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

Here, we are providing the solutions for Chapter 9 Arithmetic and Geometric Progression from ML Aggarwal Textbook for Class 10 ICSE Mathematics. Solutions of the ninth chapter has been provided in detail. This will help the students in understanding the chapter more clearly. Class 10 Chapter 9 Arithmetic and Geometric Progression ML Aggarwal Solutions for ICSE is one of the most important chapter for the board exams which is based on finding the first, last and any of the terms of arithmetic progressions, sum of the progressions, number of terms, common difference of A.P., summation of geometric series and terms of geometric series. We have also added chapter test and multiple choice questions.

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,……

(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

(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) 12, 32, 52, 72,…

(vii) 1, 3, 9, 27,…..

(viii)

(ix) a, 2a, 3a, 4a, …….

(x) a, 2a + 1, 3a + 2, 4a  + 3, ……

(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) 12, 32, 52, 72, ….

= 1, 9, 25, 49, …..

Here, first term (a) = 12 = 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.

(viii)

⇒ √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 20th term.

Tn = 7 – 3n

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

T1 = 7 – 3 × 1 = 7 – 3 = 4

T2 = 7 – 3 × 2 = 7 – 6 = 1

T3 = 7 – 3 × 3 = 7 – 9 = - 2

T4 = 7 – 3 × 4 = 7 – 12 = - 5

T20 = 7 – 3 × 20 = 7 – 60 = - 53

A.P. is 4, 1, -2, -5, …..

20th term = - 53

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

(i) 1, 6, 11, 16, ….; a20

(ii) -4, -7, - 10, - 13, …, a25, an

(i) 1, 6, 11, 6, …..

Here, a = 1, d = 6 – 1 – 5

a20 = 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

a25 = a + (n – 1)d

= - 4 + (25 – 1) × - 3

= - 4 + 24 × (-3)

= - 4 – 72

= - 76

And an = a + (n – 1)d

= - 4 + (n – 1)(- 3)

= - 4 – 3n + 3

= - 1 – 3n

= - 3n – 1

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

5, -2, -1, - 4, …..

Here, a = 5

d = 2 – 5 = - 3

(i) Tn = a + (n – 1)d

= 5 + (n – 1)(-3)

= 5 – 3n + 3

= 8 – 3n

(ii) T12 = a + 11d

= 5 + 11(-3)

= 5 – 33

= - 28

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

First term (a) = 7

And common difference (d) = 3

A.P. = 7, 10, 13, 16, 19,….

T8 = 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 18th term is – 5, then find its first term.

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

(i) Common difference (d) = 3

T18 = - 5

a + (n – 1)d = Tn

a + (18 – 1) (-3) = - 5

⇒ a + 17(-3) = - 5

⇒ a – 51 = - 5

⇒ a = - 5 + 51 = 46

∴ First term = 46

(ii) First term (a) = - 18

T10 = 0

a + (n – 1)d = Tn

- 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?

(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 16th 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 34th 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 27th 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.

(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

Tn = 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 17th 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 20th term from the last term of the A.P. 3, 8, 13, …… 253.

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

(i) A.P. is 3, 8, 13, ……253

12th 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 20th term from the last term

Then m be the 20th term from the last term

Then m = l – (n – 1)d

= 253 – (20 – 1) × 5

= 253 – 19 × 5

= 253 – 95

= 158

∴ 20th term from the end = 158

(ii) A.P. = -2, -4, -6, ……, - 100

a = -2, d = -4 – (-2)

= - 4 + 2

= - 2

l = - 100

∴ Tn = a + (n – 1)d

⇒ - 100 = - 2 + (n – 1) ×(-2)

-100 = -2 – 2n + 2

⇒ + 2n = 100

⇒ n = 100/2

= 50

Let mth term is the 12th 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

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

∴ Tn = 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

= 9th and 10th term

∴ a9 + a10

= 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 ?

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

Here, a = 53, d = 48 – 53 = - 5

∴ Tn = a + (n – 1)d

= 53 + (n – 1) × (-5)

= 53 – 5n + 5

= 58 – 5n

5n = 58

n = 58/5

= 11.3

∴ 12th 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.

In an AP.,

T5 = 19

T13 – T8 = 20

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

∴ T5 = a + 4d = 19

T13 – T8 = (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 7th term exceeds the 5th term by 12.

T3 = 16

T7 – T5 = 12

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

T3 = a + 2d = 16

T7 – T5 = (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 20th term of the A.P. whose 7th term is 24 less than the 11th term, first term being 12.

T1 – T7 = 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

∴ T20 = a + 19d = 12 + 19 × 6

= 12 + 114

= 126

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

T11 = 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

∴ T31 = 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 63rd term.

a7 = 1/9

⇒ a + 6d = 1/9 …(i)

a9 = 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

∴ a63 = a + 62d

= 1/63 + 62(1/63)

= (1 + 62)/63

= 63/63

= 1

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

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

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

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

We have,

a10 = 41

⇒ a + 9d = 41 …(i)

and a15 = 2a7 + 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, a2 = a1 + d

= 1 + 5 = 6

a3 = a2 + d

= 6 + 5 = 11

a4 = a3 + d

= 11 + 5 = 16

∴ The A.P. formed is 1, 6, 11, 16, …

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

T8 = 0

To prove that T38 = 3 ×T18

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

∴ T8 = a+ 7d = 0

⇒ a = -7d

Now, T38 = a + 37d

= -7d + 37d

= 30d

And T18 = a + 17d

= -7d + 17d

= 10d

It is clear that T38 is triple of T18

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

A.P. is 3, 10, 17,…

Here, a = 3,

d – 10 – 3 = 7

T13 = a+ 12d

= 3 = 12 × 7

= 3 + 84

= 87

Let nth term is 84 more then its 13th rem

∴ Tn = 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

∴ 25th 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

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

Tn = a + (n – 1)(-2)

= 9 – 2n + 2

= 11 – 2n

And in second A.P. 24, 21, 18 ,…..

a1 = 24, d1 = 21 – 24 = - 3

Tn = 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 T16 = 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 ?

(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

∴ Tn = 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

an = 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

∴ Tn = 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.

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.

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

⇒ 12 – d2 = - 35

⇒ 1 – d2 = - 35

⇒ d2 = 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.

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.

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

a2 – d2 = a + 29

⇒ (11)2 – d2 = 11 + 29

⇒ 121 – d2 = 40

d2 = 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) n2 + 1

(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) n2 + 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

(i) 2, 7, 12, …… to 10 terms

Here, a = 2, d = 7 – 2 = 5 and n = 10

S10 = 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

∴ S11 = 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?

A.P. = 27, 24, 21, ….

a = 27

d = 24 – 27 = - 3

Sn = 0

Let n terms be there in A.P.

Sn = 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)

(i) 34 + 32 + 30 + …. + 10

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

Tn = a + (n – 1)d

10 = 34 + (n – 1)(-2)

- 24 = -2(n – 1)

= (-24/-2)

= n – 1

⇒ n – 1 = 12

∴ n = 12 + 1 = 13

Sn = 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

∴ Sn = 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, an = 50, find n and Sn

(ii) given a = 7, a13 = 35, find d and S13

(iii) given d = 5, S9 = 75, find a and a9

(iv) given a = 8, an = 62, Sn = 210, find n and d

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

(i) a = 5, d = 3, an = 50

an = 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 Sn = n/2[2a + (n – 1)d]

= 16/2[2 × 5 + (16 – 1) × 3]

= 8[10 + 45]

= 8 × 55

= 440

(ii) a = 7, a13 = 35

an = a + (n – 1)d

35 = 7 + (13 – 1)d

⇒ 35 – 7 = 12d

⇒ 28 = 12d

⇒ d = 28/12 = 7/3 = 2.1/3

and S13 = 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, S9 = 75

an = a + (n – 1)d

a9 = a + (9 – 1) × 5

= a + 40 …..(i)

S9 = 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),

a9 = a + 40 = (-35)/3 + 40

= (- 35 + 120)/3

= 85/3

∴ a = (- 35)/3, a9 = 85/3

(iv) a = 8, an = 62, Sn = 210

an = a + (n – 1)d

62 = 8 + (n – 1)d

(n – 1)d = 62 – 8 = 54 …(i)

Sn = 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

Sn = 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 20th term.

(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)

Sn = 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,

Sn = n/2[2a + (n – 1)d]

⇒ S15 = 15/2[2a + (15 – 1)d]

⇒ 750 = 15/2(2a + 14d)

⇒ a + 7d = 50

⇒ 15 + 7d = 50

⇒ 7d = 35

⇒ d = 5

Now, 20th term = a20 = 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 ?

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

and last term (l) = 350

d = 9

l = Tn = 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 Sn = 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 Sn = 6973

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

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

Sn = n/2[2a + (n – 1)d]

287 = n/2[2 × 1 + (n – 1)3]

574 = n(2 – 3n – 3)

⇒ 3n2 – n – 574 = 0

⇒ 3n2 – 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.

(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 – 3n2

⇒ 3n2 – 53n + 232 = 0

{∵ 232 × 3 = 696}

∴ 696 = - 24 × (-29)

- 53 = - 24 – 29 }

⇒ 3n2 – 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

Sn = n/2[2a + (n – 1)d]

⇒ 78 = n/2[2 × 24 + (n – 1)(-3)]

⇒ 156 = n(48 – 3n + 3)

⇒ 156 = 51n – 3n2

⇒ 3n2 – 51n + 156 = 0

⇒ 3n2 – 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

n4 = a + (n – 1)d

= 24 + 3(-3)

= 24 – 9

= 15

n13 = 24 + 12(-3)

= 24 – 36

= - 12

∴ Sum of 5th 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 a22 is 149.

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

a22 = 149 and n = 22

149 = a + (n – 1)d

= a + 21 × 7

149 = a + 17

⇒ a = 149 – 147 = 2

∴ S22 = 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 6th term is – 11, find the sum of its first 32 terms.

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

T2 = 14, T3 = 18

∴ d = T3 – T2 = 18 – 14

= 4

And a = T1

= 14 – 4

= 10, n = 51

Now, S51 = 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) T3 = 1, T6 = -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

S32 = 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.

S6 = 36

S16 = 256

Sn = n/2[2a + (n – 1)d]

∴ S6 = 6/2[2a + (6 – 1)d] = 36

⇒ 3[2a + 5d] = 36

⇒ 2a + 5d = 12 …(i)

and S16 = 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, S10 = 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 a1, a2, a3, …. From an A.P. where an is defined as an = 3 + 4n. Also find the sum of first 15 terms.

an = 3 + 4n

a1 = 3 + 4 × 1 = 3 + 4 = 7

a2 = 3 + 4 × 2 = 3 + 8 = 11

a3 = 3 + 4 × 3 = 3 + 12 = 15

a4 = 3 + 4 × 4 = 3 + 16 = 19

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

S15 = 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 an = 3 – 4n, show that a1, a2, a3, …. form an A.P. Also find S20.

(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.

(i) an = 3 - 4n

a1 = 3 – 4 × 1 = 3 – 4 = - 1

a2 = 3 – 4 × 2 = 3 – 8 = - 5

a3 = 3 – 4 × 3 = 3 – 12 = - 9

a4 = 3 – 4 × 4 = 3 – 16 = - 13 and so on

Here, a = - 1,

d = - 5 – (-1)

= - 5 + 1

= - 4

Now, S20 = 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

a1 + a2 + a3 + a4 = 1/2(a5 + a6 + a7 + a8)

∴ 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?

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 15th 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 10th and nth terms of the list of numbers 5, 25, 125, ……

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

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

(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

∴ a15 = arn - 1

= √3(1/3)15-1

= √3 × (1/3)14

= √3 × 1/314

(iv) 1/√2, - 2, 4√2, - 16, ……

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

an = arn-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

a10 = arn – 1

= 5 × (5)10-1

= 5 × 59

= 59 + 1

= 510

an = arn-1

= 5 ×5n-1

= 5n – 1 + 1

= 5n

(vi) 3/2, 3/4, 3/8, …..

Here, a = 3/2, r = 3/4 ÷ 3/2

= 3/4 × 2/3

= 1/2

∴ an = ar n-1

= 3/2 × 1/2n-1

= 3 × 1/2 × (1/2)n-1

= 3 × (1/2)n -1 + 1

= 3 × (1/2)n

= 3/(2n)

a6 = 3/2n

= 3/26

= 3/64

(vii) 3, -6, -12, - 24, …..12288

6th term from the end of the list

Here, a = 3, r = - 6 ÷ 3

= - 2,

l = 12288

Now, 6th 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) ?

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

an = 128 = arn-1

⇒ 128 = 2(√2)n-1

⇒ 2(√2)n-1 = 27

(√2)n-1 = 27 – 1 = 26

(√2)n-1 = (√2)12

Comparing, we get

n – 1 = 12

⇒ n = 12 + 1 = 13

∴ 128 is the 13th 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

an = 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 6th 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

an = 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 9th term

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

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 11th term

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

In a G.P.

a8 = 192 and r = 2

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

a8 = arn-1

192 = a(2)n-1

= a28-1

= a27

a = (192/27) = 192/128 = 3/2

a12 = a(r)n-1

a12 = 3/2(2)12-1

= 3/2 × 211

= 3/2 × 2048

= 3072

a12 = 3072

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

In a G.P.

a3 = 24 and a6 = 192,  a10 = ?

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

a6 = ar 6-1

= ar5 = 192 { an = arn-1}

a3 = ar3-1 = ar2 = 24

Dividing, we get

ar5/ar2 = 192/24

r3 = 8 = (2)3

r = 2

Now, ar2 = 24

a × 22 = 24

a = 24/22 = 24/4 = 6

a = 6

Now, a10 = ar10-1 = ar9

= 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.

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

And common ratio (r) = 2

Last term = 384

Let number of terms is n then an = arn-1

384 = 3/4(2)n – 1

2n-1 = (384 × 4)/3 = 512

= 29

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.

Find the value of x

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

∴ x2 = -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

x2 – 12x + 36 = 4x + 36

x2 – 12x – 4x + 36 – 36 = 0

x2 – 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)

x2 + 6x + 9 = x2 + 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.

In a G.P.

a4 = x, a7 = y, a10 = z

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

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

a4 = ar n-1

= ar 4-1

= ar3

= x

Similarly,

d7 = ar6 = y

a10 = ar9 = z

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

y2 = xz

Now, xz = ar3 × ar9

= a2r3+9

= a2r12

y2 = (ar6)2 = a2r12

L.H.S = R.H.S

x, y and z are in G.P.

10. The 5th, 8th and 11th terms of a G.P. are p, q and s respectively. Show that q2 = ps.

In a G.P.

a5 = p, a8 = q and a11 = s

To show that q2 = px

Let a be the first term and r be the common

a5 = arn-1

= ar5-1

= ar4

= p

Similarly, a8 = ar7 = q and

a11 = ar10 = s

q2 = (ar7)2 = ar14

and px = ar4 × ar10

= a2r 4+ 10

= a2r14

Hence, q2 = ps

11. If a, b, c are in G.P., then show that a2, b2, c2 are also in G.P.

a, b, c are in G.P.

Show that a2, b2, c2 are also in G.P.

∵ a, b, c are in G.P.., then

b2 = ac …(i)

a2, b2, c2 will be in G.P.

if (b2)2 = a2 × c2

⇒ (ac)2 = a2c2 [From (i)]

⇒ a2c2 = a2c2 which is true.

Hence, proved.

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

a, b and c are in A.P.

then, 2b = a + c

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

if (3b)2 = 3a.3c

if 32b = 3a+ 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 10ax + 10, 10bx + 10 , 10cx + 10, x ≠ 0, are in G.P.

a, b, c are in A.P.

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

∵ a, b, c are A.P.

∵ 2b = a + c …(i)

Now,

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

If (10bx + 10)2 = (10ax + 10) × (10cx + 10)

If 102bx + 20 = 10ax+10+cx+10

If 102bx + 20 = 10ax+cx+20

Comparing,

If 2bx + 20 = ax + cx + 20

If 2bx = ax + cx

If 2b = a + c

Which is given.

14. If a, a2 + 2 and a3 + 10 are in G.P., then find the values(s) of a.

a, a2 + 2 and a3 + 10 are in G.P.

∵ (a2 + 2)2 = a(a3 + 10)

⇒ a4 + 4a2 + 4 = a4 + 10a

⇒ 4a2 – 10a + 4 = 0

⇒ 2a2 – 5a + 2 = 0

⇒ 2a2 – 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 xm. If its 8th term is x52. Then find the value of m.

In a G.P.,

First term (a1) = x-4 …(i)

Second term (a2) = xm

Eighth term = (a8) = x52

r = a2/a1 = xm/x-4 = xm + 4 …..(ii)

Now, a8 = arn-1 = ar 8-1 = ar7

x52 = x-4 × r7 = x-4 × x7(m+4)

x52 = x-4 + 7m + 28 = x7m + 24

Comparing,

52 = 7m + 24

⇒ 7m = 52 – 24 = 28

⇒ m = 28/7 = 4

Hence m = 4

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

In a G.P.

4th term (a4) = 54

And 7th term (a7) = 1458

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

ar3 = 54 and ar6 = 1458

Dividing,

(ar6/ar3) = 1458/54

⇒ r3 = 27 = (33)

∴ r = 3

Now, ar3 = 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.

In a G.P.

an is square of a2 i.e., an = (a2)2

a1 = - 3

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

a4 = ar n-1

= ar4-1

= ar3

and a2 = ar2-1

= ar

∴ ar3 = (ar)2

⇒ ar3 = a2r2

r3/r2 = a2/a = r = a = - 3 (∵ a1 = - 3)

Now, a7 = ar 7 -1

= ar6

= (-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.

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 a3 = 1

⇒ a = 1

Substitute the value of a

∴ 1(1/r + 1 + r) = 39/10

(1 + r + r2)/r = 39/10

10 + 10r + 10r2 = 39r

⇒ 10r2 + 10r – 39r + 10 = 0

⇒ 10r2 – 29r + 10 = 0

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

⇒ 10r2 – 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.

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.

∴ b2 = ac

∴ (a + 4)2 = (a – d + 1) (a + d + 19)

⇒ a2 + 8a + 16 = a2 + ad + 19a – ad – d2 – 19d + a + d + 19

⇒ a2 + 8a + 16 = a2 – d2 – 18d + 20a + 19

⇒ 8a + 16 = 20a – 18d – d2 + 19

⇒ 8a + 16 – 20a + 18d + d2 – 19 = 0

⇒ d2 + 18d – 12a – 3 = 0

⇒ d2 + 18d – 12 × 5 – 3 = 0

⇒ d2 + 18d – 60 – 3 = 0

⇒ d2 + 18d – 63 = 0

{∵ - 63 = + 21 × -3 + 18 = + 21 – 3}

⇒ d2 + 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.

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 + r2

⇒ r2 – 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.

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 + 2r2

⇒ 2r2 – 5r + 2 = 0

⇒ 2r2 – 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 ab-c + bc-a + ca-b

(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.

∴ b2 = 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 b2 = ac

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

Now, ab-c + bc-a + ca – b

Since, a = b = c

∴ b – c = 0, c – a = 0 and a – b = 0

∴ a0 + b0 + c0 = 1 + 1 + 1 (∵ x0 = 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.

In a G.P.,

The first term = a

And common ratio = r

G.P. is a, ar, ar2

Squaring we get

a2, a2r2, a2r4 are in G.P.

if b2 = 4ac

⇒ (a2r2)2 = a2 × a2r4

⇒ a4r4 = a4r4

Which is true

The first term is a2

And common ratio is r2

The nth term will be

an = arn-1

= a2(rn-1)2

= a2r2n-2

24. Show that the products of the corresponding terms of two G.P,’s a, ar, ar2, ……, arn-1 and A, AR, AR2,……AR n-1 form a G.P. and find the common ratio.

It has to be proved that the sequence

aA, arAR, ar2AR2, ….. arn-1 ARn-1 and forms a G.P.

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

and (Third term/Second term) = (ar2 AR2)/(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 Ka, Kb, Kc 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.

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

∴ b2 = ac

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

If (1/b)2 = 1/a × 1/c

⇒ 1/b2 = 1/ac

⇒ ac = b2 (By cross multiplication)

Which is given

Hence proved.

(ii) k is any positive number Ka, Kb, Kc are in G.P.

Then (Kb)2 = Ka × Kc

⇒ K2b = Ka + 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 = ARr-1

These will be in G.P.

If (ARq-1)2 = ARp-1 × ARr-1

If A2R2q-2 = A2Rp-1+r -1

If A2R2q-2 = A2Rp+r-2

If R2q-2 = Rp+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) a3, b3, c3

(ii) a2 + b2, ab + bc, b2 + c2

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

∴ b2 = ac

a3, b3, c3 are in G.P.

If (b3)2 = a3 × c3

If (b2)3 = (a × c)3

If b2 = ac

Which is given.

Hence proved.

(ii) a2 + b2, ab + bc, b2 + c2 will be in G.P.

a2 + b2, ab + bc, b2 + c2 are in G.P.

If (ab + bc)/(a2 + b2) = (b2 + c2)/(ab + bc)

i.e., if {a(ar) + ar(ar2)}/{a2 + (ar)2} = {(ar)2 + (ar2)2 }/{a(ar) + ar(ar2)}

⇒ If {a2r(1 + r2)/a2(1 + r2)} = a2r2 (1 + r2)}/{a2r(1 + r2)}

⇒ If r = r which is true.

∴ a2 + b2, ab + bc, b2 + c2 are in G.P.

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

(i) a2 + b2, b2 + c2, c2 + d2 are in G.P.

(ii) (b – c)2 + (c – a)2 + (d – b)2 = (a – d)2

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

Let r be the common ratio, then a = a

b = ar, c = ar2, d = ar3

(i) a2 + b2, b2 + c2, c2 + d2 are in G.P.

∴ a2 + b2 = a2 + a2r2 = a2(1 + r2)

b2 + c2 = a2r2 + a2r4 = a2r2(1 + r2)

c2 + d2 = a2r4 + a2r6 = a2r4(1 + r2)

a2 + b2, b2 + c2, c2 + d2 will be in G.P.

if (b2 + c2)2 = (a2 + b2)(c2 + d2)

Now, (b2 + c2)2 = [a2r2(1 + r2)]2

= a4r4(1 + r2)2 …(i)

(a2 + b2)(c2 + d2) = [a2(1 + r2)][a2r4(1 + r2)]2

= a4r4(1 + r2)2 …(ii)

From (i) and (ii),

(b2 + c2)2 = (a2 + b2)(c2 + d2)

Hence, a2 + b2, b2 + c2, c2 + d2 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 – ar2)2 + (ar2 – a)2 + (ar3 – ar)2

= a2r2(1 – r)2 + a2(r2 – 1)2 + a2r2(r2 – 1)2

= a2[r2(1 – r2 – 2ar) + r4 – 2r2 + 1 + r2(r4 – 2r2 + 1)]

= a2[r2 – r4 – 2ar3 + r4 – 2r2 + 1 + r6 – 2r4 + r2)]

= a2(r6 – 2r3 + 1)

R.H.S. = (a – d)2 = (a – ar3)2

= a2(1 – r3)2

= a2(1 + r6 – 2r2)

= a2[r6 – 2r3 + 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, a2, - a3, (a ≠ - 1)

(vii) n terms of the G.P. x3, x5, x7, ……(x ≠ ±1)

(i) 2 + 6 + 18 + … 20 terms

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

S20 = a(rn – 1)/(r – 1)

= 2(320 – 1)/(3 – 1)

= {2(320 – 1)}/2

= 320 – 1

(ii) 1 + √3 + 3 + … 10 terms

Here, a = 1, r = √3, n = 10, r > 1

S10 = a(rn – 1)/(r – 1)

= 1[(√3)10 – 1]/(√3 – 1)

= {(√310 – 1)(√3 + 1)}/{(√3 – 1)(√3 + 1)}

(Rationalising the denominator)

= {(35 – 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

S6 = a(1 – rn)/(1 – r) = 1[1 – (-2/3)6]/(1 + 2/3)

= 3/5 {1 – (-2)6/36}

= 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)

Sn = a(1 – rn)/(1 – r) = 1[1 – (2/3)n]/(1 – 2/3)

= (1 × 3)/1 [1 – (2/3)n]

= 3[1 – (2/3)n]

and S5 = 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

Sn = a(rn – 1)/(r – 1)

⇒ Sn = √7[(√3)n – 1]/(√3 – 1)

Rationalising the denominator,

⇒ Sn = √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, a2, -a3, ….(a ≠ -1) upto n terms

Here, a = 1, r = - a,

Sn = a(1 – rn)/(1 – r)

= 1[1 – (-a)n]/[1 – (-a)]

= [1 – (-a)n]/(1 + a)

(vii) x3, x5, x7, …..(x ≠ ±1)

Here, a = x3, r = x2

∴ Sn = a(1 – rn)/(1 – r)

= x3[1 – (x2)n]/(1 – x2) if r < 1

= x3(1 – x2n)/(1 – x2)

or Sn = a(rn – 1)/(1 – r)

= x3[(x2)n – 1]/(x2 – 1)

= x3(x2n – 1)/(x2 – 1)

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

Here, a = √2, r = √3, r > 1

∴ S10 = a(rn – 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)

Given :

81 – 27 + 9 ……. – (1/27)

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

Sn = (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.

In a G.P.

Tn = 128

Sn = 255

r = 2,

Let a be the first term, then

Tn = ar n-1

⇒ 128 = a2n-1

⇒ a = 128/(2n-1) …(i)

Sn = a(rn – 1)/(r – 1)

⇒ 255 = a(2n – 1)/(2 – 1)

⇒ 255 = a(2n – 1)

⇒ a = 255/(2n – 1) ….(ii)

From (i) and (ii),

255/(2n – 1) = 128/(2n – 1)

⇒ 255 × 2n-1

= 128(2n – 1)

⇒ 255 × 2n-1 = 128 × 2n – 128

(255 × 2n)/2 = 128 × 2n – 128

⇒ 255 × 2n = 256 × 2n – 256

⇒ 256 × 2n – 255 × 2n = 256

⇒ 2n = 256 = 28

Comparing, we get

n = 8

Now, 128 = a.27

⇒ 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.

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

∴ S6 = 9 × a3

⇒ Sn = a(rn – 1)/(r – 1)

S6 = a(r6 – 1)/(r – 1) and S3 = a(r3 – 1)/(r – 1)

∴ a(r6 – 1)/(r – 1) = 9 × a(r3 – 1)/(r – 1)

⇒ r6 – 1 = 9(r3 – 1)

⇒ (r6 – 1)/(r3 – 1) = 9

⇒ (r3 + 1)(r3 – 1)}/(r3 – 1) = 9

⇒ r3 + 1 = 9

⇒ r3 = 9 – 1 = 8 = (2)2

∴ r = 2

∴ Common ratio = 2

6. (i) How many terms of the G.P. 3, 32, 33, …… 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 ?

In G.P.

(i) 3, 32, 33, ….

Sum = 120, Here, a = 3, r = 32/3 = 3, r > 1

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

Sn = a(rn – 1)/(r – 1) = 120

⇒ 3(3n – 1)/(3 – 1) = 120

⇒ 3(3n – 1)/2 = 120

⇒ 3n – 1 = (120 × 2)/3 = 80

3n = 80 + 1 = 81 = 34

∴ 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

Sn = a(rn – 1)/(r – 1) = 341

⇒ 1(4n – 1)/(4 – 1) = 341

⇒ 1(4n – 1)/(4 – 1) = 341

⇒ (4n – 1)/3 = 341

⇒ 4n – 1 = 341 × 3 = 1023

4n = 1023 + 1

= 1024

= 210

= 45

∴ n = 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) ?

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

Sn = a(rn – 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

= 210 or (√2)20

Comparing, we get

n = 20

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

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

∴ Sn = a(1 – rn)/(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 2nd and 5th terms of a geometric series are – (1/2) and sum 1/16 respectively. Find the sum of the series upto 8 terms.

In a G.P.

a2 = -(1/2) and a5 = 1/16

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

∴ a2 = arn- 1

= ar 2-1

ar = -1/2 ….(i)

a5 = ar5-1

= ar4 = 1/16 ...(ii)

Dividing (ii) by (i),

r3 = 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, S8 = a(1 – rn)/(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 8th term is 1/81. Find the sum of its first 10 terms.

From the question it is given that,

First term (a) = 27

8th term a8 = 1/81

Then, an = arn-1

a8 = ar8-1 = 1/81

arn-1 = ar8 -1

= ar7

= 1/81

ar7 = 1/81

⇒ 27r7 = 1/81

⇒ r7 = 1/(81 × 27)

r7 = 1/2187

r7 = 1/(3)7

∴ r = 1/3

Now, S10 = a(1 - rn)/(1 – r)

= 27[1 – (1/3)10]/(1 – 1/3)

= 27 [1 – 1/310]/((3 – 1)/3)

= ((27 × 3)/2)[1 – 1/310]

= 81/2 [1 – 1/310]

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.

Common ratio of G.P. = 3

And last term = 486

And sum of terms = 728

Sn = a(rn – 1)/(r – 1) = a(3n – 1)/(3 – 1)

= a(3n – 1)/2

= 728

a(3n – 1) = 728 × 2 = 1456 ...(i)

l = 486

⇒ arn-1 = a3n-1 = 486

⇒ a[3n/3] = 486

⇒ a3n = 486 × 3 = 1458

But a(3n – 1) = 1456 [From (i)]

a3n – 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.

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

⇒ 7rn-1 = 448

rn-1 = 448/7 = 64 …(i)

and sum = a(rn – 1)/(r – 1) = 889

⇒ 7(rn – 1)/(r – 1) = 889

⇒ (rn – 1)/(r – 1) = 889/7 = 127 …(ii)

From (i),

rn/r = 64

⇒ rn = 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.

In a G.P.

Common ratio = 3

S7 = 2186 = a(rn – 1)/(r – 1)

⇒ 2186 = a(3n – 1)/(3 – 1)

⇒ 2186 = a(3n – 1)/2

⇒ 4372 = a(37 – 1)

⇒ 4372 = a(2187 – 1)

⇒ 4372 = 2186a

⇒ a = 4372/2186 = 2

a3 = ar3-1

= ar2

= 2 ×32

= 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.

In a G.P.

First term (a) = 5

S3 = 31/5

= a(r3 – 1)/(r – 1)

= 5(r3 – 1)/(r - 1)

(r3 – 1) /(r – 1) = 31/(5 × 5) = 31/25

⇒ {(r – 1)(r2 + r + 1)}/(r – 1) = 31/25

r2 + r + 1 = 31/25

⇒ 25r2 + 25r + 25 = 31

⇒ 25r2 + 25r – 6 = 0

⇒ 25r2 + 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.

S3 ÷ S6 = 125 : 152

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

a(r3 – 1)/(r – 1) : a(r6 – 1)/(r - 1) = 125 : 152

(r3 – 1) : (r6 – 1) = 125 : 152

(r3 – 1) : (r3 + 1)(r3 – 1) = 125 : 152

1 : (r3 + 1) = 125 : 152

(r3 + 1) × 125 = 152 × 1

125r3 + 125 = 152

125r3 = 152 – 125 = 27

r3 = 27/125 = (3/5)3

∴ r = 3/5

∴ Common ratio = 3/5

16. Evaluate

Here n = 1, 2, 3, …….., 50

Sn = 21 - 1 + 22 – 1 + 23 – 1 + 24 – 1… 250 – 1

Sn = (21 + 22 + 23 + 24 + … 250) – 1 × 50

Sn = 2 + 4 + 8 + 16 … 250 – 50

Sn = a(rn – 1)/(r – 1) – 50

= 2(250 – 1)/(2 – 1) – 50

Sn = 2 × 250 – 2 – 50

Sn = 251 – 52

17. Sum the series x(x + y) + x2(x2 + y2) + x3(x3 + y3) … to n terms.

Given

Sn = x(x + y) + x2(x2 + y2) + x3(x3 + y3) … n terms

18. Find the sum of the series 1 + (1 + x) + (1 + x + x2) + …… to n terms, x ≠ 1.

1 + (1 + x) + (1 + x + x2) + …… 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 + ….

(i) Sn = 7 + 77 + 777 + … n terms

(ii) Sn = 8 + 88 + 888 + … n terms
(iii) Sn = 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.

(b) an A.P. with d = 4

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

= - 6 + 10

= 4

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

(a) 32

(b) 35

(c) 38

(d) 185

(a) 32

10th 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 30th term of the A.P. 10, 7, 4, ….. is

(a) 87

(b) 77

(c) – 77

(d) – 87

(c) -77

30th term of A.P. 10, 7, 4, …. is

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

= 10 + (30 – 1) × (-3)

= 10 + 29(-3)

= 10 – 87

= - 77

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

(a) 28

(b) 22

(c) – 38

(d) – 48

(b) 22

Given -3, -(1/2), 2, …

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

11th term = a + (n – 1)d

= - 3 + (11 – 1) × 5/2

= - 3 + 10 × 5/2

= - 3 + 25

= 22

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

(a) 37

(b) 40

(c) 43

(d) 58

(b)

4th 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, 4th term from the end = l – (n – 1)d

= 49 – (4 – 1) × 3

= 49 – 9

= 40

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

(a) 49

(b) 85

(c) 88

(d) 110

(c) 88

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

Here, a = 7, d = 10 – 7 = 3, l = 130

15th 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 a18 – a13 is

(a) 5

(b) 20

(c) 25

(d) 30

(c) 25

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

a18 – a13 = a + 17d – a – 12d

= 5d

= 5 × 5

= 25

8. In an A.P., if a18 – a14 = 32 then the common difference is

(a) 8

(b) – 8

(c) – 4

(d) 4

(a) 8

If a18 – a14 = 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, an = 4, then a is

(a) 6

(b) 7

(c) 20

(d) 28

(d) 28

In an A.P., d = - 4, x = 7, an = 4 then a = ?

an = a(n – 1)d = 4

a7 = 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 an will be

(a) 0

(b) 3.5

(c) 103.5

(d) 104.5

(b) 3.5

In an A.P.

a = 3.5, d = 0, n = 101, then an = ?

an = a101 = 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, an = 7.2, then n is

(a) 1

(b) 3

(c) 4

(d) 5

In an A.P.

a = - 7.2, d = 3.6, an = 7.2, n = ?

an = 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) 9th

(b) 10th

(c) 11th

(d) 12th

(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 10th 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

(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

(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 21st term of an A.P. whose first two terms are – 3 and 4 is

(a) 17

(b) 137

(c) 143

(d) – 143

(b) 137

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

a = -3, d = 4 – (-3)

= 4 + 3 = 7

21st term = a + 20d

= - 3 + 20(7)

= -3 + 140

= 137

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

(a) 30

(b) 33

(c) 37

(d) 38

(b) 33

In an A.P.

2nd term = 13

⇒ a + d = 13 ...(i)

5th 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

7th 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

(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

(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, an = 20 and Sn = 399, then n is

(a) 19

(b) 21

(c) 38

(d) 42

(c) 38

In an A.P., a = 1, an = 20, Sn = 399, n is ?

an = 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 Sn = 200, then n is equal to

(a) 50

(b) 40

(c) 32

(d) 25

(d) 25

In an A.P.

a = - 5, l = 21, Sn = 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 S8 = 192, then d is

(a) 8

(b) 7

(c) 6

(d) 4

(c) 6

In an A.P.

a = 3, S8 = 192, d = ?

22. The sum of first five multiples of 3 is

(a) 45

(b) 55

(c) 65

(d) 75

(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

(c) 30

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

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

l = an = 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

(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

(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.

(c) G.P. with r = 3

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

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

(a) 64

(b) – 64

(c) 128

(d) – 128

(c) 128

11th of the G.P.

1/8, -(1/4), 2, -1,…… is

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

(a) 162

(b) 486

(c) 54

(d) 1458

(a) 162

5th 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

(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) 12th

(b) 11th

(c) 10th

(d) 9th

(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

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

(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

(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.

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, ……

(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

a2 = 4 – 1/3 = 11/3

a3 = 11/3 – 1/3 = 10/3

a4 = 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.

Tn of an A.P. = 6n + 2

T1 = 6 × 1 + 2 = 6 + 2 = 8

T2 = 6 × 2 + 2 = 12 + 2 = 14

T3 = 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 16th term and the nth.

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.

T16 = a + (n – 1)d

= 9 + (16 – 1) × 3

= 9 + 15 × 3

= 9 + 45

= 54

and Tn = a + (n – 1)d

= 9 + (n – 1) × 3

= 9 + 3n – 3

= 3n + 6

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

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, 6th term from the end = l – (n – 1)d)

= 40 – (6 – 1)(-3)

= - 40 – (5)(-3)

= - 40 + 15

= - 25

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

In an A.P.

a8 = 31, a15 = a11 + 6

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

a8 = a + (n – 1)d = 31

⇒ a + 7d = 31 …(i)

Similarly,

a15 = 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 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43, then find the wth term.

In an A.P.

a17 = 2 × a8 + 5

a11 = 43, find an

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

a11 = a + (n – 1)d = a + (11 – 1)d

= a + 10d = 43 …(i)

Similarly,

a17 = 2 × a+ 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, an = a + (n – 1)d

= 3 + 4(n – 1)

= 3 + 4n – 4

= 4n – 1

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

In an A.P.

a19 = 3 × a6 and a= 19

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

a9 = a + (n – 1)d

= a + (9 – 1)d = a + 8d

a + 8d = 19 …(i)

Similarly,

a19 = 3 × a6

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

In an A.P.

a= 4, a9 = - 8, which term of A.P. will be zero

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

a3 = 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 nth 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?

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

∴ an = 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 21st term.

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

In an A.P.

24th term = 2 × 10th term

To show that 72nd term = 4 × 15th term

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

24th term = a + (24 – 1)d = a + 23d

And 10th term = a + 9d

∴ a + 23d = 2(a + 9d)

⇒ a + 23d = 2a + 18d

⇒ 2a – a = 23d – 18d

⇒ a = 5d …(i)

And 72nd term = a + 17d

And 15th 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 72nd term is 4 times the 15th term.

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

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

∴ an = a + (n – 1)d

Let nth term be first negative term, then

an = 20 + (n – 1) (-3/4)

⇒ an = 20 + (n – 1)(-3/4)

⇒ an = 20 – 3/4n + 3/4

Now, an < 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

∴ 28th 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)

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?

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

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

∴ l = an = 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.

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(a2 – d2) = 8

⇒ -1[(-1)2 – d2] = 8

⇒ 1 – d2 = 8/-1 = - 8

d2 = 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.

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 n2 + n + 1. Justify your answer.

nth term of an A.P. can’t be n2 + n + 1

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

We find then

a1 = 12 + 1 + 1 = 1 + 1 + 1 = 3

a2 = 22 + 2 + 1 = 4 + 2 + 1 = 7

a3 = 32 + 3 + 1 = 9 + 3 + 1 = 13

a4 = 42 + 4 + 1 = 16 + 4 + 1 = 21

We see that,

d = a2 – a1 = 7 – 3 = 4

d = a3 – a2 = 13 – 7 = 6

d = a4 – a3 = 21 – 13 = 8

We see that d is not constant

∴ It is not an A.P.

Hence, an ≠ n2 + n + 1

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

nth term is 15 – 4n

So, an = 15 – 4n

Giving values in place of n from 1 to 20

We get,

a1 = 15 – (4 × 1) = 15 – 4 = 11

a2 = 15 – (4 × 2) = 15 – 8 = 7

a3 = 15 – (4 × 3) = 15 – 12 = 3

a4 = 15 – (4 × 4) = 15 – 16 = - 1

Then, a20 = 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)

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

l = - 49.1/2 = - 99/2

an = 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, Sn = n/2[2a + (n – 1)d]

S28 = 28/2 [2 × 18 + (28 – 1)(-5/2)]

S28 = 14[36 + (27 × -5/2)]

= 14[36 – 135/2]

S28 = 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.

(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

∴ Sn = 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]

1/2n2 – 25/n + 50 = 0

⇒ n2 – 25n + 100 = 0

{∵ 100 = - 20 × -5, - 25 = - 20 – 5}

⇒ n2 – 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)

Sn = n/2[2a + (n – 1)d]

⇒ 155 = n/2[2 × 2 + (n – 1) × 3]

⇒ 155 × 2 = n[4 + 3n – 3]

⇒ 310 = n(3n + 1) = 3n2 + n

∴ 3n2 + n – 310 = 0

3n2 – 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 6th term to the 10th term is 7 : 13, then find the sum of first 20 terms of this A.P.

3rd term of an A.P. = 5

Ratio in 6th term and 10th term = 7 : 13

Find S20

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

∴ a3 = a + (n – 1)d

⇒ a + (3 – 1)d = 5

⇒ a + 2d = 5 …(i)

Similarly,

a6 = a + 5d and a10 = 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.

In an A.P.

First term (a) = 2

Last term (l) = 29

Sum of terms = 155

l = an = a + (n – 1)d

⇒ 29 = 2 + (n – 1)d

⇒ 29 – 2 = d(n – 1)

⇒ d(n – 1) = 27 …(i)

Sn = 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 25th term.

Sum of first 14 terms = 1505

First term (a) = 10

Find 25th term S14 = 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, a25 = 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 25th term is this A.P.

Sn = 3n2 + 4n

Sn – 1 = 3(n – 1)2 + 4(n – 1)

= 3(n2 – 2n + 1) + 4(n – 1)

= 3n2 – 6n + 3 + 4n – 4

= 3n2 – 2n – 1

Now, an = Sn – Sn -1

= (3n2 + 4n) – (3n2 – 2n – 1)

= 3n2 + 4n – 3n2 + 2n + 1

= 6n + 1

a25 = 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.

In an A.P.

Sum of first 10 terms = - 150

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

Sum of first 10 terms = - 150

S10 = n/2[2a + (n – 1)d]

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

⇒ 10a + 45d = - 150

S20 = 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 4m2  – m. If its nth term is 107, find the value of n. Also find the 21st term of this A.P.

Sm = 4m2 – m

Sn = 4n2 – n

and Sn – 1 = 4(n – 1)2 – (n – 1)

= 4[n2 – 2n + 1] – n + 1

= 4n2 – 8n + 4 – n + 1

= 4n2 – 9n + 5

Now, an = Sn – Sn-1

= 4n2 – n – 4n2 + 9n - 5

= 8a – 5

Now, an = 107

8n – 5 = 107

8n = 107 + 5 = 112

n = 112/8 = 14

and an = 8n – 5

a21 = 8 × 21 – 5

= 168 – 5

= 163

27. Find the geometric progression whose 4th term is 54 and 7th term is 1458.

In  a G.P.

a4 = 54

a7 = 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 7th term.

In G.P.

a4 = (a2)2,   a1 = - 3

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

an = ar n-1

a4 = ar3

a2 = ar

ar3 = (ar)2

ar3 = a2r2

r = a

a1 = - 3

d = - 3

a7 = ar 7 – 1

= ar6

= - 3 × (-3)6

= - 3 × 729

= - 2187

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

In a G.P.

a4 = x,  a10 = y,   a16 = 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,

an = arn – 1

a4 = ar4 – 1

a4 = ar3 = x

a10 = ar9 = y

a16 = ar15 = z

x, y, z are in G.P.

If y2 = xy

Substitute the value of x and y,

y2 = (ar9)2

y2 = a2 r18

Then, xz = ar3 × ar15

= a1+1 r3+15 [from am × an = am+n]

= a2 r18

So, y2 = 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 ?

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

Sn = 3069/512

Here, a = 3, r = 1/2

Let n be the number of terms

Sn = a(1 – rn)/(1 – r)

Comparing, we get n =10

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

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.

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

Required sum = S50

= 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, Sn = a(1 - rn)/(1 - r)

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