63 }= a + 62d

= 1/63 + 62(1/63)

= (1 + 62)/63

= 63/63

= 1

**16. (i) The 15 ^{th} term of an A.P. is 3 more than twice its 7^{th} term. If the 10^{th} term of the A.P. is 41, find its nth term.**

**(ii) The sum of 5 ^{th} and 7^{th} terms of an A.P. is 52 and the 10^{th} term is 46. Find the A.P.**

**(iii) The sum of 2 ^{nd} and 7^{th} terms of an A.P. is 30. If its 15^{th} term is 1 less than twice its 8^{th} term, find the A.P.**

**Answer**

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

We have,

a_{10} = 41

⇒ a + 9d = 41 **…(i)**

and a_{15} = 2a_{7} + 3

⇒ a + 14d = 2(a + 6d) + 3

⇒ a + 14d = 2a + 12d + 3

⇒ a – 2d = - 3 **…(ii)**

Subtracting (i) from (ii),

4d = 20

⇒ d = 20/4 = 5

⇒ d = 5

Now, put the value of d in eq. (i)

a + 5 × 5 = 26

⇒ a = 26 – 25

⇒ a = 1

Hence, a_{2} = a_{1} + d

= 1 + 5 = 6

a_{3} = a_{2} + d

= 6 + 5 = 11

a_{4} = a_{3} + d

= 11 + 5 = 16

∴ The A.P. formed is 1, 6, 11, 16, …

**17. If 8 ^{th} term of an A.P. is zero, prove that its 38^{th} term is triple of its 18^{th} term. **

**Answer**

T_{8 }= 0

To prove that T38 = 3 ×T_{18}

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

∴ T_{8} = a+ 7d = 0

⇒ a = -7d

Now, T_{38 }= a + 37d

= -7d + 37d

= 30d

And T_{18} = a + 17d

= -7d + 17d

= 10d

It is clear that T_{38} is triple of T_{18}

_{}

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

**Answer**

A.P. is 3, 10, 17,…

Here, a = 3,

d – 10 – 3 = 7

T_{13} = a+ 12d

= 3 = 12 × 7

= 3 + 84

= 87

Let nth term is 84 more then its 13^{th} rem

∴ T_{n} = 84 + 87 = 171

⇒ a + (n – 1)d = 171

⇒ 3 + (n – 1) × 7 = 171

(n – 1) × 7 = 171 – 3 = 168

n – 1 = 168/7 = 24

n = 24 + 1 = 25

∴ 25^{th} term is the required term.

**19. If the nth terms of the two A.P.s, 9, 7, 5, …..
and 24, 21, 18, …… are the same, find the value of n. Also, find that term **

**Answer**

nth term of two A.P.s 9, 7, 5, …. and 24, 21, 18, …. are same

In the first A.P. 9, 7, 5, ….

a = 9 and d = 7 – 9 = - 2

T_{n }= a + (n – 1)(-2)

= 9 – 2n + 2

= 11 – 2n

And in second A.P. 24, 21, 18 ,…..

a_{1} = 24, d_{1} =
21 – 24 = - 3

T_{n} = 24 + (n – 1)(-3)

= 24 – 3n + 3

= 27 – 3n

∵ The nth terms of both A.P.s is same

∴ 11 – 2n = 27 – 3n

- 2n + 3n = 27 – 11

⇒ n = 16

And T_{16} = a + (n – 1)d

= 9 + 15 × (-2)

= 9 – 30

= - 21

**20. (i) How many two digit numbers are divisible by 3? **

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

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

**Answer**

**(i) **Two digits numbers divisible by 3 are

12, 15, 18, 21, … 99

Here, a = 13, d = 15 – 12

= 3 and l = 99

Let number divisible by 3 and n

∴ T_{n} = l = a + (n – 1)d

99 = 12 + (n – 1) × 3

⇒ 99 – 12 = 3(n – 1)

⇒ 3(n – 1) = 87

⇒ n – 1 = 87/3 = 29

∴ n = 29 + 1= 30

**(ii)** Numbers divisible by both 2 and 5 are 110, 120, 130, ….., 990

Here a = 110, d = 120 – 110 = 10

a_{n} = 990

⇒ a + (n – 1)d = 990

⇒ 110 + (n – 1)(10) = 990

⇒ (n – 1)(10) = 990 – 110 = 880

⇒ (n – 1) = 880/10 = 88

∴ n = 88 + 1

= 89

Hence, number between 101 and 999 which are divisible by both 2 and 5 are 89.

**(iii)** Numbers between 10 and 300, which when divided by 4 leave a remainder 3 will be 11, 15, 19, 23,…. 299

Here, a = 11, d = 15 – 11 = 4, l = 299

∴ T_{n} = l = a + (n – 1)d

299 = 11 + (n – 1) × 4

⇒ 299 – 11 = (n – 1)4

4(n – 1) = 288

⇒ n – 1 = 288/4 = 72

∴ n = 72 + 1 = 73

**21. If the numbers n – 2, 4n – 1 and 5n + 2 are in A.P., find the value of n. **

**Answer**

n – 2, 4n – 1 and 5n + 2 are in A.P.

∴ 2(4n – 1) = n – 2 + 5n + 2

8n – 2 = 6n

⇒ 8n – 6n = 2

⇒ 2n = 1

⇒ n = 2/2 = 1

∴ n = 1

**22. Sum of three numbers in A.P. is 3 and their product is – 35. Find the numbers. **

**Answer**

Sum of three numbers which are in A.P. = 3

Their product = - 35

Let three numbers which are in A.P.

a – d, a, a + d

a – d + a + a + d = 3

⇒ 3a = 3,

⇒ a = 3/3 = 1

and (a – d) × a × (a + d) = - 35

⇒ (1 – d) × 1 × (1 + d) = - 35

⇒ 1^{2} – d^{2} = - 35

⇒ 1 – d^{2} = - 35

⇒ d^{2} = 35 + 1 = 36

∴ d = ± 6

If d = 6

∴ Numbers are 1 – 6, 1 + 6

= -5, 1, 7

If d = - 6

1 + 6, 1 – 6

⇒ 7, 1, - 5

Hence, numbers in A.P. are -5, 1, 7 or 7, 1, - 5

**23. The sum of three numbers in A.P. is 30 and the ratio of first number to the third number is 3 : 7. Find the numbers.**

**Answer**

Sum of three numbers in A.P. = 30

Ratio between first and the third number = 3 : 7

Let numbers are

a – d, a + d, then

a – d + a + a + d = 30

⇒ 3a = 30

⇒ a = 30/3 = 10

And (a – d)/(a + d) = 3/7

⇒ 7a – 7d = 3a + 3d

⇒ 7a – 3a = 3d + 7d

⇒ 4a = 10d

⇒ 10d = 4 × 10 = 40

⇒ d = 40/10 = 4

∴ Numbers are 10 – 4, 10, 10 + 4

⇒ 6, 10, 14

**24. The sum of the first three terms of an A.P. is 33. If the product of the first and the third terms exceeds the second term by 29, find the A.P.**

**Answer**

Let the three numbers in A.P. are

a – d, a, a + d

Now, a – d + a + a + d = 33

⇒ 3a = 33

⇒ a = 33/3 = 11

And (a – d) (a + d) = a + 29

a^{2} – d^{2} = a + 29

⇒ (11)^{2} – d^{2} = 11 + 29

⇒ 121 – d^{2} = 40

d^{2} = 121 – 40 = 81 = (±9)^{2}

∴ d = ± 9

If d = 9, then

∴ Numbers are 11 – 9, 11 + 9

⇒ 2, 11, 20

If d = - 9, then

11 + 9, 11, 11- 9

⇒ 20, 11, 2

Hence numbers are 2, 11, 20 or 20, 11, 2

**25.**** Justify whether it is true to say that the following are the nth terms of an A.P. **

**(i) 2n – 3**

**(ii) n ^{2 }+ 1**

**Answer**

**(i)** 2n – 3

Giving the some difference values to n such as 1, 2, 3, 4, …. then

2 × 1 – 3 = 2 – 3 = - 1

2 × 2 – 3 = 4 – 3 = 1

2 × 3 – 3 = 6 – 3 = 3

2 × 4 – 3 = 8 – 3 = 5

We see that -1, 1, 3, 5, …. Are in A.P. whose first term = - 1 and,

d = 1 – (-1) = 1 + 1 = 2

**(ii)** n^{2} + 1

Giving some difference values to n such as 1, 2, 3, 4,……

(1)^{2} + 1 = 1 + 1 = 2

(2)^{2} + 1 = 4 + 1 = 5

(3)^{3} + 1 = 9 + 1 = 10

(4)^{2} + 1 = 16 + 1 = 17

We see that a = 2,

d = 5 – 2 = 3

= 10 – 5 = 5

= 17 – 10

= 7

The common difference is not same.

∴ No, It is not an A.P.

**Exercise 9.3 **

**1. Find the sum of the following A.P.s : **

**(i) 2, 7, 12, ….. to 10 terms **

**(ii) 1/15, 1/12, 1/10, …. to 11 terms **

**Answer**

**(i)** 2, 7, 12, …… to 10 terms

Here, a = 2, d = 7 – 2 = 5 and n = 10

S_{10} = n/2[2a + (n – 1)d]

= 10/2 [2 × 2 + (10 - 1) × 5]

= 5(4 + 45)

= 5 × 49

= 245

**(ii)** 1/15, 1/12, 1/10, ……. to 11 terms

a = 1/15,

d = 1/12 – 1/15

= (5 – 4)/60

= 1/60 or 1/10 – 1/12

= (6 – 5)/60

= 1/60

n = 11

∴ S_{11} = n/2 × [2a + (n – 1)d]

= 11/2 × [2 × 1/15 + (11 – 1) × 1/60]

= 11/2 × [2/15 + 1/6]

= 11/2 × [(4 + 5)/30]

= 11/2 × 9/30

= 33/20 or 1.13

**2. How many terms of the A.P. 27, 24, 21, ……, should be taken so that their sum is zero? **

**Answer**

A.P. = 27, 24, 21, ….

a = 27

d = 24 – 27 = - 3

S_{n} = 0

Let n terms be there in A.P.

S_{n} = n/2 [2a + (n – 1)d]

⇒ 0 = n/2[(2 × 27) + (n – 1)(-3)]

⇒ 0 = n(54 – 3n + 3)]

⇒ n[57 – 3n] = 0

⇒ (57 – 3n) = 0/n = 0

⇒ 3n = 57

∴ n = 57/3 = 19

**3. Find the sums given below: **

**(i) 34 + 32 + 30 + ….. + 10 **

**(ii) -5 + (-8) + (- 11) + …. + (-230) **

**Answer**

**(i)** 34 + 32 + 30 + …. + 10

Here, a = 34, d = 32 – 34 = - 2, l = 10

T_{n} = a + (n – 1)d

10 = 34 + (n – 1)(-2)

- 24 = -2(n – 1)

= (-24/-2)

= n – 1

⇒ n – 1 = 12

∴ n = 12 + 1 = 13

S_{n} = n/2[a + 1]

= 13/2 [34 + 10]

= 13/2 × 44

= 286

**(ii)** – 5 + (-8) + (-11) + ……. + (- 230)

Here, a = - 5, d = - 8 – (-5)

= - 8 + 5

= - 3

l = - 230

∴ l = a + (n – 1)d

⇒ - 230 = - 5 + (n – 1)(-3)

- 230 + 5 = - 3(n – 1)

⇒ - 225 = - 3(n – 1)

⇒ (- 225/-3) = n – 1

⇒ n – 1 = 75

⇒ n = 75 + 1 = 76

∴ S_{n} = n/2 [a + l]

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

= 38[-5 – 230]

= 38 × (-235)

= - 8930

**4. In an A.P. (with usual notations): **

**(i) given a = 5, d = 3, a _{n} = 50, find n and S_{n} **

**(ii) given a = 7, a _{13} = 35, find d and S_{13} **

**(iii) given d = 5, S _{9} = 75, find a and a_{9} **

**(iv) given a = 8, a _{n} = 62, S_{n} = 210, find n and d **

**(v) given a = 3, n = 8, S = 192, find d. **

**Answer**

**(i) **a = 5, d = 3, a_{n} = 50

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

50 = 5 + (n – 1) × 3

⇒ 50 – 5 = 3(n – 1)

⇒ 45 = 3(n – 1)

⇒ 45/3 = n – 1

⇒ n – 1 = 15

⇒ n = 15 + 1 = 16

∴ n = 16

And S_{n} = n/2[2a + (n – 1)d]

= 16/2[2 × 5 + (16 – 1) × 3]

= 8[10 + 45]

= 8 × 55

= 440

**(ii)** a = 7, a_{13} = 35

a_{n} = a + (n – 1)d

35 = 7 + (13 – 1)d

⇒ 35 – 7 = 12d

⇒ 28 = 12d

⇒ d = 28/12 = 7/3 = 2.1/3

and S_{13} = n/2[2a + (n – 1)d]

= 13/2 [2 × 7 + (13 – 1) × 7/3]

= 13/2 [14 + 28]

= 13/2 × (42)

= 13 × 21

= 273

**(iii)** d = 5, S_{9} = 75

a_{n} = a + (n – 1)d

a_{9} = a + (9 – 1) × 5

= a + 40 …..(i)

S_{9 }= n/2 [2a + (n – 1)d]

75 = 9/2[2a + 8 × 5]

150/9 = 2a + 40

2a = 150/9 – 40

= 50/3 – 40

2a = -70/3

⇒ a = -70/(2 × 3)

a = - 35/3

From (i),

a_{9 }= a + 40 = (-35)/3 + 40

= (- 35 + 120)/3

= 85/3

∴ a = (- 35)/3, a_{9} = 85/3

**(iv)** a = 8, a_{n} = 62, S_{n} = 210

a_{n} = a + (n – 1)d

62 = 8 + (n – 1)d

(n – 1)d = 62 – 8 = 54 **…(i)**

S_{n} = n/2[2a + (n – 1)d]

⇒ 210 = n/2[2 × 8 + 54] **[From (i)]**

⇒ 420 = n(16 + 54)

⇒ 420 = 70n

⇒ n = 420/70 = 6

∴ (6 – 1)d = 54

⇒ 5d = 54

⇒ d = 54/5

Hence d = 54/5 and n = 6

**(v)** a = 3, n = 8, S = 192

S_{n }= n/2[2a + (n – 1)d]

⇒ 192 = 8/2[2 × 3 + 7 × d]

⇒ 192 = 4[6 + 7d]

⇒ 192/4 = 6 + 7d

⇒ 48 = 6 + 7d

⇒ 7d = 48 – 6 = 42

d = 42/7 = 6

∴ d = 6

**5. (i) 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. **

**(ii) The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its 20 ^{th} term. **

**Answer**

**(i)** First term of an A.P. (a) = 5

Last term (l) = 45

Sum = 400

l = a + (n – 1)d

⇒ 45 = 5 + (n – 1)d

⇒ (n – 1)d = 45 – 5

= 40 **….(i)**

S_{n} = n/2 [2a + (n – 1)d]

400 = n/2 [2 × 5 + 40]

⇒ 800 = n(10 + 40)

50n = 800

⇒ n = 800/50 = 16

From (i),

(16 – 1)d = 40

⇒ 15d = 40

⇒ d = 40/15

∴ d = 8/3 and n = 16

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

Now, a = 15

Sum of first n terms of an A.P. is given by,

S_{n} = n/2[2a + (n – 1)d]

⇒ S_{15} = 15/2[2a + (15 – 1)d]

⇒ 750 = 15/2(2a + 14d)

⇒ a + 7d = 50

⇒ 15 + 7d = 50

⇒ 7d = 35

⇒ d = 5

Now, 20^{th} term = a_{20} = a + 19d = 15 + 19 × 5

= 15 + 95

= 110

**6. 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**

First term of an A.P. (a) = 17

and last term (l) = 350

d = 9

l = T_{n} = a + (n – 1)d

350 = 17 + (n – 1) × 9

⇒ 350 – 17 = 9(n – 1)

⇒ 333 = 9(n – 1)

⇒ n – 1 = 333/9 = 37

n = 37 + 1 = 38

And S_{n} = n/2[2a + (n – 1)d]

= 38/2 [2 × 17 + (38 – 1) × 9]

= 19[34 + 37 × 9]

= 19[34 + 333]

= 19 × 367

= 6973

Hence, n = 38 and S_{n} = 6973

**7. Solve for x : 1 + 4 + 7 + 10 + ….. + x = 287 **

**Answer**

1 + 4 + 7 + 10 + ….. + x = 287

Here, a = 1,

d = 4 – 1 = 3,

n = x

l = x = a+ (n – 1)d

= 1 + (n – 1) × 3

⇒ x – 1 = (n – 1)d

S_{n} = n/2[2a + (n – 1)d]

287 = n/2[2 × 1 + (n – 1)3]

574 = n(2 – 3n – 3)

⇒ 3n^{2} – n – 574 = 0

⇒ 3n^{2} – 42n + 41n – 574 = 0

⇒ 3n(n – 14) + 41(n – 14) = 0

⇒ (n – 14)(3n + 41) = 0

Either n – 14 = 0, then n = 14

Or, 3n + 41 = 0, then 3n = - 41

⇒ n = -41/3

Which is not possible being negative.

∴ n = 14

Now, x = a + (n – 1)d

= 1 + (14 – 1) × 3

= 1 + 13 × 3

= 1 + 39

= 40

∴ x = 40

**8. (i) How many terms of the A.P. 25, 22, 19, … are needed to the sum 116 ? Also find the last term. **

**(ii) How many terms of the A.P. 24, 21, 18, …. must be taken so that the sum is 78? Explain the double answer.**

**Answer**

**(i) **A.P. is 25, 22, 19, …….

Sum = 116

Here, a = 25, d = 22 – 25 = 3

Let the number of terms be n, then

116 = n/2[2a + (n – 1)d]

⇒ 232 = n[2 × 25 + (n – 1)(-3)

⇒ 232 = n[50 – 3n + 3]

⇒ 232 = n(53 – 3n)

⇒ 232 = 53n – 3n^{2}

⇒ 3n^{2} – 53n + 232 = 0

**{∵ 232 × 3 = 696}**

∴ 696 = - 24 × (-29)

- 53 = - 24 – 29 }

⇒ 3n^{2} – 24n – 29n + 232 = 0

⇒ 3n(n – 8) – 29(n – 8) = 0

⇒ (n – 8)(3n – 29) = 0

Either n – 8 = 0, then n = 8

Or 3n – 29 = 0, then 3n = 29

⇒ n = 29/3 which is not possible because of fraction

∴ n = 8

Now, T = a + (n – 1)d

= 25 + 7 × (-3)

= 25 – 21

= 4

**(ii)** A.P. is 24, 21, 18, ……

Sum = 78

Here, a = 24, d = 21 – 24 = - 3

S_{n }= n/2[2a + (n – 1)d]

⇒ 78 = n/2[2 × 24 + (n – 1)(-3)]

⇒ 156 = n(48 – 3n + 3)

⇒ 156 = 51n – 3n^{2}

⇒ 3n^{2} – 51n + 156 = 0

⇒ 3n^{2 }– 12n – 39n + 156 = 0

{∵ 156 × 3 = 468

∴ 468 = - 12 × - 39

- 51 = - 12 – 39}

⇒ 3n(n – 4) – 39(n – 4) = 0

⇒ (n – 4)(3n – 39) = 0

Either n – 4 = 0, then n = 4

or 3n – 39 = 0, then 3n = 39

⇒ n = 13

∴ n = 4 and 13

n_{4 }= a + (n – 1)d

= 24 + 3(-3)

= 24 – 9

= 15

n_{13} = 24 + 12(-3)

= 24 – 36

= - 12

∴ Sum of 5^{th} term to 13 term = 0

∵ 12 + 9 + 6 + 3 + 0 + (-3) + (-6) + (-9) + (-12) = 0

**9. Find the sum of first 22 terms, of an A.P. in which d = 7 and a _{22} is 149. **

**Answer**

Sum of first 22 terms of an A.P. whose d = 7

a_{22} = 149 and n = 22

149 = a + (n – 1)d

= a + 21 × 7

149 = a + 17

⇒ a = 149 – 147 = 2

∴ S_{22} = n/2[2a + (n – 1)d]

= 22/2 [2 × 2 + (22 – 1) (7)]

= 11[4 + 21 × 7]

= 11 × [4 + 147]

= 11 × 151

= 1661

**10. (i) Find the sum of first 51 terms of the A.P. whose second third terms are 14 and 18 respectively.**

**(ii) If the third term of an A.P. is 1 and 6 ^{th} term is – 11, find the sum of its first 32 terms. **

**Answer**

**(i)** Sum of first 51 terms of an A.P. in which

T_{2} = 14, T_{3} = 18

∴ d = T_{3} – T_{2} = 18 – 14

= 4

And a = T_{1}

= 14 – 4

= 10, n = 51

Now, S_{51} = n/2 [2a + (n – 1)d]

= 51/2 [2 × 10 + (51 – 1) × 4]

= 51/2 [20 + 50 × 4]

= 51/2 [20 + 200]

= 51/2 × 220

= 5610

**(ii)** T_{3} = 1, T_{6} = -11, n = 32

a + 2d = 1** …(i)**

a + 5d = - 11 **…(ii)**

Subtracting (i) and (ii),

- 3d = 12

⇒ d = 12/-3

= - 4

Substitute the value of d in eq. (i)

a + 2(-4) = 1

⇒ a – 8 = 1

a = 1 + 8 = 9

∴ a = 9, d = - 4

S_{32} = n/2[2a + (n - 1)d]

= 32/2 [2 × 9 + (32 – 1) × (-4)]

= 16[18 + 31 × (-4)]

= 16[18 – 124]

= 16 × (-106)

= - 1696

**11. If the sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of first 10 terms. **

**Answer**

S_{6} = 36

S_{16} = 256

S_{n} = n/2[2a + (n – 1)d]

∴ S_{6 }= 6/2[2a + (6 – 1)d] = 36

⇒ 3[2a + 5d] = 36

⇒ 2a + 5d = 12 **…(i)**

and S_{16 }= 16/2[2a + (16 – 1)d] = 256

8[2a + 15d] = 256

2a + 15d = 32 **…(ii)**

⇒ 2a + 5d = 12

2a + 15d = 32

Subtracting (i) from (ii),

- 10d = - 20

⇒ d = (- 20/- 10) = 2

Substitute the value of d in eq. (i),

2a + 5d = 12

⇒ 2a + 5 × 2 = 12

⇒ 2a + 10 = 12

⇒ 2a = 12 – 10 = 2

⇒ a = 2/2 = 1

∴ a = 1, d = 2

Now, S_{10} = n/2 [2a + (n – 1)d]

= 10/2 [2 × 1 + (10 – 1) × 2]

= 5[2 + 9 × 2]

= 5[2 + 18]

= 5 × 20

= 100

**12. Show that a _{1}, a_{2}, a_{3}, …. From an A.P. where a_{n }is defined as a_{n} = 3 + 4n. Also find the sum of first 15 terms. **

**Answer**

a_{n} = 3 + 4n

a_{1} = 3 + 4 × 1 = 3 + 4 = 7

a_{2} = 3 + 4 × 2 = 3 + 8 = 11

a_{3} = 3 + 4 × 3 = 3 + 12 = 15

a_{4} = 3 + 4 × 4 = 3 + 16 = 19

and so on here, a = 1 and d = 11 – 7 = 4

S_{15} = n/2[2a + (n – 1)d]

= 15/2 [2 × 7 + (15 – 1) × 4]

= 15/2 [14 + 14 × 4]

= 15/2 [14 + 56]

= 15/2 × 70

= 525

**13. (i) If a _{n} = 3 – 4n, show that a_{1}, a_{2}, a_{3}, …. form an A.P. Also find S_{20}.**

**(ii) Find the common difference of an A.P. whose first term is 5 and the sum of first four terms is half the sum of next four terms. **

**Answer**

**(i) **a_{n} = 3 - 4n

a_{1 }= 3 – 4 × 1 = 3 – 4 = - 1

a_{2} = 3 – 4 × 2 = 3 – 8 = - 5

a_{3} = 3 – 4 × 3 = 3 – 12 = - 9

a_{4} = 3 – 4 × 4 = 3 – 16 = - 13 and so on

Here, a = - 1,

d = - 5 – (-1)

= - 5 + 1

= - 4

Now, S_{20} = n/2[2a + (n – 1)d]

= 20/2[2 × (-1) + (20 – 1) × (-4)]

= 10[-2 + 19 × (-4)]

= 10[- 2 – 76]

= 10 × (-78)

= - 780

**(ii)** Let a and d be the first term and common difference of A.P. respectively.

Given, a = 5

a_{1} + a_{2} + a_{3} + a_{4} = 1/2(a_{5} + a_{6} + a_{7} + a_{8})

∴ a + (a + d) + (a + 2d) + (a + 3d) = 1/2[(a + 4d) + (a + 5d) + (a + 6d) + (a + 7d)]

⇒ 2(4a + 6d) = (4a + 22d)

⇒ 2(20 + 6d) = (20 + 22d) (∵ a = 5)

⇒ 40 + 12d = 20 + 22d

⇒ 10d = 20

⇒ d = 2

Thus, the common difference of A.P. is 2.

**Exercise 9.4 **

**1. Can 0 be a term of a geometric progression? **

**Answer**

No, 0 is not a term of geometric progression.

**2. (i) Find the next term of the list of numbers = 1/6, 1/3, 2/3, ….. **

**(ii) Find the next term of the list of numbers 3/16, - (3/8), 3/4, -(3/2), ……**

**(iii) Find the 15 ^{th} term of the series √3 + 1/√3 + 1/(3√3) + …..**

**(iv) Find the nth term of the list of numbers 1/√2, - 2, 4√2, - 16,……**

**(v) Find the 10 ^{th} and nth terms of the list of numbers 5, 25, 125, ……**

**(vi) Find the 6 ^{th} and nth terms of the list of numbers 3/2, 3/4, 3/8, …… **

**(vii) Find the 6 ^{th} term from the end of the list of numbers 3, - 6, 12, - 24, ……, 12288. **

**Answer**

**(i) **Given

⇒ 1/6, 1/3, 2/3, …..,

Here, a = 1/6,

r = 1/3 ÷ 1/6

= 1/3 × 6/1

= 2

∴ Next term = 2/3 × 2 = 4/3

**(ii)** 3/16, -(3/8), 3/4, -(3/2), ……

Here, a = 3/16,

r = -3/8 ÷ 3/16

= -3/8 × 16/3

= - 2

∴ Next term = -3/2 × (-2) = 3

**(iii)** √3 + 1/√3 + 1/3√3 + …….

Here, a = √3,

r = 1/√3 ÷ √3 = 1/√3 × 1/√3

= 1/3

∴ a_{15 }= ar^{n - 1}

= √3(1/3)^{15-1}

= √3 × (1/3)^{14}

= √3 × 1/3^{14}

**(iv)** 1/√2, - 2, 4√2, - 16, ……

Here, a = 1/√2, r = - 2 ÷ 1/√2 = - 2 × √2 = -2√2

a_{n} = ar^{n-1}

= 1/√2 × (-2√2)^{n-1}

= 1/√2 × (-1)^{n-1} × [(√2)^{2} × √2]^{n-1}

= (-1)^{n-1} × 1/√2 × [(√2)^{ 3}]^{n – 1}

= (-1)^{n-1 }× (1/√2) × (√2)^{3n - 3}

= (-1)^{n-1 } (√2)^{3n – 3 - 1}

= (-1)^{n -1} (√2)^{3n – 4}

= (-1)^{n – 1} × 2^{(3n – 4)/2}

**(v)** 5, 25, 125, …..,

Here, a = 5, r = 25 ÷ 5 = 5

a_{10} = ar^{n – 1 }

= 5 × (5)^{10-1}

= 5 × 59

= 5^{9 + 1}

= 5^{10 }

a_{n} = ar^{n-1}

= 5 ×5^{n-1}

= 5^{n – 1 + 1}

= 5^{n}

**(vi)** 3/2, 3/4, 3/8, …..

Here, a = 3/2, r = 3/4 ÷ 3/2

= 3/4 × 2/3

= 1/2

∴ a_{n} = ar ^{n-1}

= 3/2 × 1/2^{n-1}

= 3 × 1/2 × (1/2)^{n-1}

= 3 × (1/2)^{n -1 + 1}

= 3 × (1/2)^{n}

= 3/(2^{n})

a_{6} = 3/2^{n}

= 3/2^{6}

= 3/64

**(vii)** 3, -6, -12, - 24, …..12288

6^{th} term from the end of the list

Here, a = 3, r = - 6 ÷ 3

= - 2,

l = 12288

Now, 6^{th} term from the end

= l × (l/r)^{n-1}

= 12288 × (1/-2)^{6-1}

= 12288 × 1/(-2)^{5}

= 12288/(-32)

= - 384

**3. Which term of the G.P. **

**(i) 2, 2√2, 4, ……. is 128 ? **

**(ii) 1, 1/3, 1/9, …. Is 1/243 ? **

**(iii) 1/3, 1/9, 1/27, …… is 1/(19683) ? **

**Answer**

Given

**(i)** 2, 2√2, 4, ….. is 128 ?

Here a = 2, r = (2√2)/2 = √2, 1 = 128

Let 128 be the nth term, then

a_{n }= 128 = ar^{n-1}

⇒ 128 = 2(√2)^{n-1 }

⇒ 2(√2)^{n-1} = 2^{7}

(√2)^{n-1} = 2^{7 – 1} = 2^{6}

(√2)^{n-1} = (√2)^{12}

Comparing, we get

n – 1 = 12

⇒ n = 12 + 1 = 13

∴ 128 is the 13^{th} term

**(ii)** 1, 1/3, 1/9, …. is 1/243

Here, a = 1, r = 1/3 ÷ 1 = 1/3, l = 1/243

Let 1/243 is the nth term, then

a_{n} = 1/(243)

= ar ^{n-1}

= 1 × (1/3)^{n-1}

⇒ (1/3)^{n-1} = (1/3)^{5}

Comparing, we get

n – 1 = 5

⇒ n = 5 + 1 = 6

∴ 1/(243) is the 6^{th} term

**(iii)** 1/3, 1/9, 1/27, …. is 1/(19683)

Here, a = 1/3, r = 1/9 ÷ 1/3

= 1/9 × 3/1

= 1/3,

l = 1/(19683)

Let 1/(19683) is the nth term, then

a_{n} = 1/(19683)

= ar ^{n-1}

= 1/3(1/3)^{n-1}

= (1/3) ^{n – 1 + 1}

= (1/3)^{n}

⇒ (1/3)^{n} = (1/3)^{9 }

Comparing, we get

n = 9

Hence, 1/(19683) is the 9^{th} term

**4. Which term of the G.P. 3, -3√3, 9, - 9√3, …… is 729 ? **

**Answer**

G.P. 3, -3√3, 9, - 9√3, …… is 729?

Here a = 3, r = (-3√3)/3

Comparing, we get

n - 1 = 10

⇒ n = 10 + 1 = 11

∴ 729 is the 11^{th} term

**5. Determine the 12 ^{th} term of a G.P.
whose 8^{th} term is 192 and common ratio is 2. **

**Answer**

In a G.P.

a_{8 }= 192 and r = 2

Let a be the first term and r be
the common ratio then.

a_{8} = ar^{n-1}

⇒ 192 =
a(2)^{n-1}

= a2^{8-1}

= a2^{7}

⇒ a =
(192/2^{7}) = 192/128 = 3/2

∴ a_{12}
= a(r)^{n-1}

⇒ a_{12}
= 3/2(2)^{12-1 }

= 3/2 × 2^{11}

= 3/2 × 2048

= 3072

∴ a_{12} = 3072

**6. In a G.P., the third term is 24 and 6 ^{th}
term is 192. Find the 10^{th} term. **

**Answer**

In a G.P.

a_{3} = 24 and a_{6}
= 192, a_{10} = ?

Let a be the first term and r be
the common ratio, therefore

a_{6} = ar ^{6-1}

= ar^{5 }= 192 **{****∵**** a _{n}
= ar^{n-1}}**

a_{3} = ar^{3-1} =
ar^{2} = 24

Dividing, we get

ar^{5}/ar^{2} =
192/24

⇒ r^{3
}= 8 = (2)^{3}

∴ r = 2

Now, ar^{2} = 24

⇒ a × 2^{2
}= 24

⇒ a =
24/2^{2 }= 24/4 = 6

∴ a = 6

Now, a_{10} = ar^{10-1} = ar^{9}

= 6 × (2)^{9}

= 6 × 512

= 3072

**7. Find the number of terms of a G.P. whose
first term is 3/4, common ratio is 2 and
he last term is 384. **

**Answer: **

First term of a G.P. = 3/4

And common ratio (r) = 2

Last term = 384

Let number of terms is n then a_{n}
= ar^{n-1}

⇒ 384 =
3/4(2)^{n – 1}

⇒ 2^{n-1}
= (384 × 4)/3 = 512

= 2^{9 }

∴ n – 1 = 9

⇒ n = 9
+ 1 = 10

∴ Number
of terms in G.P. = 10

**8. Find the value of x such that **

**(i) –2/7, x, -7/2 are three consecutive terms of a G.P. **

**(ii) x + 9, x – 6 and 4 are three consecutive terms of a G.P.
**

**(iii) x, x + 3, x + 9 are first three terms of a G.P. Find the
value of x.**

**Answer**

Find the value of x

**(i)**–2/7, x, -7/2 are three consecutive terms of a G.P.

∴ x

^{2}= -2/7 × -7/2

= 1

= (±1)

^{2}

∴ x = ± 1

Hence, x = 1 or -1

**(ii)**x + 9, x – 6 and 4 are three consecutive terms of a G.P., then

(x – 6)

^{2}= (x + 9) × 4

⇒ x^{2}
– 12x + 36 = 4x + 36

⇒ x^{2}
– 12x – 4x + 36 – 36 = 0

⇒ x^{2}
– 16x = 0

⇒ x(x –
16) = 0

Either x – 16 = 0, then x = 16

Or, x = 0

∴ x = 0, 16

**(iii)** x, x + 3, x + 9 are first
three terms of a G.P.

∴ (x + 3)^{2} = x(x + 9)

⇒ x^{2} + 6x + 9 = x^{2} + 9x

⇒ 9 = 9x – 6x = 3x

∴ x = 9/3 = 3

**9. If the fourth, seventh and tenth terms of a G.P.
are x, y, z respectively, prove that x, y, z are in G.P. **

**Answer**

In a G.P.

a_{4} = x, a_{7} =
y, a_{10} = z

To prove : x, y, z are in G.P.

Let a be the first term and r be
the common ratio, therefore,

a_{4} = ar ^{n-1}

= ar ^{4-1}

= ar^{3}

= x

Similarly,

d_{7} = ar^{6} = y

a_{10 }= ar^{9 }= z

If x, y and z are in G.P.., then

y^{2} = xz

Now, xz = ar^{3} × ar^{9}

= a^{2}r^{3+9}

= a^{2}r^{12}

y^{2 }= (ar^{6})^{2}
= a^{2}r^{12}

∵ L.H.S = R.H.S

∴ x, y and z are in G.P.

**10. The 5 ^{th}, 8^{th} and 11^{th} terms of a G.P. are p, q and s respectively. Show that q^{2} = ps.**

**Answer**

In a G.P.

a_{5} = p, a_{8} = q and a_{11} = s

To show that q^{2} = px

Let a be the first term and r be the common

a_{5} = ar^{n-1}

= ar^{5-1}

= ar^{4}

= p

Similarly, a_{8} = ar^{7} = q and

a_{11} = ar^{10} = s

q^{2} = (ar^{7})^{2} = ar^{14 }

and px = ar^{4} × ar^{10}

= a^{2}r ^{4+ 10}

= a^{2}r^{14 }

Hence, q^{2} = ps

**11. If a, b, c are in G.P., then show that a ^{2}, b^{2}, c^{2} are also in G.P. **

**Answer**

a, b, c are in G.P.

Show that a^{2}, b^{2}, c^{2} are also in G.P.

∵ a, b, c are in G.P.., then

b^{2} = ac **…(i)**

a^{2}, b^{2}, c^{2} will be in G.P.

if (b^{2})^{2} = a^{2} × c^{2 }

⇒ (ac)^{2 }= a^{2}c^{2 }** [From (i)]**

⇒ a^{2}c^{2} = a^{2}c^{2} which is true.

Hence, proved.

**12. If a, b, c are in A.P.., then show that 3 ^{a}, 3^{b}, 3^{c }are in G.P. **

**Answer**

a, b and c are in A.P.

then, 2b = a + c

Now, 3^{a}, 3^{b}, 3^{c} will be in G.P.

if (3^{b})^{2} = 3^{a}.3^{c}

if 3^{2b} = 3^{a+ c}

Comparing, we get

if 2b = a + c

Which are in A.P. is given

**13. If a, b, c are in A.P., then show that 10 ^{ax + 10}, 10^{bx + 10} , 10^{cx + 10}, x ≠ 0, are in G.P.**

**Answer**

a, b, c are in A.P.

To show that are in G.P. 10^{ax + 10}, 10^{bx + 10}, 10^{cx + 10}, x ≠ 0

∵ a, b, c are A.P.

∵ 2b = a + c **…(i)**

Now,

(10^{ax + 10}), (10^{bx + 10}), (10^{cx + 10}) will be in G.P.

If (10^{bx + 10})^{2} = (10^{ax + 10}) × (10^{cx + 10})

If 10^{2bx + 20} = 10^{ax+10+cx+10 }

If 10^{2bx + 20} = 10^{ax+cx+20}

Comparing,

If 2bx + 20 = ax + cx + 20

If 2bx = ax + cx

If 2b = a + c

Which is given.

**14. If a, a ^{2 }+ 2 and a^{3} + 10 are in G.P., then find the values(s) of a. **

**Answer**

a, a^{2 }+ 2 and a^{3} + 10 are in G.P.

∵ (a^{2} + 2)^{2} = a(a^{3} + 10)

⇒ a^{4 }+ 4a^{2} + 4 = a^{4} + 10a

⇒ 4a^{2 }– 10a + 4 = 0

⇒ 2a^{2} – 5a + 2 = 0

⇒ 2a^{2} – a – 4a + 2 = 0

⇒ a(2a – 1) – 2(2a – 1) = 0

⇒ (2a – 1) (a – 2) = 0

Either 2a – 1 = 0, then 2a = 1

⇒ a = 1/2

Or a – 2 = 0, then a = 2

Hence a = 2 or 1/2

**15. The first and the second terms of a G.P. are x ^{-4} and x^{m}. If its 8^{th} term is x^{52}. Then find the value of m. **

**Answer**

In a G.P.,

First term (a_{1}) = x^{-4} **…(i)**

Second term (a_{2}) = x^{m}

Eighth term = (a_{8}) = x^{52}

r = a_{2}/a_{1} = x^{m}/x^{-4} = x^{m + 4} **…..(ii)**

Now, a_{8} = ar^{n-1} = ar ^{8-1} = ar^{7}

x^{52} = x^{-4} × r^{7} = x^{-4} × x^{7(m+4)}

x^{52} = x^{-4 + 7m + 28} = x^{7m + 24}

Comparing,

52 = 7m + 24

⇒ 7m = 52 – 24 = 28

⇒ m = 28/7 = 4

Hence m = 4

**16. Find the geometric progression whose 4 ^{th} term is 54 and the 7^{th}. Term is 1458. **

**Answer**

In a G.P.

4^{th} term (a_{4}) = 54

And 7^{th} term (a_{7}) = 1458

Let a be the first term and r be the common ratio, then

ar^{3} = 54 and ar^{6} = 1458

Dividing,

(ar^{6}/ar^{3}) = 1458/54

⇒ r^{3} = 27 = (3^{3})

∴ r = 3

Now, ar^{3} = 54

⇒ a × 27 = 54

⇒ a = 54/27 = 2

Hence G.P. is 2, 6, 18, 54,…..

**17. The fourth term of G.P. is the square of its second term and the first term is -3. Determine its seventh term. **

**Answer**

In a G.P.

a_{n} is square of a_{2} i.e., an = (a_{2})^{2 }

a_{1} = - 3

Let a be the first term and r be the common ratio, then

a_{4} = ar ^{n-1}

= ar^{4-1}

= ar^{3}

and a_{2} = ar^{2-1}

= ar

∴ ar^{3} = (ar)^{2}

⇒ ar^{3} = a^{2}r^{2 }

r^{3}/r^{2} = a^{2}/a = r = a = - 3 (∵ a_{1} = - 3)

Now, a_{7} = ar ^{7 -1}

= ar^{6}

= (-3)(-3)^{6 }

= (-3)^{7}

= -2187

**18. The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms. **

**Answer**

Sum of first three terms of G.P. = 39/10

And their product = 1

Let a be the first term and r be the common ratio, then

Let a/r, a, ar be the three terms of G.P., then

a/r + a + ar = 39/10 and a/r × a × ar = 1

⇒ a(1/r + 1 + r) = 39/10 and a^{3} = 1

⇒ a = 1

Substitute the value of a

∴ 1(1/r + 1 + r) = 39/10

(1 + r + r^{2})/r = 39/10

10 + 10r + 10r^{2} = 39r

⇒ 10r^{2} + 10r – 39r + 10 = 0

⇒ 10r^{2} – 29r + 10 = 0

**{∵ 10 × 10 = 100, ∴ 100 = - 25 × (-4), - 29 = - 25 – 4 }**

⇒ 10r^{2} – 25r – 4r + 10 = 0

⇒ 5r(2r – 5) – 2(2r – 5) = 0

⇒ (2r – 5)(5r – 2) = 0

Either 2r – 5 = 0, then r = 5/2

Or 5r – 2 = 0, then r = 2/5

Hence r = 5/2 or 2/5

And terms will be if r = 5/2

1, 5/2, 25/4, 125/8, ……..

If r = 2/5, then terms will be 1, 2/5, 4/25, 8/125, …….

**19. Three numbers are in A.P. and their sum is 15. If 1, 4 and 19 are added to these numbers respectively, the resulting numbers are in G.P. Find the numbers.**

**Answer**

Given : Three numbers are in A.P. and their sum = 15

Let a – d, a, a + d be the three numbers in A.P.

∴ a – d + a + a + d = 15

⇒ 3a = 15

⇒ a = 15/3 = 5

By adding, 1, 4, 19 in then,

We get

a – d + 1, a + 4, a + d + 19

These are in G.P.

∴ b^{2} = ac

∴ (a + 4)^{2 }= (a – d + 1) (a + d + 19)

⇒ a^{2 }+ 8a + 16 = a^{2} + ad + 19a – ad – d^{2} – 19d + a + d + 19

⇒ a^{2 }+ 8a + 16 = a^{2 }– d^{2} – 18d + 20a + 19

⇒ 8a + 16 = 20a – 18d – d^{2} + 19

⇒ 8a + 16 – 20a + 18d + d^{2 }– 19 = 0

⇒ d^{2 }+ 18d – 12a – 3 = 0

⇒ d^{2} + 18d – 12 × 5 – 3 = 0

⇒ d^{2} + 18d – 60 – 3 = 0

⇒ d^{2} + 18d – 63 = 0

**{∵ - 63 = + 21 × -3 + 18 = + 21 – 3}**

⇒ d^{2} + 21d – 3d – 63 = 0

⇒ d(d + 21) – 3(d + 21) = 0

⇒ (d + 21)(d – 3) = 0

Either d + 21 = 0, then d = - 21

Or d – 3 = 0, then d = +3

If d = 3 and a = 5, then G.P.

5 - 3, 5, 5 + 3 i.e., 2, 5, 8

If d = - 21, then

5 + 21, 5, 5 – 21

⇒ 26, 5, - 16

**20. three numbers form an increasing G.P. If the middle term is doubled, then the new numbers are in A.P. Find the common ration of the G.P.**

**Answer**

Three numbers form an increasing G.P.

Let a/r, a, ar, be three numbers in G.P.

Double the middle term, we get

a/r, 2a, ar will be in A.P.

If 2(2a) = a/r + ar

If 4a = a(1/r + r)

If 4 = 1/r + r

⇒ 4r = 1 + r^{2}

⇒ r^{2 }– 4r + 1 = 0

= (2 ± √3)

∴ r = 2 *± √3 *

∵ The numbers are increasing.

∴ r = 2 + √3

**21. Three numbers whose sum is 70 are in G.P. If each of the extremes is multiplied by 4 and the mean by 5, the numbers will be in A.P. Find the numbers. **

**Answer**

Three numbers are in G.P.

Let numbers be r/a, a, ar

∴ a/r + a + ar = 70

⇒ a(1/r + 1 + r) = 70 **…(i)**

By multiplying the extremes by 4 and mean by 5, then

a/r × 4, a × 5, ar × 4

4a/r, 5a, 4ar

But these are in A.P.

∴ 2(5a) = 4a/r + 4ar

⇒ 10a = 4a(1/r + r)

⇒ 5 = 2(1/r + r)

⇒ 5r = 2 + 2r^{2 }

⇒ 2r^{2} – 5r + 2 = 0

⇒ 2r^{2} – r – 4r + 2 = 0

**{ ∵ 2 × 2 = 4, ∴ 4 = - 1 × (-4), - 5 = - 1 – 4}**

⇒ r(2r – 1) – 2(2r – 1) = 0

⇒ (2r – 1)(r – 2) = 0

Either 2r – 1 = 0, then r = 1/2

Or r – 2 = 0, then r = 2

From (i),

a(1/2 + 1 + 2) = 70

(7/2)a = 70

⇒ a = 70 × 2/7 = 20

∴ Numbers are (if r = 2)

20/2, 20, 20 × 2 ⇒ 10, 20, 40

If r = 1/2, then

(20/1)/2, 20, 20 × 1/2

⇒ (20 × 2)/1, 20, 20 × 1/2

⇒ 40, 20, 10

**22.(i) If a, b, c are in A.P. as well in G.P., prove that a = b = c. **

**(ii) If a, b, c are in A.P. as well as in G.P., then find the value of a ^{b-c} + b^{c-a} + c^{a-b}**

**Answer**

**(i)** a, b, c are in A.P. as well as in G.P.

To prove : a = b = c

a, b, c are in A.P.

∴ 2b = a + c

⇒ b = (a + c)/2 **...(i)**

∴ a, b, c are in G.P.

∴ b^{2} = ac

⇒ {(a + c)/2}^{2 }= ac

⇒ {(a + c)^{2}}/4 = ac

⇒ (a + c)^{2} = 4ac

⇒ (a + c)^{2 }– 4ac = 0

⇒ (a – c)^{2 }= 0

⇒ a – c = 0

⇒ a = c **…(iii)**

From (i), 2b = a + c = a + a = 2a

∴ b = a **...(iv)**

From (iii) and (iv),

Hence a = b = c

**(ii)** a, b, c are in A.P. as well as in G.P.

∴ 2b = a + c

and b^{2 }= ac

and a = b = c **[Proved in (i)]**

Now, a^{b-c} + b^{c-a} + c^{a – b}

Since, a = b = c

∴ b – c = 0, c – a = 0 and a – b = 0

∴ a^{0} + b^{0 }+ c^{0} = 1 + 1 + 1 (∵ x^{0} = 1)

= 3

**23. The terms of a G.P. with first term a and common ratio r are squared. Prove that resulting numbers form a G.P. Find its first term, common ratio and the nth term. **

**Answer**

In a G.P.,

The first term = a

And common ratio = r

G.P. is a, ar, ar^{2}

Squaring we get

a^{2}, a^{2}r^{2}, a^{2}r^{4} are in G.P.

if b^{2} = 4ac

⇒ (a^{2}r^{2})^{2} = a^{2} × a^{2}r^{4 }

⇒ a^{4}r^{4} = a^{4}r^{4}

Which is true

The first term is a^{2 }

And common ratio is r^{2}

The nth term will be

a_{n} = ar^{n-1}

= a^{2}(r^{n-1})^{2}

= a^{2}r^{2n-2}

^{}

**24. Show that the products of the corresponding terms of two G.P,’s a, ar, ar ^{2}, ……, ar^{n-1} and A, AR, AR^{2},……AR ^{n-1} form a G.P. and find the common ratio. **

**Answer**

It has to be proved that the sequence

aA, arAR, ar^{2}AR^{2}, ….. ar^{n-1} AR^{n-1} and forms a G.P.

Hence, (Second term)/(First term) = (arAR)/(aA) = rR

and (Third term/Second term) = (ar^{2} AR^{2})/(arAR) = rR

thus, the above sequence forms a G.P. and the common ratio is rR.

**25. (i) If a, b, c are in G.P. show that 1/a, 1/b, 1/c are also in G.P. **

**(ii) If K is any positive real number and K ^{a}, K^{b}, K^{c} are three consecutive terms of a G.P., prove that a, b, c are three consecutive terms of an A.P. **

**(iii) If p, q, r are in A.P.., show that pth, qth and rth terms of any G.P. are themselves in G.P.**

**Answer**

**(i)** a, b, c are in G.P.

∴ b^{2} = ac

1/a, 1/b, 1/c will be in G.P.

If (1/b)^{2} = 1/a × 1/c

⇒ 1/b^{2} = 1/ac

⇒ ac = b^{2} (By cross multiplication)

Which is given

Hence proved.

**(ii)** k is any positive number K^{a}, K^{b}, K^{c }are in G.P.

Then (K^{b})^{2} = K^{a} × K^{c}

⇒ K^{2b} = K^{a + c}

⇒ 2b = a + c

Hence, a, b, c are in A.P.

**(iii)** p, q, r are in A.P.

∴ 2q = p + r

pth term in G.P. = AR ^{p-1}

qth term = AR ^{q-1}

rth term = AR^{r-1}

These will be in G.P.

If (AR^{q-1})^{2} = AR^{p-1} × AR^{r-1}

If A^{2}R^{2q-2} = A^{2}R^{p-1+r -1}

If A^{2}R^{2q-2} = A^{2}R^{p+r-2}

If R^{2q-2} = R^{p+r-2}

Comparing, we get

2q – 2 = p + r – 2

⇒ 2q = p + r

⇒ p, q, r are in A.P. which is given

Hence, proved.

**26. If a, b, c are in G.P., prove that the following are also in G.P. **

**(i) a ^{3}, b^{3}, c^{3} **

**(ii) a ^{2} + b^{2}, ab + bc, b^{2} + c^{2}**

**Answer**

**(i)** a, b, c are in G.P.

∴ b^{2 }= ac

a^{3}, b^{3}, c^{3} are in G.P.

If (b^{3})^{2 }= a^{3} × c^{3}

If (b^{2})^{3} = (a × c)^{3}

If b^{2} = ac

Which is given.

Hence proved.

**(ii)** a^{2} + b^{2}, ab + bc, b^{2} + c^{2} will be in G.P.

a^{2} + b^{2}, ab + bc, b^{2} + c^{2} are in G.P.

If (ab + bc)/(a^{2 }+ b^{2}) = (b^{2} + c^{2})/(ab + bc)

i.e., if {a(ar) + ar(ar^{2})}/{a^{2 }+ (ar)^{2}} = {(ar)^{2} + (ar^{2})^{2} }/{a(ar) + ar(ar^{2})}

⇒ If {a^{2}r(1 + r^{2})/a^{2}(1 + r^{2})} = a^{2}r^{2 }(1 + r^{2})}/{a^{2}r(1 + r^{2})}

⇒ If r = r which is true.

∴ a^{2} + b^{2}, ab + bc, b^{2} + c^{2} are in G.P.

**27. If a, b, c, d are in G.P., show that **

**(i) a ^{2} + b^{2}, b^{2} + c^{2}, c^{2} + d^{2} are in G.P. **

**(ii) (b – c) ^{2 }+ (c – a)^{2} + (d – b)^{2 }= (a – d)^{2 }**

**Answer**

a, b, c, d are in G.P.

Let r be the common ratio, then a = a

b = ar, c = ar^{2}, d = ar^{3}

**(i)** a^{2 }+ b^{2}, b^{2 }+ c^{2}, c^{2} + d^{2} are in G.P.

∴ a^{2 }+ b^{2} = a^{2 }+ a^{2}r^{2} = a^{2}(1 + r^{2})

b^{2} + c^{2} = a^{2}r^{2} + a^{2}r^{4} = a^{2}r^{2}(1 + r^{2})

c^{2 }+ d^{2} = a^{2}r^{4 }+ a^{2}r^{6 }= a^{2}r^{4}(1 + r^{2})

a^{2} + b^{2}, b^{2} + c^{2}, c^{2} + d^{2} will be in G.P.

if (b^{2} + c^{2})^{2} = (a^{2} + b^{2})(c^{2} + d^{2})

Now, (b^{2} + c^{2})^{2} = [a^{2}r^{2}(1 + r^{2})]^{2}

= a^{4}r^{4}(1 + r^{2})^{2} **…(i)**

(a^{2 }+ b^{2})(c^{2} + d^{2}) = [a^{2}(1 + r^{2})][a^{2}r^{4}(1 + r^{2})]^{2}

= a^{4}r^{4}(1 + r^{2})^{2 }** …(ii)**

From (i) and (ii),

(b^{2 }+ c^{2})^{2} = (a^{2} + b^{2})(c^{2} + d^{2})

Hence, a^{2} + b^{2}, b^{2} + c^{2}, c^{2} + d^{2} are in G.P.

**(ii)** Show that

(b – c)^{2} + (c – a)^{2} + (d – b)^{2} = (a – d)^{2}

L.H.S. = (b – c)^{2} + (c – a)^{2} + (d – b)^{2}

= (ar – ar^{2})^{2} + (ar^{2 }– a)^{2} + (ar^{3} – ar)^{2}

= a^{2}r^{2}(1 – r)^{2} + a^{2}(r^{2 }– 1)^{2} + a^{2}r^{2}(r^{2} – 1)^{2 }

= a^{2}[r^{2}(1 – r^{2} – 2ar) + r^{4} – 2r^{2} + 1 + r^{2}(r^{4} – 2r^{2} + 1)]

= a^{2}[r^{2 }– r^{4 }– 2ar^{3} + r^{4} – 2r^{2} + 1 + r^{6} – 2r^{4} + r^{2})]

= a^{2}(r^{6 }– 2r^{3} + 1)

R.H.S. = (a – d)^{2} = (a – ar^{3})^{2 }

= a^{2}(1 – r^{3})^{2}

= a^{2}(1 + r^{6} – 2r^{2})

= a^{2}[r^{6 }– 2r^{3} + 1]

∴ L.H.S. = R.H.S

**Exercise 9.5 **

**1. Find the sum of : **

**(i) 20 terms of the series 2 + 6 + 18 + …….**

**(ii) 10 terms of series 1 + √3 + 3 + ……**

**(iii) 6 terms of the G.P. 1, - 2/3, 4/9, .……**

**(iv) 5 terms and n terms of the series 1 + 2/3 + 4/9 + ……..**

**(v) n terms of the G.P. √7, ****, 3√7, ………**

**(vi) n terms of the G.P. 1, - a, a ^{2}, - a^{3}, (a ≠ - 1) **

**(vii) n terms of the G.P. x ^{3}, x^{5}, x^{7}, ……(x ≠ **

**±1)****Answer**

**(i) **2 + 6 + 18 + … 20 terms

Here, a = 2, r = 3, n = 20, r > 1

S_{20 }= a(r^{n} – 1)/(r – 1)

= 2(3^{20} – 1)/(3 – 1)

= {2(3^{20} – 1)}/2

= 3^{20} – 1

**(ii)** 1 + √3 + 3 + … 10 terms

Here, a = 1, r = √3, n = 10, r > 1

S_{10} = a(r^{n} – 1)/(r – 1)

= 1[(√3)^{10} – 1]/(√3 – 1)

= {(√3^{10} – 1)(√3 + 1)}/{(√3 – 1)(√3 + 1**)}**

**(Rationalising the denominator) **

= {(3^{5} – 1)(√3 + 1)}/(3 – 1)

= {(243 – 1)(√3 + 1)}/2

= {242(√3 + 1)}/2

= 121(√3 + 1)

**(iii)** 1, -(2/3), 4/9, ….. 6 terms

Here, a = 1, r = (-2)/3, n = 6, r < 1

S_{6} = a(1 – r^{n})/(1 – r) = 1[1 – (-2/3)^{6}]/(1 + 2/3)

= 3/5 {1 – (-2)^{6}/3^{6}}

= 3/5[1 – 64/729]

= 3/5[(729 – 64)/729]

= 3/5 × 665/729

= 133/243

**(iv) **1 + 2/3 + 4/9 + …. n terms, 5 terms

Here, a = 1, r = 2/3 ÷ 1 = 2/3, n = 5, n, (r < 1)

S_{n} = a(1 – r^{n})/(1 – r) = 1[1 – (2/3)^{n}]/(1 – 2/3)

= (1 × 3)/1 [1 – (2/3)^{n}]

= 3[1 – (2/3)^{n}]

and S_{5} = 3[1 – (2/3)^{5}]

= 3[1 – 32/243]

= 3[(243 – 32)/243]

= 211/81

**(v)** √7, √21, 3√7, ……

Here, a = √7, r = /√7

= (√7 × √3)/√7

= √3

r > 1, n = n terms

S_{n} = a(r^{n} – 1)/(r – 1)

⇒ S_{n }= √7[(√3)^{n} – 1]/(√3 – 1)

Rationalising the denominator,

⇒ S_{n} = √7[(√3)^{n }– 1]/(√3 – 1) × {(√3 + 1)/(√3 + 1)}

= {√7[(√3)^{n} – 1](√3 + 1)}/{(√3)^{2} – (1)^{2}}

= {√7[(√3)^{n }– 1](√3 + 1)}/(3 – 1)

= √7/2 [(√3)^{n} – 1] (√3 + 1)

**(vi)** 1, -a, a^{2}, -a^{3}, ….(a ≠ -1) upto n terms

Here, a = 1, r = - a,

S_{n} = a(1 – r^{n})/(1 – r)

= 1[1 – (-a)^{n}]/[1 – (-a)]

= [1 – (-a)^{n}]/(1 + a)

**(vii)** x^{3}, x^{5}, x^{7}, …..(x ≠ ±1)

Here, a = x^{3}, r = x^{2}

∴ S_{n} = a(1 – r^{n})/(1 – r)

= x^{3}[1 – (x^{2})^{n}]/(1 – x^{2}) if r < 1

= x^{3}(1 – x^{2n})/(1 – x^{2})

or S_{n} = a(r^{n }– 1)/(1 – r)

= x^{3}[(x^{2})^{n} – 1]/(x^{2} – 1)

= x^{3}(x^{2n} – 1)/(x^{2} – 1)

**2. Find the sum of the first 10 terms of the geometric series √2 + √6 +**** ****√18 ****+ ….**

**Answer**

Here, a = √2, r = √3, r > 1

∴ S_{10} = a(r^{n} – 1)/(r – 1)

= √2[(√3)^{10} – 1]/(√3 – 1)

= √2/(√3 – 1)[(3)^{5} – 1]

= √2/(√3 – 1)(243 – 1)

= {√2/(√3 – 1)} × 242

= {√2(√3 + 1) × 242}/(√3 – 1)(√3 + 1)

(Rationalising the denominator)

= {242(√6 + √2)}/(3 – 1)

= {121(√6 + √2)}/2

= 121(√6 + √2)

**3. Find the sum of the series 81 – 27 + 9 ……. – (1/27)**

**Answer**

Given :

81 – 27 + 9 ……. – (1/27)

Here, a = 81, r = (-27)/81 = (-1)/3, l = (-1) /27, r < 1

S_{n} = (a – lr)/(1 – r)

= (81 + 1/27 × -1/3)/(1 + 1/3)

= (81 – 1/81)/(4/3)

= (6561 – 1)/(81 × 4/3)

= (6560 × 3)/(81 × 4)

= 1640/27

**4. The nth term of a G.P. is 128 and the sum of its n terms is 255. If its common ratio is 2, then find its first term. **

**Answer**

In a G.P.

T_{n} = 128

S_{n} = 255

r = 2,

Let a be the first term, then

T_{n} = ar ^{n-1}

⇒ 128 = a2^{n-1}

⇒ a = 128/(2^{n-1}) **…(i)**

S_{n} = a(r^{n} – 1)/(r – 1)

⇒ 255 = a(2^{n} – 1)/(2 – 1)

⇒ 255 = a(2^{n} – 1)

⇒ a = 255/(2^{n }– 1) **….(ii)**

From (i) and (ii),

255/(2^{n }– 1) = 128/(2^{n} ^{– 1})

⇒ 255 × 2^{n-1}

= 128(2^{n} – 1)

⇒ 255 × 2^{n-1} = 128 × 2^{n} – 128

(255 × 2^{n})/2 = 128 × 2^{n }– 128

⇒ 255 × 2^{n} = 256 × 2^{n} – 256

⇒ 256 × 2^{n} – 255 × 2^{n} = 256

⇒ 2^{n} = 256 = 28

Comparing, we get

n = 8

Now, 128 = a.2^{7 }

⇒ 128 = a × 128

⇒ a = 128/128

= 1

∴ a = 1

**5. If the sum of first six terms of any G.P. is equal to 9 times the sum of the first three terms, then find the common ratio of the G.P. **

**Answer**

Sum of first 6 terms of a G.P. = 9 × The of first 3 terms

Let a be the first term and r be the common ratio

∴ S_{6} = 9 × a_{3 }

⇒ S_{n} = a(r^{n} – 1)/(r – 1)

S_{6} = a(r^{6 }– 1)/(r – 1) and S_{3} = a(r^{3 }– 1)/(r – 1)

∴ a(r^{6} – 1)/(r – 1) = 9 × a(r^{3} – 1)/(r – 1)

⇒ r^{6} – 1 = 9(r^{3 }– 1)

⇒ (r^{6} – 1)/(r^{3 }– 1) = 9

⇒ (r^{3} + 1)(r^{3} – 1)}/(r^{3} – 1) = 9

⇒ r^{3 }+ 1 = 9

⇒ r^{3} = 9 – 1 = 8 = (2)^{2}

∴ r = 2

∴ Common ratio = 2

**6. (i) How many terms of the G.P. 3, 3 ^{2}, 3^{3}, …… are needed to give the sum 120? **

**(ii) How many terms of the G.P. 1, 4, 16, must be taken to have their sum equal to 341 ? **

**Answer**

In G.P.

**(i)** 3, 3^{2}, 3^{3}, ….

Sum = 120, Here, a = 3, r = 3^{2}/3 = 3, r > 1

Let number of terms in G.P. be n, then

S_{n} = a(r^{n} – 1)/(r – 1) = 120

⇒ 3(3^{n} – 1)/(3 – 1) = 120

⇒ 3(3^{n} – 1)/2 = 120

⇒ 3^{n} – 1 = (120 × 2)/3 = 80

3^{n} = 80 + 1 = 81 = 3^{4 }

∴ n = 4

∴ Number of terms = 4

**(ii)** G.P. is 1, 4, 16, ….

Sum = 341

Here, a = 1, r = 4/1 = 4, r > 1

Let number of terms be n, then

S_{n} = a(r^{n} – 1)/(r – 1) = 341

⇒ 1(4^{n} – 1)/(4 – 1) = 341

⇒ 1(4^{n} – 1)/(4 – 1) = 341

⇒ (4^{n }– 1)/3 = 341

⇒ 4^{n }– 1 = 341 × 3 = 1023

4^{n }= 1023 + 1

= 1024

= 2^{10}

= 4^{5 }

∴ Number of terms = 5

**7. How many terms of the G.P. 1, √2 > 2, 2√2, ….. are required to give a sum of 1023 (√2 + 1) ? **

**Answer**

G.P. 1, √2 > 2, 2√2, …..

Sum = 1023 (√2 + 1)

Here, a = 1, r = √2, r > 1

Let number of terms be n, then

S_{n} = a(r^{n} – 1)/(r – 1) = 1023(√2 + 1)

⇒ 1[(√2)^{n} – 1]/(√2 – 1) = 1023(√2 + 1)

⇒ (√2)^{n }- 1 = 1023(√2 + 1)(√2 – 1)

⇒ (√2)^{n} – 1 = 1023[(√2)^{2} – (1)^{2}]

⇒ (√2)^{n} – 1 = 1023(2 – 1) = 1023

(√2)^{n} = 1023 + 1

= 1024

= 2^{10} or (√2)^{20}

n = 20

**8. How many terms of the 2/9 – 1/3 + 1/2 + ….. will make the sum 55/72 ? **

**Answer**

G.P. is 2/9 – 1/3 + 1/2 + ……

Sum 55/72

Here, a = 2/9, r = -1/3 × 9/2 = -3/2, r < 1

Let number of terms be n

∴ S_{n} = a(1 – r^{n})/(1 – r) = 55/72

⇒ 2/9 [1 – (-3/2)^{n}]/(1 + 3/2) = 55/72

⇒ 2/9[1 – (-3/2)^{n}]/(5/2) = 55/72

⇒ 1 – (-3/2)^{n} = 55/72 × 5/2 × 9/2 = 275/32

⇒ 1 – (-1)^{n} (3/2)^{n} = 275/32

⇒ 1 + 1(3/2)^{n} = 275/32

⇒ (3/2)^{n} = (275/32 – 1)

= (275 – 32)/32 = 243/32

= (3/5)^{5}

Comparing, we get

n = 5

∴ Number of terms = 5

**9. The 2 ^{nd} and 5^{th} terms of a geometric series are – (1/2) and sum 1/16 respectively. Find the sum of the series upto 8 terms. **

**Answer**

In a G.P.

a_{2} = -(1/2) and a_{5 }= 1/16

Let a be the first term and r be the common ratio

∴ a^{2} = ar^{n- 1}

= ar ^{2-1}

ar = -1/2 **….(i)**

a_{5} = ar^{5-1}

= ar^{4} = 1/16 **...(ii)**

Dividing (ii) by (i),

r^{3} = 1/16 ÷ (-1)/2

= 1/16 × -2/1

= -1/8

= (-1)^{3}/2

∴ r = (-1/2)

and ar = -1/2

⇒ a × (-1)/2 = -1/2

⇒ a = -1/2 × -2/1 = 1

a = 1, and r = -1/2

Now, S_{8} = a(1 – r^{n})/(1 – r)

= 1[1 – (-1/2)^{8}]/(1 + 1/2)

= (1 – 1/256)/(3/2)

= 255/256 × 2/3

= 510/768

= 85/128

**10. The first term of G.P. is 27 and 8 ^{th} term is 1/81. Find the sum of its first 10 terms. **

**Answer**

From the question it is given that,

First term (a) = 27

8^{th} term a_{8 }= 1/81

Then, a_{n }= ar^{n-1}

a_{8} = ar^{8-1} = 1/81

ar^{n-1} = ar^{8 -1}

= ar^{7}

= 1/81

ar^{7} = 1/81

⇒ 27r^{7} = 1/81

⇒ r^{7} = 1/(81 × 27)

r^{7} = 1/2187

r^{7} = 1/(3)^{7}

∴ r = 1/3

Now, S_{10} = a(1 - r^{n})/(1 – r)

= 27[1 – (1/3)^{10}]/(1 – 1/3)

= 27 [1 – 1/3^{10}]/((3 – 1)/3)

= ((27 × 3)/2)[1 – 1/3^{10}]

= 81/2 [1 – 1/3^{10}]

**11. Find the first term of the G.P. whose common ratio is 3, last term is 486 and the sum of whose terms is 728. **

**Answer**

Common ratio of G.P. = 3

And last term = 486

And sum of terms = 728

S_{n} = a(r^{n} – 1)/(r – 1) = a(3^{n} – 1)/(3 – 1)

= a(3^{n} – 1)/2

= 728

a(3^{n} – 1) = 728 × 2 = 1456 **...(i)**

l = 486

⇒ ar^{n-1} = a3^{n-1} = 486

⇒ a[3^{n}/3] = 486

⇒ a3^{n} = 486 × 3 = 1458

But a(3^{n} – 1) = 1456 **[From (i)]**

a3^{n} – a = 1456

1458 – a = 1456 [from (ii)]

∴ a = 1458 – 1456 = 2

Hence first term = 2

**12. In a G.P. the first term is 7, the last term is 448, and the sum is 889. Find the common ratio. **

**Answer**

In a G.P.

First term (a) = 7, last term (l) = 448

And sum = 889

Let r be the common ratio, then

l = ar ^{n-1}

⇒ 7r^{n-1} = 448

r^{n-1} = 448/7 = 64 **…(i)**

and sum = a(r^{n} – 1)/(r – 1) = 889

⇒ 7(r^{n }– 1)/(r – 1) = 889

⇒ (r^{n} – 1)/(r – 1) = 889/7 = 127 **…(ii)**

From (i),

r^{n}/r = 64

⇒ r^{n} = 64r

From (ii)

(64r – 1)/(r – 1) = 127

⇒ 64r – 1 = 127r – 127

⇒ 127r – 64r = -1 + 127

⇒ 63r = 126

⇒ r = 126/23 = 2

Hence common ratio = 2

**13. Find the third term of a G.P. whose common ratio is 3 and the sum of whose first seven terms is 2186. **

**Answer**

In a G.P.

Common ratio = 3

S_{7} = 2186 = a(r^{n} – 1)/(r – 1)

⇒ 2186 = a(3^{n} – 1)/(3 – 1)

⇒ 2186 = a(3n – 1)/2

⇒ 4372 = a(37 – 1)

⇒ 4372 = a(2187 – 1)

⇒ 4372 = 2186a

⇒ a = 4372/2186 = 2

a^{3} = ar^{3-1}

= ar^{2}

= 2 ×3^{2}

= 2 × 9

= 18

**14. If the first term of a G.P. is 5 and the sum of first three terms is 31/5, find the common ratio. **

**Answer**

In a G.P.

First term (a) = 5

S_{3} = 31/5

= a(r^{3} – 1)/(r – 1)

= 5(r^{3} – 1)/(r - 1)

(r^{3} – 1) /(r – 1) = 31/(5 × 5) = 31/25

⇒ {(r – 1)(r^{2} + r + 1)}/(r – 1) = 31/25

r^{2} + r + 1 = 31/25

⇒ 25r^{2} + 25r + 25 = 31

⇒ 25r^{2} + 25r – 6 = 0

⇒ 25r^{2} + 30r – 5r – 6 = 0

⇒ 5r(5r + 6) – 1(5r + 6) = 0

⇒ (5r + 6)(5r – 1) = 0

Either 5r + 6 = 0, then r = -6/5

Or 5r – 1 = 0, then r = 1/5

Hence common ratio = 1/5 or -6/5

**15. The sum of first three terms of a G.P. is to the sum of first six terms as 125 : 152. Find the common ratio of the G.P. **

**Answer**

S_{3} ÷ S_{6} = 125 : 152

Let r be the common ratio and a be the first number, then

a(r^{3 }– 1)/(r – 1) : a(r^{6} – 1)/(r - 1) = 125 : 152

(r^{3} – 1) : (r^{6} – 1) = 125 : 152

(r^{3 }– 1) : (r^{3} + 1)(r^{3} – 1) = 125 : 152

1 : (r^{3} + 1) = 125 : 152

(r^{3} + 1) × 125 = 152 × 1

125r^{3 }+ 125 = 152

125r^{3} = 152 – 125 = 27

r^{3} = 27/125 = (3/5)^{3}

∴ r = 3/5

∴ Common ratio = 3/5

**Answer**

S_{n} = 2^{1} - 1 + 2^{2} – 1 + 2^{3} – 1 + 2^{4} – 1… 2^{50} – 1

S_{n} = (2^{1} + 2^{2} + 2^{3} + 2^{4} + … 2^{50}) – 1 × 50

S_{n} = 2 + 4 + 8 + 16 … 2^{50} – 50

S_{n} = a(r^{n} – 1)/(r – 1) – 50

= 2(2^{50} – 1)/(2 – 1) – 50

S_{n} = 2 × 2^{50} – 2 – 50

S_{n} = 2^{51} – 52

**17. Sum the series x(x + y) + x ^{2}(x^{2 }+ y^{2}) + x^{3}(x^{3} + y^{3}) … to n terms. **

**Answer**

Given

S_{n }= x(x + y) + x^{2}(x^{2} + y^{2}) + x^{3}(x^{3} + y^{3}) … n terms

**18. Find the sum of the series 1 + (1 + x) + (1 + x + x ^{2}) + …… to n terms, x ≠ 1.**

**Answer**

1 + (1 + x) + (1 + x + x^{2}) + …… n terms, x ≠ 1

Multiply and divide by (1 – x)

**19. Find the sum of the following series to n terms:**

**(i) 7 + 77 + 777 + …**

**(ii) 8 + 88 + 888 + …**

**(iii) 0.5 + 0.55 + 0.555 + ….**

**Answer**

**(i)** S_{n} = 7 + 77 + 777 + … n terms

**(ii)**S

_{n}= 8 + 88 + 888 + … n terms

**(iii)**S

_{n}= 0.5 + 0.55 + 0.555 + …. n terms

**Multiple Choice Questions**

**Choose the correct answer from the given four options (1 to 33): **

**1. The list of numbers – 10, -6, - 2, 2, …… is **

**(a) an A.P. with d = - 16 **

**(b) an A.P. with d = 4 **

**(c) an A.P. with d = - 4 **

**(d) not an A.P. **

**Answer**

(b) an A.P. with d = 4

- 10, -6, -2, 2, …. is an A.P. with d = - 6 – (-10)

= - 6 + 10

= 4

**2. The 10 ^{th} term of the A.P. 5, 8, 11, 14, …. Is **

**(a) 32 **

**(b) 35 **

**(c) 38 **

**(d) 185 **

**Answer**

(a) 32

10^{th} term of A.P. 5, 8, 11, 14, …….

{∵ a = 5, d = 3}

a + (n – 1)d = 5 + (10 – 1) × 3

= 5 + 9 × 3

= 5 + 27

= 32

**3. The 30 ^{th} term of the A.P. 10, 7, 4, ….. is**

**(a) 87 **

**(b) 77 **

**(c) – 77 **

**(d) – 87 **

**Answer**

(c) -77

30^{th} term of A.P. 10, 7, 4, …. is

30^{th} term = a + (n – 1)d {∵ a = 10, d = 7 – 10 = - 3}

= 10 + (30 – 1) × (-3)

= 10 + 29(-3)

= 10 – 87

= - 77

**4. The 11 ^{th} term of the A.P. -3, -(1/2), 2, …… is **

**(a) 28 **

**(b) 22 **

**(c) – 38 **

**(d) – 48 **

**Answer**

(b) 22

Given -3, -(1/2), 2, …

a = - 3, d = - (1/2) – (-3) = - 1/2 + 3 = 5/2

11^{th} term = a + (n – 1)d

= - 3 + (11 – 1) × 5/2

= - 3 + 10 × 5/2

= - 3 + 25

= 22

**5. The 4 ^{th} term from the end of the A.P. -11, -8, -5, …… 49 is **

**(a) 37 **

**(b) 40 **

**(c) 43 **

**(d) 58 **

**Answer**

(b)

4^{th} term from the end of the A.P. – 11, -8, -5, ……. 49 is

Here, a = -11, d = - 8 – (-11)

= - 8 + 11

= 3 and l = 49 40

∴ l = 49 = a + (n – 1)d

⇒ 49 = - 11 + (n – 1) × 3

⇒ 49 – 11 = 3(n – 1)

⇒ 60/3 = n – 1

⇒ n = 20 + 1

= 21

Now, 4^{th} term from the end = l – (n – 1)d

= 49 – (4 – 1) × 3

= 49 – 9

= 40

**6. The 15 ^{th} term from the last of the A.P. 7, 10, 13, ………, 130 is **

**(a) 49 **

**(b) 85 **

**(c) 88 **

**(d) 110 **

**Answer**

(c) 88

15^{th} term from the end of A.P. 7, 10, 13, ….. , 130

Here, a = 7, d = 10 – 7 = 3, l = 130

15^{th} term from the end = l – (n – 1)d

= 130 – (15 – 1) × 3

= 130 – 42

= 88

**7. If the common difference of an A.P. is 5, then a _{18} – a_{13} is **

**(a) 5 **

**(b) 20 **

**(c) 25 **

**(d) 30 **

**Answer**

(c) 25

Common difference of an A.P. (d) = 5

a_{18} – a_{13} = a + 17d – a – 12d

= 5d

= 5 × 5

= 25

**8. In an A.P., if a _{18} – a_{14} = 32 then the common difference is **

**(a) 8 **

**(b) – 8 **

**(c) – 4 **

**(d) 4 **

**Answer**

(a) 8

If a_{18} – a_{14} = 32, then d = ?

(a + 17d) – a – 13d = 32

⇒ a + 17d – a – 13d = 32

⇒ 4d = 32

⇒ d = 32/4 = 8

**9. In an A.P., if d = - 4, n = 7, a _{n} = 4, then a is **

**(a) 6 **

**(b) 7 **

**(c) 20 **

**(d) 28 **

**Answer**

(d) 28

In an A.P., d = - 4, x = 7, a_{n} = 4 then a = ?

a_{n} = a(n – 1)d = 4

a_{7} = a + (7 – 1)d = 4

⇒ a + 6d = 4

⇒ a + 6 × (-4) = 4

a – 24 = 4

⇒ a = 4 + 24 = 28

**10. In an A.P., if a = 3.5, d = 0, n = 101, then a _{n} will be **

**(a) 0 **

**(b) 3.5 **

**(c) 103.5 **

**(d) 104.5 **

**Answer**

(b) 3.5

In an A.P.

a = 3.5, d = 0, n = 101, then a_{n }= ?

a_{n} = a_{101} = a + (101 – 1)d

= 3.5 + 100d

= 3.5 + 100 × 0

= 3.5 + 0

= 3.5

**11. In an A.P., if a = - 7.2, d = 3.6, a _{n }= 7.2, then n is **

**(a) 1 **

**(b) 3 **

**(c) 4 **

**(d) 5 **

**Answer**

In an A.P.

a = - 7.2, d = 3.6, a_{n} = 7.2, n = ?

a_{n} = 7.2

a + (n – 1)d = 7.2

⇒ -7.2 + (n – 1)3.6 = 7.2

⇒ (n – 1) × 3.6 = 7.2 + 7.2 = 14.4

⇒ (n – 1) = (14.4)/(3.6) = 4

⇒ n = 4 + 1 = 5

**12. 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**

(b) 10th term

Which term of an A.P. 21, 42, 63, 84, …. Is 210

Let 210 be the nth term, then

Here, a = 21, d = 42 – 21 = 21

210 = a + (n – 1)d

⇒ 210 = 21 + (n – 1) ×21

⇒ 210 – 21 = 21(n – 1)

⇒ 189/21 = n – 1

⇒ 9 = n – 1

⇒ n = 9 + 1 = 10

∴ It is 10^{th} term.

**13. If the last term of the A.P. 5, 3, 1, -1, …… is -41, then the A.P. consist of **

**(a) 46 terms **

**(b) 25 terms **

**(c) 24 terms **

**(d) 23 terms **

**Answer**

(c) 24 terms

Last term of an A.P. 5, 3, 1, -1, …… is – 41

Then A.P. will consist of terms

Here, a = 5, d = 3 – 5 = -2 and n = ?

l = - 41

l = - 41 = a + (n – 1)d

⇒ -41 = 5 + (n – 1)(-2)

⇒ - 41 – 5 = (n – 1)(-2)

⇒ (- 46/-2) = n – 1

⇒ n – 1 = 23

⇒ n = 23 + 1 = 24

A.P. consist of 24 terms.

**14. If k – 1, k + 1 and 2k + 3 are in A.P., then the value of k is**

**(a) – 2 **

**(b) 0 **

**(c) 2 **

**(d) 4 **

**Answer**

(b) 0

k – 1, k + 1 and 2k + 3 are in A.P.

2(k + 1) = (k – 1) + (2k + 3)

⇒ 2k + 2 = k – 1 + 2k + 3

⇒ 2k + 2 – 3k + 2

⇒ 3k – 2k = 2 – 2

⇒ k = 0

**15. The 21 ^{st} term of an A.P. whose first two terms are – 3 and 4 is **

**(a) 17 **

**(b) 137 **

**(c) 143 **

**(d) – 143 **

**Answer**

(b) 137

First two terms of an A.P. are – 3 and 4

a = -3, d = 4 – (-3)

= 4 + 3 = 7

21^{st} term = a + 20d

= - 3 + 20(7)

= -3 + 140

= 137

**16. If the 2 ^{nd} term of an A.P. is 13 and the 5^{th} term is 25, then its 7^{th} term is **

**(a) 30 **

**(b) 33 **

**(c) 37 **

**(d) 38 **

**Answer**

(b) 33

In an A.P.

2^{nd} term = 13

⇒ a + d = 13 **...(i)**

5^{th} term = 25

⇒ a + 4d = 25 **…(ii)**

Subtracting (i) and (ii),

3d = 12

⇒ d = 1/3

Substitute the value of d in eq. (i), we get

a = 13 – 4

= 9

7^{th} term = a + 6d

= 9 + 6 × 4

= 9 + 24

= 33

**17. If the first term of an A.P. is -5 and the common difference is 2, then the sum of its first 6 terms is **

**(a) 0 **

**(b) ****5 **

**(c) 6 **

**(d) 15 **

**Answer**

(a) 0

First term (a) of an A.P. = - 5

Common difference (d) is 2

Sum of first 6 terms = n/2[2a + (n – 1)d]

= 6/2[2 × (-5) + (6 – 1) × 2]

= 3[-10 + 5 × 2]

= 3 × [- 10 – 10]

= 3 × 0

= 0

**18. The sum of 25 terms of the A.P. , -2/3, -2/3, -2/3 is **

**(a) 0 **

**(b) – 2/3**

**(c) –50/3**

**(d) – 50 **

**Answer**

(c) -50/3

a = -2/3, d = 0, n = 25

Sum of 25 terms of an AP = n/2[2a + (n – 1)d]

= 25/2 [2× -2/3 + (25-1)×0]

= 25/2 [-4/3 + 0]

= 25/2 × -4/3

= -100/6

= -50/3

**19. In an A.P., if a = 1, a _{n} = 20 and S_{n} = 399, then n is **

**(a) 19 **

**(b) 21 **

**(c) 38 **

**(d) 42 **

**Answer**

(c) 38

In an A.P., a = 1, a_{n} = 20, S_{n} = 399, n is ?

a_{n} = a + (n – 1)d = 20

1 + (n – 1)d = 20

(n – 1)d = 20 – 1 = 19 **…(i)**

**20. In an A.P., if a = - 5, l = 21. and S _{n} = 200, then n is equal to **

**(a) 50 **

**(b) 40 **

**(c) 32 **

**(d) 25 **

**Answer**

(d) 25

In an A.P.

a = - 5, l = 21, S_{n} = 200, n = ?

l = a + (n – 1)d = - 5 + (n – 1)d

21 = -5 + (n – 1)d

**21. In an A.P., if a = 3 and ****S _{8}**

**= 192, then d is**

**(a) 8 **

**(b) 7 **

**(c) 6 **

**(d) 4 **

**Answer**

(c) 6

In an A.P.

a = 3, S_{8} = 192, d = ?

**22. The sum of first five multiples of 3 is **

**(a) 45 **

**(b) 55 **

**(c) 65 **

**(d) 75 **

**Answer**

(a) 45

First 5 multiples of 3:

3, 6, 9, 12, 15

Here, a = 3, d = 6 – 3 = 3

**23. The number of two digit numbers which are divisible by 3 is **

**(a) 33 **

**(b) 31 **

**(c) 30 **

**(d) 29 **

**Answer**

(c) 30

Two digit number which are divisible by 3 is 12, 15, 18, 21, ….. 99

Here, a = 12, d = 3, l = 99

l = a_{n} = a + (n – 1)d

⇒ 12 + (n – 1) × 3 = 99

⇒ (n – 1)3 = 99 – 12 = 87

⇒ n – 1 = 87/3 = 29

⇒ n = 29 + 1 = 30

**24. The number of multiples of 4 that lie between 10 and 250 is **

**(a) 62 **

**(b) 60 **

**(c) 59 **

**(d) 55 **

**Answer**

(b) 60

Multiples of 4 lying between 10 and 250: 12, 16, 20, 24, … 248

Here, a = 12, d = 16 – 12 = 4, l = 248

**25. The sum of first 10 even whole numbers is **

**(a) 110 **

**(b) 90 **

**(c) 55 **

**(d) 45 **

**Answer**

(b) 90

Sum of first 10 even numbers

Even numbers are 0, 2, 4, 6, 8, 10, 12, 14, 16, 18

Here, a = 0, d = 2, n = 10

Here, a = 0, d = 2, n = 10

**26. The list of number 1/9, 1/3, 1, -3, ….. is a **

**(a) G.P. with r = - 3 **

**(b) G.P. with r = - (1/3) **

**(c) G.P. with r = 3 **

**(d) not a G.P. **

**Answer**

(c) G.P. with r = 3

The given list of numbers 1/9, 1/3, 1, - 3, ……

**27. The 11 ^{th} of the G.P. 1/8, - (1/4), 2, -1, ……. Is **

**(a) 64 **

**(b) – 64 **

**(c) 128 **

**(d) – 128 **

**Answer**

(c) 128

11^{th} of the G.P.

1/8, -(1/4), 2, -1,…… is

**28. The 5 ^{th} term from the end of the G.P. 2, 6, 18, ….. 13122 is **

**(a) 162 **

**(b) 486 **

**(c) 54 **

**(d) 1458 **

**Answer**

(a) 162

5^{th} term from the end of the G.P. 2, 6, 18, … 131222 is

Here, a = 2, r = 6/2 = 3, l = 13122

**29. If k, 2(k + 1), 3(k + 1) are three consecutive terms of a G.P., the value of k is **

**(a) – 1 **

**(b) – 4 **

**(c) 1 **

**(d) 4 **

**Answer**

(b) -4

k, 2(k + 1), 3(k + 1) are in G.P.

[2(k + 1)]^{2} = k × 3(k + 1)

⇒ 4(k + 1)^{2 }= 3k(k + 1)

⇒ 4(k + 1) = 3k

(Dividing by k + 1 if k + 1 ≠ 0)

⇒ 4k + 4 = 3k

⇒ 4k – 3k = - 4

⇒ k = -4

**30. Which term of the G.P. 18, -12, 8, ….. is 512/729 ? **

**(a) 12 ^{th} **

**(b) 11 ^{th} **

**(c) 10 ^{th} **

**(d) 9 ^{th} **

**Answer**

(d) 9th

Which term of the G.P.

18, -12, 8, ….. 512/729

Let it be nth term

**31. The sum of the first 8 terms of the series 1 + √3 + 3 + ……. is **

**Answer**

Sum of first 8 terms of 1 + √3 + 3 + …. Is

Here, a = 1, r = √3/1 = √3, n = 8

**32. The sum of first 6 terms of the G.P. 1, -(2/3), 4/9, ….. is **

**(a) – (133/243) **

**(b) 133/243**

**(c) 793/1215 **

**(d) none of these**

**Answer**

(b) 133/243

Sum of first 6 terms of G.P.

1, -(2/3), 4/9, …

**33. If the sum of the G.P., 1, 4, 16, ……. Is 341, then the number of terms in the G.P. is **

**(a) 10 **

**(b) 8 **

**(c) 6 **

**(d) 5 **

**Answer**

(d) 5

The sum of G.P. 1, 4, 16, …. Is 341

Let n be the numbers of terms,

**Chapter Test**

**1. Write the first four terms of the A.P. when its first term is – 5 and the common difference is -3. **

**Answer: **

First 4 terms of A.P. whose first term (a) = - 5

And common difference (d) = - 3

= -5, -8, - 11, -14

**2. Verify that each of the following lists of numbers is an A.P., and the write its next three terms : **

**(i) 0, 1/4, 1/2, 3/4, ….. **

**(ii) 5, 14/3, 13/3, 4, ……**

**Answer**

**(i)** 0, 1/4, 1/2, 3/4, ….

Here a = 0, d = 1/4

∴ Next three terms will be 1, 5/4, 3/2

**(ii)** 5, 14/3, 13/3, 4, …

Here, a = 5, d = 14/3 – 5 = (14 – 15)/3 = (-1)/3

∴ Next three terms will be

a_{2} = 4 – 1/3 = 11/3

a_{3} = 11/3 – 1/3 = 10/3

a_{4} = 10/3 – 1/3 = 9/3 = 3

i.e., 11/3, 10/3, 3

**3. The nth term of the A.P. is 6n + 2. Find the common difference. **

**Answer: **

T_{n} of an A.P. = 6n + 2

T_{1 }= 6 × 1 + 2 = 6 + 2 = 8

T_{2} = 6 × 2 + 2 = 12 + 2 = 14

T_{3} = 6 × 3 + 2 = 18 + 2 = 20

d = 14 – 8 = 6

**4. Show that the list of numbers 9, 12, 15, 18, ….. form an A.P. Find its 16 ^{th} term and the nth. **

**Answer: **

9, 12, 15, 18, ……

Here, a = 9, d = 12 – 9 = 3

Or, 15 – 12 = 3

Or, 18 – 15 = 3

Yes, it form an A.P.

T_{16} = a + (n – 1)d

= 9 + (16 – 1) × 3

= 9 + 15 × 3

= 9 + 45

= 54

and T_{n} = a + (n – 1)d

= 9 + (n – 1) × 3

= 9 + 3n – 3

= 3n + 6

**5. Find the 6 ^{th} term from the end of the A.P. 17, 14, 11, …., - 40. **

**Answer**

From the question it is given that,

First term a = 17

Common difference = 14 – 17 = - 3

Last term l = - 40

L = a + (n – 1)d

- 40 – 17 = - 3n + 3

⇒ - 57 – 3 = - 3n

⇒ n = -60/-3

⇒ n = 20

Therefore, 6^{th} term from the end = l – (n – 1)d)

= 40 – (6 – 1)(-3)

= - 40 – (5)(-3)

= - 40 + 15

= - 25

**6. If the 8 ^{th} term of an A.P. is 31 and the 15^{th} term is 16 more than its 11^{th} term, then find the A.P. **

**Answer**

In an A.P.

a_{8} = 31, a_{15} = a_{11} + 6

Let a be the first term and d be a common difference, then

a_{8} = a + (n – 1)d = 31

⇒ a + 7d = 31 **…(i)**

Similarly,

a_{15 }= a + 14d = a + 10d + 16

14d – 10d = 16

⇒ 4d = 16

⇒ d = 16/4 = 4

From (i) a + 7 × 4 = 31

⇒ a + 28 = 31

⇒ a = 31 – 28 = 3

∴ a = 3, d = 4

Now, A.P. will be 3, 7, 11, 15, …

**7. 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, then find the w^{th} term.**

**Answer**

In an A.P.

a_{17} = 2 × a_{8} + 5

a_{11 }= 43, find a_{n}

Let a be the first term and d be the common difference, then

a_{11} = a + (n – 1)d = a + (11 – 1)d

= a + 10d = 43 **…(i)**

Similarly,

a_{17} = 2 × a_{8 }+ 5

a + 16d = 2(a + 7d) + 5

a + 16d = 2a + 14d + 5

- 5 + 16a – 14d = 2a – a

⇒ a = 2d – 5 **…(ii)**

From (i) and (ii)

2d – 5 + 10d = 43

⇒ 12d = 43 + 5 = 48

d = 48/12 = 4

But a + 10d = 43

∴ a + 10 × 4 = 43

⇒ a + 40 = 43

⇒ a = 43 – 40 = 3

∴ a = 3, d = 4

Now, a_{n} = a + (n – 1)d

= 3 + 4(n – 1)

= 3 + 4n – 4

= 4n – 1

**8. The 19 ^{th} term of an A.P. is equal to three times its 6^{th} term. If its 9^{th} term is 19, find the A.P.**

**Answer**

In an A.P.

a_{19} = 3 × a_{6} and a_{9 }= 19

Let a be the first term and d be the common difference, then

a_{9} = a + (n – 1)d

= a + (9 – 1)d = a + 8d

a + 8d = 19 **…(i)**

Similarly,

a_{19} = 3 × a_{6}

⇒ a + 18d = 3(a + 5d)

a + 18d = 3a + 15d

⇒ 3a – a = 18d – 15d

⇒ 2a = 3d **…(ii)**

a = 3/2.d

From (i),

3/2d + 8d = 19

⇒ 19/2 d = 19

⇒ d = (19 × 2)/19 = 2

And a = 3/2d = 3/2 × 2 = 3

∴ a = 3, d = 2 and A.P. is 3, 5, 7, 9, …..

**9. If the 3 ^{rd} and the 9^{th} terms of an A.P. are 4 and – 8 respectively, then which term of this A.P. is zero?**

**Answer**

In an A.P.

a_{3 }= 4, a_{9} = - 8, which term of A.P. will be zero

Let a be the first term and d be a common difference, then

a_{3} = a + (n – 1)d = a + (3 – 1)d

⇒ a + 2d = 4 **…(i)**

Similarly, a + 8d = - 8

Subtracting, we get

6d = - 12

⇒ d = (-12)/6 = - 2

And a + 2d = 4

⇒ a + 2 × (-2) = 4

⇒ a – 4 = 4

⇒ a = 4 + 4 = 8

Let n^{th} term be zero, then

a + (n – 1)d = 0

⇒ 8 + (n – 1) × (-2) = 0

⇒ - 2n + 2 = - 8

⇒ - 2n = - 8 – 2 = - 10

⇒ n = (-10/-2)

n = 5

∴ 0 will be the fifth term.

**10. Which term of the list of numbers 5, 2, -1, - 4, …. Is – 55? **

**Answer**

A.P. is 5, 2, -1, -4, …

Which term of A.P. is – 55

Let it be nth term

Here, a = 5, d = 2 – 5 = - 3

Here, a = 5, d = 2 – 5 = - 3

∴ a_{n} = a + (n – 1)d

⇒ -55 = 5 + (n – 1) × (-3)

- 55 – 5 = - 3(n – 1)

⇒ (- 60)/(-3) = n – 1

⇒ n – 1= 20

⇒ n = 20 + 1 = 21

∴ - 55 is the 21^{st} term.

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

**Answer**

In an A.P.

24^{th} term = 2 × 10^{th} term

To show that 72^{nd} term = 4 × 15^{th} term

Let a be the first term and d be a common difference, then

24^{th} term = a + (24 – 1)d = a + 23d

And 10^{th} term = a + 9d

∴ a + 23d = 2(a + 9d)

⇒ a + 23d = 2a + 18d

⇒ 2a – a = 23d – 18d

⇒ a = 5d **…(i)**

And 72^{nd} term = a + 17d

And 15^{th} term = a + 14d

Substitute the value of (i), we get

a + 71d = 5d + 71d = 76d

and a + 14d = 5d + 14d

= 19d

∴ 76d = 4 × 19d

Hence 72^{nd} term is 4 times the 15^{th} term.

**12. Which term of the list of numbers 20, 19.1/4, 18.1/2, 17.3/4, ……… is the first negative term? **

**Answer**

In A.P., which is the first negative term 20, 19.1/4, 18.1/2, 17.3/4, ……

Here, a = 20, d = 19.1/4 – 20 = -3/4

Let nth term be first negative term

∴ a_{n }= a + (n – 1)d

Let nth term be first negative term, then

a_{n} = 20 + (n – 1) (-3/4)

⇒ a_{n }= 20 + (n – 1)(-3/4)

⇒ a_{n} = 20 – 3/4n + 3/4

Now, a_{n} < 0 is the first negative term

⇒ 20 + 3/4 – 3/4n < 0

⇒ 83/4 – 3/4n < 0

⇒ 83/4 < 3/4n

⇒ 83 < 3n

⇒ 83/3 < n

⇒ 28 < n

∴ 28^{th} is the first negative term.

**13. If the pth term of an A.P. is q and the qth term is p, show that its nth term is (p + q – n)**

**Answer**

In an A.P.

pth term = q

qth term = p

Show that (p + q – n) is nth term

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

∴ pth term = a + (p – 1)d = q **…(i)**

And qth term = a + (q – 1)d = p **…(ii)**

Subtracting, we get

q – p = (p -1 – q + 1)d

⇒ q – p = (p – q)d

⇒ d = (q – p)/(p – q)

⇒ d = {-(p – q)/(p – q)} = - 1

a – p + 1 = q

a = q + p – 1

L.H.S

nth term = a + (n – 1)d = (p + q – 1) + (n – 1)(-1)

= p + q – 1 – n + 1

= p + q – n

= R.H.S.

**14. How many three digit numbers are divisible by 9?**

**Answer**

3 – digit numbers which are divisible by 9 are 108, 117, 126, 135, ….., 999

Here, a = 108, d = 9 and l = 999

∴ l = a_{n }= a + (n – 1)d

⇒ 999 = 108 + (n – 1)9

⇒ 999 – 108 = 9(n – 1)

⇒ 891 = 9(n – 1)

⇒ 891/9 = n – 1

⇒ n – 1 = 99

⇒ n = 99 + 1 = 100

∴ There are 100 numbers or terms.

**15. The sum of three numbers in A.P. is -3 and the product is 8. Find the numbers.**

**Answer**

Sum of three numbers of an A.P. = - 3

And their product = 8

Let the numbers be a – d, a, a + d, then

a – d + a + a + d = - 3

⇒ 3a = - 3

⇒ a = -3/3 = -1

And (a – d)a(a + d) = 8

a(a^{2} – d^{2}) = 8

⇒ -1[(-1)^{2} – d^{2}] = 8

⇒ 1 – d^{2} = 8/-1 = - 8

d^{2} = 1 + 8 = 9 = (+3)^{2}

∴ d = ± 3

If d = 3

Hence the numbers are -1, -3, - 1, - 1 + 3

⇒ - 4, -1, 2

And if d = - 3, then

- 1 – (-3), -1, -1 – 3

⇒ - 1 + 3, -1, -4

⇒ 2, -1, -4

**16. ****The angles of a quadrilateral are in A.P. if the greatest angle is double of the smallest angle, find all the four angles. **

**Answer**

From the question it is given that,

The angles of a quadrilateral are in A.P.

Greatest angle is double of the smallest angle

Let us assume the greatest angle of the quadrilateral is a + 3d,

Then, the other angles are a + d, a – d, a – 3d

So, a – 3d is the smallest

Therefore, a + 3d = 2(a – 3d)

a + 3d = 2a – 6d

⇒ 6d + 3d = 2a – a

⇒ 9d = a **[equation (i)]**

We know that the sum of all angles of quadrilateral is 360°.

a – 3d + a – d + a + d + a+ 3d = 360°

⇒ 4a = 360°

⇒ a = 360/4

⇒ a = 90°

Now, substitute the value of a in equation (i) we get,

9d = 90

⇒ d = 90/9

⇒ d = 10

Substitute the value of a and d in assumed angles,

Greatest angle = a + 3d = 90 + (3 × 10) = 90 + 30 = 120°

Then, other angles are = a + d = 90° + 10° = 100°

a – d = 90° – 10° = 80°

⇒ a – 3d = 90° – (3 × 10) = 90 – 30 = 60°

Therefore, the angles of quadrilateral are 120°, 100°, 80° and 60°

**17. The nth term of an A.P. cannot be ****n ^{2} + n + 1.**

**Justify your answer.**

**Answer**

nth term of an A.P. can’t be n^{2} + n + 1

Giving some different values to n such as 1, 2, 3, 4, ….

We find then

a_{1} = 1^{2 }+ 1 + 1 = 1 + 1 + 1 = 3

a_{2} = 2^{2} + 2 + 1 = 4 + 2 + 1 = 7

a_{3} = 3^{2 }+ 3 + 1 = 9 + 3 + 1 = 13

a_{4} = 4^{2} + 4 + 1 = 16 + 4 + 1 = 21

We see that,

d = a_{2} – a_{1} = 7 – 3 = 4

d = a_{3 }– a_{2} = 13 – 7 = 6

d = a_{4} – a_{3} = 21 – 13 = 8

We see that d is not constant

∴ It is not an A.P.

Hence, a_{n }≠ n^{2} + n + 1

**18. Find the sum of first 20 terms of an A.P. whose nth term is 15 – 4n. **

**Answer**

n^{th} term is 15 – 4n

So, a_{n} = 15 – 4n

Giving values in place of n from 1 to 20

We get,

a_{1} = 15 – (4 × 1) = 15 – 4 = 11

a_{2} = 15 – (4 × 2) = 15 – 8 = 7

a_{3} = 15 – (4 × 3) = 15 – 12 = 3

a_{4} = 15 – (4 × 4) = 15 – 16 = - 1

Then, a_{20} = 15 – (4 × 20) = 15 – 80 = -65

So, 11, 7, 3, -1, … -65 are in A.P.

Therefore, first term a = 11

Common difference = -4

n = 20

**19. Find the sum: 18 + 15.1/2 + 13 + ……(- 49.1/2) **

**Answer**

Here, a = 18, d = 15.1/2 – 18 = -2.1/2 = -5/2

l = - 49.1/2 = - 99/2

a_{n} = a + (n – 1)d

- 99/2 = 18 + (n – 1)(-5/2)

- 99/2 – 18/1 = -5/2(n – 1)

(- 99 – 36)/2 = -5/2 (n – 1)

⇒ (-135)/2 = -5/2(n – 1)

⇒ - 135/2 = -5/2(n – 1)

⇒ - 135/2 × 2/-5 = n – 1

⇒ n - 1 = 27

⇒ n = 27 + 1 = 28

Now, S_{n} = n/2[2a + (n – 1)d]

S_{28} = 28/2 [2 × 18 + (28 – 1)(-5/2)]

S_{28} = 14[36 + (27 × -5/2)]

= 14[36 – 135/2]

S_{28 }= 14(72 – 135)/2

= 14 × (-63)/2

= - 441

**20. (i) How many terms of the A.P. – 6, - (11/2 – 5), …. Make the sum – 25 ? **

**(ii) Solve the equations 2 + 5 + 8 + …. + x = 155.**

**Answer**

**(i)** Sum = - 25

A.P. = -6, -(11/2), - 5, ….

Here, a = -6, d = -11/2, -5, …..

Here, a = -6, d = -11/2 + 6 = 1/2

Sum = - 25

Let n term be added to get the sum – 25

∴ S_{n} = n/2[2a + (n – 1)d]

-25 = n/2[2 × (-6) + (n – 1)(1/2)]

- 25 × 2 = n[- 12 + 1/2n – 1/2]

- 50 = n[- 25/2 + 1/2n]

1/2n^{2} – 25/n + 50 = 0

⇒ n^{2} – 25n + 100 = 0

{∵ 100 = - 20 × -5, - 25 = - 20 – 5}

⇒ n^{2} – 5n – 20n + 100 = 0

⇒ n(n – 5) – 20(n – 5) = 0

⇒ (n – 5)(n – 20) = 0

Either n – 5 = 0, then n = 5

Or n – 20 = 0, then n = 20

∴ Number of terms are 5 or 20

**(ii)** Solve the equation 2 + 5 + 8 + ….. + x = 155

Here, a = 2, d = 5 – 2 = 3, l = x

Sum = 155

l = a + (n – 1)d

x = 2 + (n – 1)3 = 2 + 3n – 3

⇒ x = 3n – 1 …..(i)

S_{n} = n/2[2a + (n – 1)d]

⇒ 155 = n/2[2 × 2 + (n – 1) × 3]

⇒ 155 × 2 = n[4 + 3n – 3]

⇒ 310 = n(3n + 1) = 3n^{2} + n

∴ 3n^{2} + n – 310 = 0

3n^{2} – 30n + 31n – 310 = 0

⇒ 3n(n – 10) + 31(n – 10) = 0

⇒ (n – 10)(3n + 31) = 0

Either n – 10 = 0, then n = 10

Or 3n + 31 = 0, then 3n = - 31

⇒ n = - 31/3

Which is not possible being negative

∴ n = 10

Now, x = 3n – 1

= 3 × 10 – 1

= 30 – 1

= 29 [From (i)]

**21. If the third term of an A.P. is 5 and the ratio of its 6 ^{th} term to the 10^{th} term is 7 : 13, then find the sum of first 20 terms of this A.P. **

**Answer**

3^{rd} term of an A.P. = 5

Ratio in 6^{th} term and 10^{th} term = 7 : 13

Find S_{20}

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

∴ a_{3} = a + (n – 1)d

⇒ a + (3 – 1)d = 5

⇒ a + 2d = 5 **…(i)**

Similarly,

a_{6} = a + 5d and a_{10} = a + 9d

∴ (a + 5d)/(a + 9d) = 7/13

⇒ 7a + 63d = 13a + 65d

⇒ 13a + 65d – 7a – 63d = 0

⇒ 6a + 2d = 0

⇒ 3a + d = 0

⇒ d = - 3a **…(ii)**

From (i) and (ii),

a + 2d = 5

⇒ a + 2(-3a) = 5

⇒ a – 6a = 5

⇒ - 5a = 5

⇒ a = 5/-5 = - 1

And d = - 3a = - 3 × (-1) = 3

Now sum of first 20 terms = n/2[2a + (n – 1)d]

= 20/2 [2 × (-1) + (20 – 1) × 3]

= 10[- 2 + 57]

= 10 × 55

= 550

**22. In an A.P., the first term is 2 and the last term is 29. If the sum of the terms is 155, then find the common difference of the A.P. **

**Answer**

In an A.P.

First term (a) = 2

Last term (l) = 29

Sum of terms = 155

l = a_{n} = a + (n – 1)d

⇒ 29 = 2 + (n – 1)d

⇒ 29 – 2 = d(n – 1)

⇒ d(n – 1) = 27 **…(i)**

S_{n} = n/2[2a + (n – 1)d]

155 = n/2[2 × 2 + 27]

= n/2[4 + 27]

155 = 31/2.n

⇒ n = (155 × 2)/31 = 10

d(n – 1) = 27

⇒ d(10 – 1) = 27

d × 9 = 27

⇒ d = 27/9 = 3

**23. The sum of first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25 ^{th} term. **

**Answer**

Sum of first 14 terms = 1505

First term (a) = 10

Find 25^{th} term S_{14} = n/2[2a + (n – 1)d]

1505 = 14/2[2 × 10 + (14 – 1)d]

⇒ 1505 = 7[20 + 13d]

⇒ 20 + 13d = 1505/7

⇒ 13d = - 20 + 215 = 195

⇒ d = 195/13 = 15

Now, a_{25 }= a + (n – 1)d

= 10 + (25 – 1)(15)

= 10 + 24(15)

= 10 + 360

= 370

**24. The sum of first n terms of an A.P. is 3n2 + 4n. Find the 25 ^{th} term is this A.P.**

**Answer**

S_{n} = 3n^{2 }+ 4n

S_{n} – 1 = 3(n – 1)^{2} + 4(n – 1)

= 3(n^{2 }– 2n + 1) + 4(n – 1)

= 3n^{2} – 6n + 3 + 4n – 4

= 3n^{2} – 2n – 1

Now, an = S_{n} – S_{n -1}

= (3n^{2} + 4n) – (3n^{2} – 2n – 1)

= 3n^{2 }+ 4n – 3n^{2} + 2n + 1

= 6n + 1

a_{25} = 6(25) + 1

= 150 + 1

= 151

**25. In an A.P., the sum of first 10 terms is – 150 and the sum of next 10 terms is – 550. Find the A.P.**

**Answer: **

In an A.P.

Sum of first 10 terms = - 150

Sum of next 10 terms = - 550, A.P. = ?

Sum of first 10 terms = - 150

S_{10} = n/2[2a + (n – 1)d]

- 150 = 10/2[2a + 9d] = 5(2a + 9d)

⇒ 10a + 45d = - 150

S_{20} = S10 + S10

= - 150 – 150

= - 700

= 20/2 [2a + 19d]

= 10(2a + 19d)

⇒ 20a + 190d = - 700 **…(ii)**

20a + 90d = - 300 [Multiplying (i) by 2]

Subtracting, 100d = - 400

d = - 400/100 = - 4

From (i),

10a + 45 × (-4) = - 150

10a – 180 = - 150

10a = - 150 + 180

= 30

a = 30/10 = 3

A.P. is 3, -1, -5, -9,……

**26. The sum of first m terms of an A.P. is 4m ^{2}**

**– m. If its nth term is 107, find the value of n. Also find the 21**

^{st}term of this A.P.**Answer: **

S_{m} = 4m^{2} – m

S_{n} = 4n^{2 }– n

and S_{n – 1} = 4(n – 1)^{2}
– (n – 1)

= 4[n^{2} – 2n + 1] – n + 1

= 4n^{2 }– 8n + 4 – n + 1

= 4n^{2} – 9n + 5

Now, a_{n }= S_{n}
– S_{n-1}

= 4n^{2} – n – 4n^{2 }+
9n - 5

= 8a – 5

Now, a_{n} = 107

∴ 8n – 5 = 107

⇒ 8n =
107 + 5 = 112

n = 112/8 = 14

and a_{n} = 8n – 5

a_{21} = 8 × 21 – 5

= 168 – 5

= 163

**27. Find the geometric progression whose 4 ^{th}
term is 54 and 7^{th} term is 1458.**

**Answer: **

In
a G.P.

a_{4} = 54

a_{7} = 1458

Let a be the first term and r be
the common difference

**28. The fourth term of a G.P. is the square of its
second term and the first term is -3. Find its 7 ^{th} term. **

**Answer: **

In G.P.

a_{4} = (a^{2})^{2}, a_{1} = - 3

Let a be the first term an dr be
the common ratio

∴
a_{n} = ar ^{n-1}

a_{4} = ar^{3 }

a_{2 }= ar

ar^{3} = (ar)^{2}

⇒ ar^{3}
= a^{2}r^{2}

⇒ r = a

⇒ a_{1}
= - 3

∴ d = - 3

∴ a_{7} = ar ^{7 – 1}

= ar^{6 }

= - 3 × (-3)^{6 }

= - 3 × 729

= - 2187

**29. If the 4 ^{th}, 10^{th} and 16^{th}
terms of a G.P. are x, y and z respectively, prove that x, y and z are in G.P. **

**Answer**

In a G.P.

a_{4} = x, a_{10} = y, a_{16} = z

Show that x, y, z are in G.P.

Let a be the first term and r be the common
ratio, then

We know that,

a_{n} = ar^{n – 1}

a_{4} = ar^{4 – 1}

a_{4} = ar^{3} = x

a_{10} = ar^{9} = y

a_{16} = ar^{15} = z

x, y, z are in G.P.

If y^{2} = xy

Substitute the value of x and y,

y^{2} = (ar^{9})^{2}

y^{2} = a^{2} r^{18}

Then, xz = ar^{3} × ar^{15}

= a^{1+1 }r^{3+15} **[from a ^{m} × a^{n} = a^{m+n}]**

= a^{2} r^{18}

So, y^{2} = xy

Therefore, it is proved that x, y, z are in G.P.

**30. How many terms of the G.P. 3, 3/2, 3/4, are needed to give the sum 3069/512 ? **

**Answer**

G.P. 3, 3/2, 3/4

S_{n} = 3069/512

Here, a = 3, r = 1/2

Let n be the number of terms

S_{n} = a(1 – r^{n})/(1 – r)

**31. Find the sum of first n terms of the series: 3 + 33 + 333 + ……..**

**Answer: **

Series is 3 + 33 + 333 + …… n terms,

= 3[1 + 11 + 111 + …… n terms]

= 3/9[9 + 99 + 999 + ….. n terms]

= 3/9 [(10 – 1) + (110 – 1) + (1000 – 1) + … n terms]

= 3/9[10 + 100 + 1000 + …n terms – n × 1]

**32. Find the sum of the series 7 + 7.7 + 7.77 + 7.777 + …….. to 50 terms.**

**Answer: **

The given sequence is 7, 7.7, 7.77, 7.777, …

Required sum = S_{50}

= 7 + 7.7 + 7.77 + …… 50 terms

= 7(1 + 1.1 + 1.11 + 50 terms)

= 7/9 (9 + 9.9 + 9.99 + 9.999 + … 50 terms)

= 7/9 ((10 - 1) + (10 – 0.1) + (10 – 0.01) + (10 – 0.001) + … 50 terms)

= 7/9 (10 + 10 + 10 + 10 + … n terms – (0.1 + 0.01 + 0.001 + …. 50 terms))

We know that, S_{n} = a(1 - r^{n})/(1 - r)

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