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

ML Aggarwal Solutions for Class 10 Maths Chapter 9 Arithmetic and Geometric Progression are provided here. BYJU’S tutor team has prepared solutions, which help students attain good marks in Maths. From the exam point of view, the solutions are solved in a simpler manner where students can secure an excellent score by solving the ML Aggarwal textbook. Solutions that are provided here will help you in getting acquainted with a wide variety of questions and thus, develop problem-solving skills. Download pdf of ML Aggarwal Solutions for Class 10 Maths Chapter 9 in their respective links

Chapter 9 – Arithmetic and Geometric Progression. Arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant. Difference here means the second minus the first. Geometric progression is defined as a sequence of numbers, where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In these ML Aggarwal Solutions for Class 10 Maths Chapter 9, students are going to learn solving different types of problems and topics of Arithmetic and Geometric Progression.

ML Aggarwal Solutions for Class 10 Maths Chapter 9 :-Click Here to Download PDF

 

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Access answers to ML Aggarwal Solutions for Class 10 Maths Chapter 9 Arithmetic and Geometric Progression

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, …

Solution:-

From the question,

The first term a = 3

Then, difference d = 1 – 3 = – 2

– 1 – 1 = – 2

– 3 – (-1 ) = – 3 + 1 = -2

Therefore, common difference d = – 2

(ii) 1/3, 5/3, 9/3, 13/3, ….

Solution:-

From the question,

The first term a = 1/3

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

9/3 – 5/3 = (9 – 5)/3 = 4/3

13/3 – 9/3 = (13 – 9)/3 = 4/3

Therefore, common difference d = 4/3

(iii) -3.2, -3, -2.8, -2.6, ….

Solution:-

From the question,

The first term a = -3.2

Then, difference d = -3 – (-3.2) = -3 + 3.2 = 0.2

-2.8 – (-3) = -2.8 + 3 = 0.2

-2.6 – (-2.8) = -2.6 + 2.8 = 0.2

Therefore, common difference d = 0.2

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

(i) a = 10, d = 10

Solution:-

From the question it is given that,

First term a = 10

Common difference d = 10

Then the first four terms are = 10 + 10 = 20

20 + 10 = 30

30 + 10 = 40

Therefore, first four terms are 10, 20, 30 and 40.

(ii) a = -2, d = 0

Solution:-

From the question it is given that,

First term a = -2

Common difference d = 0

Then the first four terms are = -2 + 0 = -2

-2 + 0 = -2

-2 + 0 = -2

Therefore, first four terms are -2, -2, -2 and -2.

(iii) a = 4, d = -3

Solution:-

From the question it is given that,

First term a = 4

Common difference d = -3

Then the first four terms are = 4 + (-3) = 4 – 3 = 1

1 + (-3) = 1 – 3 = – 2

-2 + (-3) = -2 – 3 = – 5

Therefore, first four terms are 4, 1, -2 and -5.

(iv) a = ½, d = -1/6

Solution:-

From the question it is given that,

First term a = ½

Common difference d = -1/6

Then the first four terms are = ½ + (-1/6) = ½ – 1/6 = (3 – 1)/6 = 2/6 = 1/3

1/3 + (-1/6) = 1/3 – 1/6 = (2 – 1)/6 = 1/6

1/6 + (-1/6) = 1/6 – 1/6 = 0

Therefore, first four terms are ½, 1/3, 1/6 and 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, …

Solution:-

From the question it is given that,

First term a = 4

Then, difference d = 10 – 4 = 6

16 – 10 = 6

22 – 16 = 6

Therefore, common difference d = 6

Hence, the numbers are form A.P.

(ii) -2, 2, -2, 2, …

Solution:-

From the question it is given that,

First term a = -2

Then, difference d = -2 – 2 = – 4

-2 – 2 = -4

2 – (-2) = 2 + 2 = 4

Therefore, common difference d is not same in the given numbers.

Hence, the numbers are not form A.P.

(iii) 2, 4, 8, 16, …

Solution:-

From the question it is given that,

First term a = 2

Then, difference d = 4 – 2 = 2

8 – 4 = 4

16 – 8 = 8

Therefore, common difference d is not same in the given numbers.

Hence, the numbers are not form A.P.

(iv) 2, 5/2, 3, 7/2, …

Solution:-

From the question it is given that,

First term a = 2

Then, difference d = 5/2 – 2 = (5 – 4)/2 = ½

3 – 5/2 = (6 – 5)/2 = ½

7/2 – 3 = (7 – 6)/2 = ½

Therefore, common difference d = ½

Hence, the numbers are form A.P.

(v) – 10, -6, -2, 2, …

Solution:-

From the question it is given that,

First term a = -10

Then, difference d = -6 – (- 10) = – 6 + 10 = 4

-2 – (-6) = – 2 + 6 = 4

2 – (-2) = 2 + 2 = 4

Therefore, common difference d = 4

Hence, the numbers are form A.P.

(vi) 12, 32, 52, 72, …

Solution:-

From the question it is given that,

First term a = 12 = 1

Then, difference d = 32 – 12 = 9 – 1 = 8

52 – 32 = 25 – 9 = 16

72 – 52 = 49 – 25 = 24

Therefore, common difference d is not same in the given numbers.

Hence, the numbers are not form A.P.

(vii) 1, 3, 9, 27, …

Solution:-

From the question it is given that,

First term a = 1 = 1

Then, difference d = 3 – 1 = 2

9 – 3 = 6

27 – 9 = 18

Therefore, common difference d is not same in the given numbers.

Hence, the numbers are not form A.P.

(viii) √2, √8, √18, √32, …

Solution:-

Given numbers can be written as, √2, 2√2, 3√2, 4√2, …

From the question it is given that,

First term a = √2

Then, difference d = 2√2 – √2 = √2

3√2 – 2√2 = √2

4√2 – 3√2 = √2

Therefore, common difference d = √2

Hence, the numbers are form A.P.

(ix) a, 2a, 3a, 4a, …

From the question it is given that,

First term a = a

Then, difference d = 2a – a = a

3a – 2a = a

4a – 3a = a

Therefore, common difference d = a

Hence, the numbers are form A.P.

(x) a, 2a + 1, 3a + 2, 4a + 3, …

From the question it is given that,

First term a = a

Then, difference d = (2a + 1) – a = 2a + 1 – a = a + 1

(3a + 2) – (2a + 1) = 3a + 2 – 2a – 1 = a + 1

(4a + 3) – (3a + 2) = 4a + 3 – 3a – 2 = a + 1

Therefore, common difference d = a +1

Hence, the numbers are form A.P.

Exercise 9.2

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

Solution:-

From the question it is given that,

nth term is 7 – 3k

So, Tn = 7 – 3n

Now, we start giving values, 1, 2, 3, … in the place of 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

Therefore, A.P. is 4, 1, -2, -5, …

So, 20th term is – 53

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

(i) 1, 6, 11, 16, …; a20

Solution:-

From the question,

The first term a = 1

Then, difference d = 6 – 1 = 5

11 – 6 = 5

16 – 11 = 5

Therefore, common difference d = 5

From the formula, an = a + (n – 1)d

So, a20 = a + (20 – 1)d

= 1 + (20 – 1)5

= 1 + (19)5

= 1 + 95

= 96

Therefore, a20 = 96

(ii) -4, -7, -10, -13, …, a25, an

Solution:-

From the question,

The first term a = -4

Then, difference d = -7 – (-4) = – 7 + 4 = -3

-10 – (-7) = -10 + 7 = -3

-13 – (-10) = -13 + 10 = -3

Therefore, common difference d = -3

From the formula, an = a + (n – 1)d

So, a25 = a + (25 – 1)d

= -4 + (25 – 1)(-3)

= -4 + (24)-3

= – 4 – 72

= -76

Therefore, a25 = -76

Now, an = a + (n – 1)d

an = -4 + (n – 1)-3

= -4 – 3n + 3

= -1 – 3n

3. Find the nth term and the 12th term of the list of numbers: 5, 2, – 1, – 4, …

Solution:-

From the question,

The first term a = 5

Then, difference d = 2 – 5 = – 3

-1 – 3 = -3

– 4 – (-1) = -4 + 1 = -3

Therefore, common difference d = -3

From the formula, an = a + (n – 1)d

Tn = a + (n – 1)d

= 5 + (n – 1)-3

= 5 – 3n + 3

= 8 – 3n

So, T12 = a + (12 – 1)d

= 5 + (12 – 1)(-3)

= 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.

Solution:-

From the question it is given that,

The first term a = 7

Then, common difference d = 3

Tn = a + (n – 1)d

So, T8 = a + (8 – 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.

Solution:-

From the question it is given that,

The 18th term = -5

Then, common difference d = -3

Tn = a + (n – 1)d

So, T18 = a + (18 – 1)d

-5 = a + (18 – 1)(-3)

-5 = a + (17)(-3)

-5 = a – 51

a = 51 – 5

a = 46

Therefore, first term a = 46


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

Solution:-

From the question it is given that,

The 10th term = 0

Then, first term a = -18

Tn = a + (n – 1)d

So, T10 = a + (10 – 1)d

0 = -18 + (10 – 1)d

0 = -18 + 9d

9d = 18

d = 18/9

d = 2

Therefore, common difference d = 2

6. Which term of the A.P.

(i) 3, 8, 13, 18, … is 78?

Solution:-

Let us assume 78 as nth term.

From the question,

The first term a = 3

Then, difference d = 8 – 3 = 5

13 – 8 = 5

18 – 13 = 5

Therefore, common difference d = 5

Tn = a + (n – 1)d

So, 78 = a + (n – 1)d

78 = 3 + (n – 1)5

78 = 3 + 5n – 5

78= -2 + 5n

5n = 78 + 2

5n = 80

n = 80/5

n = 16

Therefore, 78 is 16th term.

(ii) 7, 13, 19, … is 205 ?

Solution:-

Let us assume 205 as nth term.

From the question,

The first term a = 7

Then, difference d = 13 – 7 = 6

19 – 13 = 6

Therefore, common difference d = 6

Tn = a + (n – 1)d

So, 205 = a + (n – 1)d

205 = 7 + (n – 1)6

205 = 7 + 6n – 6

205 = 1 + 6n

6n = 205 – 1

6n = 204

n = 204/6

n = 34

Therefore, 205 is 34th term.

(iii) 18, 15½, 13, … is – 47 ?

Solution:-

Convert mixed fraction into improper fraction = 15½ = 31/2

Let us assume -47 as nth term.

From the question,

The first term a = 18

Then, difference d = 31/2 – 18 = (31 – 36)/2 = -5/2

13 – 31/2 = (26 – 31)/2 = -5/2

Therefore, common difference d = -5/2

Tn = a + (n – 1)d

So, -47 = a + (n – 1)d

-47 = 18 + (n – 1)(-5/2)

-47 = 18 – 5/2n + 5/2

-47 – 18 = -5/2n + 5/2

-65 = -5/2n + 5/2

-65 – 5/2 = – 5/2n

(-130 – 5)/2 = -5/2n

-135/2 = -5/2n

n = (-135/2) × (-2/5)

n = -135/-5

n = 27

Therefore, -47 is 27th term.

7.

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

Solution:-

From the question it is given that,

The first term a = 11

Then, difference d = 8 – 11 = -3

5 – 8 = -3

2 – 5 = -3

Then, common difference d = – 3

Let us assume -150 as nth term,

Tn = a + (n – 1)d

So, -150 = 11 + (n – 1)(-3)

-150 = 11 – 3n + 3

-150 = 14 – 3n

3n = 150 + 14

3n = 164

n = 164/3

n =
ML Aggarwal Solutions for Class 10 Maths Chapter 9 Image 1

Therefore, – 150 is not a term of the A.P. 11, 8, 5, 2, …


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

Solution:-

From the question it is given that,

The first term a = 7

Then, difference d = 10 – 7 = 3

13 – 10 = 3

Then, common difference d = 3

Let us assume 55 as nth term,

Tn = a + (n – 1)d

So, 55 = 7 + (n – 1)3

55 = 7 + 3n – 3

55 = 4 + 3n

3n = 55 – 4

3n = 51

n = 51/3

n = 17

Therefore, 55 is 17th term of the A.P. 7, 10, 13, …

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

Solution:-

From the question it is given that,

The first term a = 31

Then, difference d = 28 – 31 = -3

25 – 28 = -3

Then, common difference d = – 3

Let us assume 0 as nth term,

Tn = a + (n – 1)d

So, 0 = 31 + (n – 1)(-3)

0 = 31 – 3n + 3

0 = 34 – 3n

3n = 34

n = 34/3

n =
ML Aggarwal Solutions for Class 10 Maths Chapter 9 Image 2

Therefore, 0 is not a term of the A.P. 31, 28, 25, …

8.

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

Solution:-

Let us assume 253 as nth term.

From the question,

The first term a = 3

Then, difference d = 8 – 3 = 5

13 – 8 = 5

Therefore, common difference d = 5

Tn = a + (n – 1)d

So, 253 = a + (n – 1)d

253 = 3 + (n – 1)5

253 = 3 + 5n – 5

253 = -2 + 5n

5n = 253 + 2

5n = 255

n = 255/5

n = 51

Therefore, 253 is 51th term.

Now, assume ‘P’ be the 20th term from the last.

Then, P = L – (n – 1)d

= 253 – (20 – 1) 5

= 253 – (19) 5

= 253 – 95

P = 158

Therefore, 158 is the 20th term from the last.


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

Solution:-

Let us assume -100 as nth term.

From the question,

The first term a = -2

Then, difference d = – 4 – (-2) = – 4 + 2 = -2

-6 – (-4) = -6 + 4 = – 2

Therefore, common difference d = -2

Tn = a + (n – 1)d

So, – 100 = a + (n – 1)d

– 100 = -2 + (n – 1)(-2)

– 100 = -2 – 2n + 2

– 100 = -2n

n = -100/-2

n = 50

Therefore, -100 is 50th term.

Now, assume ‘P’ be the 12th term from the last.

Then, P = L – (n – 1)d

= -100 – (12 – 1) (-2)

= -100 – (11) (-2)

= -100 + 22

P = – 78

Therefore, -78 is the 12th term from the last of the A.P. – 2, – 4, – 6, …

9. Find the sum of the two middle most terms of the A.P.

-4/3, -1, -2/3, …,
ML Aggarwal Solutions for Class 10 Maths Chapter 9 Image 3

Solution:-

From the question,

Last term (nth) =
ML Aggarwal Solutions for Class 10 Maths Chapter 9 Image 4 = 13/3

First term a = -4/3

Then, difference d = -1 – (-4/3) = -1 + 4/3 = (-3 + 4)/3 = 1/3

= -2/3 – (-1) = -2/3 + 1 = (-2 + 3)/3 = 1/3

Therefore, common difference d = 1/3

We know that,

Tn = a + (n – 1)d

So, 13/3 = -4/3 + (n – 1)(1/3)

13/3 + 4/3 = 1/3n – 1/3

13/3 + 4/3 + 1/3 = 1/3n

(13 + 4 + 1)/3 = 1/3n

18/3 = 1/3n

6 = 1/3n

n = 6 × 3

n = 18

So, middle term is 18/2 and (18/2) + 1 = 9th and 10th term

Then, a9 + a10 = a + 8d + a + 9d

= 2a + 17d

Now substitute the value of ‘a’ and ‘d’.

= 2(-4/3) + 17(1/3)

= -8/3 + 17/3

= (-8 + 17)/3

= 9/3

= 3

Therefore, the sum of the two middle most terms of the A.P is 3.

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

Solution:-

From the question,

The first term a = 53

Then, difference d = 48 – 53 = -5

= 43 – 48 = -5

Therefore, common difference d = -5

Tn = a + (n – 1)d

= 53 + (n – 1)(-5)

= 53 – 5n + 5

= 58 – 5n

5n = 58

n= 58/5

n = 11.6 ≈ 12

Therefore, 12th term is the first negative term of the A.P. 53, 48, 43,…

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

Solution:-

From the question it is given that,

T5 = 19

T8 – T13 = 20

We know that, Tn = a + (n – 1)d

So, T5 = a + 4d = 19 … [equation (i)]

T13 – T8 = (a + 12d) – (a + 7d) … [equation (ii)]

20 = a + 12d – a – 7d

20 = 5d

d = 20/5

d = 4

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

Then, T5 = a + 4d

19 = a + 4(4)

a = 19 – 16

a = 3

Therefore, A.P. is 3 + 4 = 7, 7 + 4 = 11, 11 + 4 = 15

Hence, the four term of 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.

Solution:-

From the question it is given that,

T3 = 16

The 7th term exceeds the 5th term by 12 = T7 – T5 = 12

We know that, Tn = a + (n – 1)d

So, T3 = a + 2d = 16 … [equation (i)]

T7 – T5 = (a + 6d) – (a + 4d) = 12 … [equation (ii)]

12 = a + 6d – a – 4d

12 = 2d

d = 12/2

d = 6

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

Then, T3 = a + 2d

16 = a + 2(6)

a = 16 – 12

a = 4

Therefore, A.P. is 4 + 6 = 10, 10 + 6 = 16, 16 + 6 = 22

Hence, the four term of A.P. is 4, 10, 16, 22, ….

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

Solution:-

From the question it is given that,

First term a = 12

7th term is 24 less than the 11th term = T11 – T7 = 24

T11 – T7 = (a + 10d) – (a + 6d) = 24

24 = a + 10d – a – 6d

24 = 4d

d = 24/4

d = 6

Now, T20 = a + 19d

Substitute the values of a and d,

T20 = 12 + 19(6)

T20 = 12 + 114

T20 = 126

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

Solution:-

From the question it is given that,

T11 = 38

T6 = 73

Let us assume ‘a’ be the first term and ‘d’ be the common difference,

So, T11 = a + 10d = 38 equation (i)

T6 = a + 5d = 73 equation (ii)

Subtracting both equation (i) and equation (i),

(a + 10d) – (a + 5d) = 73 – 38

a + 10d – a – 5d = 35

5d = 35

d = 35/5

d = 7

now, substitute the value of d in equation (i) to find out a, we get

a + 10d = 38

a + 10(7) = 38

a + 70 = 38

a = 38 – 70

a = -32

Therefore, T31 = a + 30d

= -32 + 30(7)

= -32 + 210

= 178

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

Solution:-

From the question it is given that,

T9 = 1/7

T7 = 1/9

Let us assume ‘a’ be the first term and ‘d’ be the common difference,

So, T9 = a + 8d = 1/7 equation (i)

T7 = a + 6d = 1/9 equation (ii)

Subtracting both equation (i) and equation (i),

(a + 6d) – (a + 8d) = 1/9 – 1/7

a + 6d – a – 8d = (7 – 9)/63

-2d = -2/63

d = (-2/63) × (-1/2)

d = 1/63

now, substitute the value of d in equation (ii) to find out a, we get

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

a = 1/9 – 6/63

a = (7 – 6)/63

a = 1/63

Therefore, T63 = a + 62d

= 1/63 + 62(1/63)

= 1/63 + 62/63

= (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.

Solution:-

From the question it I s given that,

T10 = 41

T10 = a + 9d = 41 … [equation (i)]

T15 = a + 14d = 2T7 + 3

= a + 14d = 2(a + 6d) + 3

= a + 14d = 2a + 12 d + 3

-3 = 2a – a + 12d – 14d

a – 2d = -3 … [equation (ii)]

Now, subtracting equation (ii) from (i), we get,

(a + 9d) – (a – 2d) = 41 – (-3)

a + 9d – a + 2d = 41 + 3

11d = 44

d = 44/11

d = 4

Then, substitute the value of d is equation (i) to find a,

a + 9(4) = 41

a + 36 = 41

a = 41 – 36

a = 5

Therefore, nth term = Tn = a + (n – 1)d

= 5 + (n – 1)4

= 5 + 4n – 4

= 1 + 4n


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

Solution:-

From the question it is given that,

a5 + a7 = 52

(a + 4d) + (a + 6d) = 52

a + 4d + a + 6d = 52

2a + 10d = 52

Divide both the side by 2 we get,

a + 5d = 26 … equation (i)

Given, a10 = a + 9d = 46

a + 9d = 46 … equation (ii)

Now subtracting equation (i) from equation (ii),

(a + 9d) – (a + 5d) = 46 – 26

a + 9d – a – 5d = 20

4d = 20

d = 20/4

d = 5

Substitute the value of d in equation (i) to find out a,

a + 5d = 26

a + 5(5) = 26

a + 25 = 26

a = 26 – 25

a = 1

Then, a2 = a + d

= 1 + 5 = 6

a3 = a2 + d

= 6 + 5

= 11

a4 = a3 + d

= 11 + 5

= 16

Therefore, 1, 6, 11, 16,… are A.P.

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

Solution:-

Froom the question it is given that,

T8 = 0

We have to prove that, 38th term is triple of its 18th term = T38 = 3T18

T8 = a + 7d = 0

T8 = a = -7d

T38 = a + 37d

= -7d + 37d

= 30d

Take, T18 = a + 17d

Substitute the value of a and d,

T18 = -7d + 17d

T18 = 10d

By comparing results of T38 and T18, 38th term is triple of its 18th term.

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

Solution:-

From the question it is given that,

First term a = 3

Common difference d = 10 – 3 = 7

Then, T13 = a + 12d

= 3 + 12(7)

= 3 + 84

= 87

Let us assume that, nth term is 84 more than its 13th term

So, Tn = 84 + 87

= 171

We know that, Tn = a + (n – 1)d = 171

3 + (n – 1)7 = 171

3 + 7n – 7 = 171

7n – 4 = 171

7n = 171 + 4

7n = 175

n = 175/7

n = 25

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.

Solution:-

First take, A.P. 9, 7, 5, ….

From the above A.P.

First term a = 9

Then, common difference d = 7 – 9 = -2

We know that, Tn = a + (n – 1)d

= 9 + (n – 1) (-2)

= 9 – 2n + 2

= 11 – 2n

Now, consider A.P. 24, 21, 18, …

From the above A.P.

First term a1 = 24

Then, common difference d = 21 – 24 = – 3

We know that, Tn = a + (n – 1)d

= 24 + (n – 1) (-3)

= 24 – 3n + 3

= 27 – 3n

Froom the question it is given that, nth term of both A.P. is same,

So, 11- 2n = 27 – 3n

-2n + 3n = 27 – 11

n = 16

Then, T16 = a + (n – 1)d

= 9 + 15 (-2)

= 9 – 30

= – 21

20.

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

Solution:-

The two digits numbers divisible by 3 are, 12, 15, 18, 21, 24,…..,99.

The above numbers are A.P.

So, first number a = 12

Common difference d = 15 – 12 = 3

Then, last number is 99

We know that, Tn (last number) = a + (n – 1)d

99 = 12 + (n – 1)3

99 = 12 + 3n – 3

99 = 9 + 3n

99 – 9 = 3n

3n = 90

n = 90/3

n = 30

Therefore, 30 two digits number are divisible by 3.


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

Solution:-

The natural numbers which are divisible by both 2 and 5 are 110, 120, 130, 140, ….,999

The above numbers are A.P.

So, first number a = 110

Common difference d = 120 – 110 = 10

Then, last number is 999

We know that, Tn (last number) = a + (n – 1)d

999 = 110 + (n – 1)10

999 = 110 + 10n – 10

999 = 100 + 10n

999 – 100 = 10n

10n = 888

n = 888/10

n = 88

The number of natural numbers which are divisible by both 2 and 5 are 88.


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

Solution:-

The numbers which are lie between 10 and 300, when divisible by 4 leave a remainder 3 are 11, 15, 19, 23, 27,….299

The above numbers are A.P.

So, first number a = 11

Common difference d = 15 – 11 = 4

Then, last number is 299

We know that, Tn (last number) = a + (n – 1)d

299 = 11 + (n – 1)4

299 = 11 + 4n – 4

299 = 7 + 4n

299 – 7 = 4n

4n = 292

n = 292/4

n = 73

The total which are lie between 10 and 300, when divisible by 4 leave a remainder 3 are 73.

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

Solution:-

From the question it is given that, n – 2, 4n – 1 and 5n + 2 are in A.P.

Multiplying by 2 to 4n – 1 then it becomes = 8n – 2

So, 8n – 2 = n – 2 + 5n + 2

8n – 2 = 6n

8n – 6n = 2

2n = 2

n = 2/2

n = 1

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

Solution:-

From the question it is given that,

The sum of three numbers in A.P. = 3

Given, Their product = -35

Let us assume the 3 numbers which are in A.P. are, a – d, a, a + d

Now adding 3 numbers = a – d + a + a + d = 3

3a = 3

a = 3/3

a = 1

From the question, product of 3 numbers is – 35

So, (a – d) × (a) × (a + d) = – 35

(1 – d) × (1) × (1 + d) = – 35

12 – d2 = – 35

d2 = 35 + 1

d2 = 36

d = √36

d = ±6

Therefore, the numbers are (a – d) = 1 – 6 = – 5

a = 1

(a + d) = 1 + 6 = 7

If d = – 6

The numbers are (a – d) = 1 – (-6) = 1 + 6 = 7

a = 1

(a + d) = 1 + (-6) = 1 – 6 = -5

Therefore, the numbers -5, 1, 7,… and 7, 1, -5,… are in A.P.

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.

Solution:-

From the question it is given that, sum of three numbers in A.P. = 30

The ratio of first number to the third number is 3: 7

Let us assume the 3 numbers which are in A.P. are, a – d, a, a + d

Now adding 3 numbers = a – d + a + a + d = 30

3a = 30

a = 30/3

a = 10

Given ratio 3 : 7 = a – d : a + d

3/7 = (a – d)/(a + d)

(a – d)7 = 3(a + d)

7a – 7d = 3a + 3d

7a – 3a = 7d + 3d

4a = 10d

4(10) = 10d

40 = 10d

d = 40/10

d = 4

Therefore, the numbers are a – d = 10 – 4 = 6

a = 10

a + d = 10 + 4 = 14

So, 6, 10, 14, … are in A.P.

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.

Solution:-

From the question it is given that, sum of the first three terms of an A.P. is 33.

Let us assume the 3 numbers which are in A.P. are, a – d, a, a + d

Now adding 3 numbers = a – d + a + a + d = 33

3a = 33

a = 33/3

a = 11

Given, the product of the first and the third terms exceeds the second term by 29.

(a – d) (a + d) = a + 29

a2 – d2 = 11 + 29

112 – d2 = 40

121 – 40 = d2

d2 = 81

d =√81

d = ±9

If d = 9

Therefore, the numbers are (a – d) = 11 – 9 = 2

a = 11

(a + d) = 11 + 9 = 20

If d = – 9

The numbers are (a – d) = 1 – (-9) = 11 + 9 = 20

a = 11

(a + d) = 11 + (-9) = 11 – 9 = 2

Therefore, the numbers 2, 11, 20,… and 20, 11, 2,… are in A.P.

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

Solution:-

(i) 2n – 3

From the question it is given that,

nth term is 2n – 3

So, Tn = 2n – 3

Now, we start giving values, 1, 2, 3, … in the place of n, we get,

(2 × 1) – 3 = 2 – 3 = – 1

(2 × 2) – 3 = 4 – 3 = 1

(2 × 3) – 3 = 6 – 3 = 3

(2 × 4) – 3 = 8 – 3 = 5

From the above results, -1, 1, 3, 5, …. Are in A.P.

So, first term a = – 1

Common difference d = 1 – (-1) = 1 + 1 = 2

= 3 – 2 = 2

(ii) n2 + 1

From the question it is given that,

nth term is n2 + 1

So, Tn = n2 + 1

Now, we start giving values, 1, 2, 3, … in the place of n, we get,

12 + 1 = 1 + 1 = 2

22 + 1 = 4 + 1 = 5

32 + 1 = 9 + 1 = 10

42 + 1 = 16 + 1 = 17

From the above results, 2, 5, 10, 17, ….

So, first term a = 2

Common difference d = 5 – 2 = 3

= 10 – 5 = 5

The common difference is not same.

Therefore, 2, 5, 10, 17,… are not in A.P.

Exercise 9.3

1. Find the sum of the following A.P.s :
(i) 2, 7, 12, … to 10 terms

Solution:-

From the question,

First term a = 2

Then, d = 7 – 2 = 5

12 – 7 = 5

So, common difference d = 5

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

Solution:-

From the question,

First term a = 1/15

Then, d = 1/12 – 1/15

= (5 – 4)/60

= 1/60

So, common difference d = 1/60

n = 11

S11 = 11/2(2a + (n – 1)d)

= 11/2 ((2 × (1/15)) + (11 – 1)(1/60))

= 11/2 ((2/15) + (10/60))

= 11/2 (2/15 + 1/6)

= 11/2 (4 + 5)/30

= 11/2 (9/30)

= 11/2(3/10)

= 33/20

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

Solution:-

From the question,

The first term a = 27

Difference d = 24 – 27 = – 3

= 21 – 24 = – 3

So, common difference d = -3

Sn = 0

Let us assume n be there in A.P.

So, Sn = (n/2) (2a + (n – 1)d)

0 = n/2 ((2 × 27) + (n – 1)(-3))

0 = n/2(54 – 3n + 3)

0 = n/2(57 – 3n)

0 × (2/n) = 57 – 3n

0 = 57 – 3n

3n = 57

n = 57/3

n = 19

3. Find the sums given below :
(i) 34 + 32 + 30 + … + 10

Solution:-

From the question,

First term a = 34,

Difference d = 32 – 34 = -2

So, common difference d = – 2

Last term Tn = 10

We know that, Tn = a + (n – 1)d

10 = 34 + (n – 1)(-2)

-24 = – 2(n – 1)

-24 = – 2n + 2

2n = 24 + 2

2n = 26

n = 26/2

n = 13

Sn = n/2(a + 1)

= 13/2 (34 + 10)

= 13/2 (44)

= 13 (22)

= 286


(ii) – 5 + ( – 8) + ( – 11) + … + ( – 230)

Solution:-

From the question,

First term a = -5,

Difference d = -8 – (-5) = -8 + 5 = -3

So, common difference d = – 3

Last term Tn = -230

We know that, Tn = a + (n – 1)d

-230 = -5 + (n – 1)(-3)

-230 = – 5 – 3n + 3

-230 = – 2 – 3n

3n = 230 -2

3n = 228

n = 228/3

n = 76

Therefore, 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

Solution:-

From the question,

First term a = 5

Then common difference d = 3

an = 50,

We know that, an = a + (n – 1)d

50 = 5 + (n – 1)3

50 = 5 + 3n – 3

50 = 2 + 3n

3n = 50 – 2

3n = 48

n = 48/3

n = 16

So, Sn = (n/2)(2a + (n – 1)d)

= (16/2) ((2 × 5) + (16 – 1) × 3)

= 8(10 + 45)

= 8(55)

= 440


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

Solution:-

From the question,

First term a = 7

a13 = 35,

We know that, an = a + (n – 1)d

35 = 7 + (13 – 1)d

35 = 7 + 13d – d

35 = 7 + 12d

12d = 35 – 7

12d = 28

d = 28/12 … [divide by 4]

d = 7/3

So, S13 = (n/2)(2a + (n – 1)d)

= (13/2) ((2 × 7) + ((13 – 1) × (7/3))

= (13/2) ((14 + (12 × 7/3))

= (13/2) (14 + 28)

= (13/2) (42)

= 13 × 21

= 273


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

Solution:-

From the question it is given that,

Common difference d = 5

S9 = 75

We know that, an = a + (n – 1)d

a9 = a + (9 – 1)5

a9 = a + 45 – 5

a9 = a + 40 … [equation (i)]

Then, S9 = (n/2) (2a + (n – 1)d)

75 = (9/2) (2a + (9 – 1)5)

75 = (9/2) (2a + (8)5)

(75 × 2)/9 = 2a + 40

150/9 = 2a + 40

2a = 150/9 – 40

2a = 50/3 – 40

2a = (50 – 120)/3

2a = -70/3

a = -70/(3 × 2)

a = – 35/3

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

a9 =a + 40

= -35/3 + 40

= (-35 + 120)/3

= 85/3


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

Solution:-

From the question it is give that,

First term a = 8,

an = 62 and Sn = 210

We know that, an = a + (n – 1)d

62 = 8 + (n – 1)d

(n – 1)d = 62 – 8

(n – 1)d = 54 … [equation (i)]

Then, Sn = (n/2) (2a + (n – 1)d)

210 = (n/2) ((2 × 8) + 54) … [from equation (i) (n – 1)d = 54]

210 = (n/2) (16 + 54)

420 = n(70)

n = 420/70

n = 6

Now, substitute the value of n in equation (i),

(n – 1)d = 54

(6 – 1)d = 54

5d = 54

d = 54/5

Therefore, d = 54/5 and n = 6


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

Solution:-

From the question it is given that,

First term a = 3

n = 8

S = 192

We know that, Sn = (n/2) (2a + (n – 1)d)

192 = (8/2) ((2 × 3) + (8 – 1)d)

192 = 4 (6 + 7d)

192/4 = 6 + 7d

48 = 6 + 7d

48 – 6 = 7d

42 = 7d

d = 42/7

d = 6

Therefore, common difference d is 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.

Solution:-

From the question it is give that,

First term a = 5

Last term = 45

Then, sum = 400

We know that, last term = a + (n – 1)d

45 = 5 + (n – 1)d

(n – 1)d = 45 – 5

(n – 1)d = 40 … [equation (i)]

So, Sn = (n/2) (2a + (n – 1)d)

400 = (n/2) ((2 × 5) + 40) … [from equation (i) (n – 1)d = 40]

800 = n(10 + 40)

800 = 50n

n = 800/50

n = 16


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

Solution:-

From the question it is give that,

First term a = 15

Therefore, 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)

(750 × 2)/15 = 2a + 14d

100 = 2a + 14d

Dividing both the side by 2 we get,

50 = a + 7d

Now, substitute the value a,

50 = 15 + 7d

7d = 50 – 15

7d = 35

d = 35/7

d = 5

So, 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?

Solution:-

From the question it is give that,

First term a = 17

Last term (l) = 350

Common difference d = 9

We know that, l = Tn = a + (n – 1)d

350 = 17 + (n – 1) × 9

350 – 17 = 9n – 9

333 + 9 = 9n

342 = 9n

n = 342/9

n = 38

So, Sn = (n/2) (2a + (n – 1)d)

= (38/2) ((2 × 17) + (38 – 1)d)

= 19(34 + (37 × 9))

= 19(34 + 333)

= 19 × 367

= 6973

Therefore, n = 38 and Sn = 6973

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

Solution:-

From the question,

First term a = 1

Difference d = 4 – 1 = 3

n = x

x = a = (n – 1)d

x – 1 = (n – 1)d

Sn = (n/2) (2a + (n – 1)d)

287 = (n/2) ((2 × 1)+ (n – 1)3)

= n (2 + 3n – 3)

574 = n(2 + 3n – 3)

574 = 2n + 3n2 – 3n

574 = – n + 3n2

3n2 – n – 574 = 0

3n2 – 42n + 41 – 574 = 0

3n(n – 14) + 41(n – 14) = 0

(n – 14) (3n + 41) = 0

If n – 14 = 0

n = 14

or 3n + 41 = 0

3n = -41

n = -41/3

We have to take positive number so n = 14

Then, = a + (n – 1)d

= 1 + (14 – 1) 3

= 1 + (13)3

= 1 + 39

= 40

Therefore, x = 40


8.

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

Solution:-

From the question it is given that,

First term a = 25

Common difference d = 22 – 25 = – 3

Sum = 116

Sn = (n/2) (2a + (n – 1)d)

116 = (n/2) (2a + (n – 1)d)

By cross multiplication,

232 = n ((2 × 25) + (n – 1) (-3))

232 = n (50 – 3n + 3)

232 = n (53 – 3n)

232 = 53n – 3n2

3n2 – 53n + 232 = 0

3n2 – 24n – 29n + 232 = 0

3n (n – 8) – 29 (n – 8) = 0

(n – 8) (3n – 29) = 0

If n – 8 = 0

n = 8

or 3n – 29 = 0

3n = 29

n = 29/3

not possible to take fraction,

So, n = 8

Then, T = a + (n – 1)d

= 25 +(8 – 1) (-3)

= 25 + 7 (-3)

= 25 – 21

= 4


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

Solution:-

From the question it is given that,

First term a = 24

Common difference d = 21 – 24 = – 3

Sum = 78

Sn = (n/2) (2a + (n – 1)d)

78 = (n/2) (2a + (n – 1)d)

By cross multiplication,

156 = n ((2 × 24) + (n – 1) (-3))

156 = n (48 – 3n + 3)

156 = n (51 – 3n)

156 = 51n – 3n2

3n2 – 51n + 156 = 0

3n2 – 12n – 39n + 156 = 0

3n (n – 4) – 39 (n – 4) = 0

(n – 4) (3n – 39) = 0

If n – 4 = 0

n = 4

or 3n – 39 = 0

3n = 39

n = 39/3

n = 13

now we have to consider both values

So, n = 4

Then, T = a + (n – 1)d

= 24 +(4 – 1) (-3)

= 24 + 3 (-3)

= 24 – 9

= 15

n = 13

Then, T = a + (n – 1)d

= 24 +(13 – 1) (-3)

= 24 + 12 (-3)

= 24 – 36

= -12

So, (12 + 9 + 6 + 3 + 0 + (-3)+ (-6) + (-9) + (-12)) = 0

Hence, the sum of 5th term to 13th term = 0

9. Find the sum of first 22 terms, of an A.P. in which d = 7 and a22 is 149.

Solution:-

From the question it is given that,

Common difference d = 7

a22 = 149

n = 22

we know that,

a22 = (n – 1)d

149 = a + (22 – 1)7

149 = a + (22)7

149 = a + 147

a = 149 – 147

a = 2

So, 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 and third terms are 14 and 18 respectively.

Solution:-

From the question it is given that,

T2 = 14, T3 = 18

So, common difference d = T3 – T2

= 18 – 14

= 4

Where, a = T1 = 14 – 4 = 10

n = 51

We know that,

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

Solution:-

From the question it is given that,

T3 = 1, T6 = – 11 and n = 32

We know that,

T3 = a + 2d = 1 … [equation (i)]

T6 = a + 5d = – 11 … [equation (ii)]

Now, subtracting equation (ii) from equation (i), we get,

(a + 2d) – (a + 5d) = 1 – (-11)

a + 2d – a – 5d = 1 + 11

-3d = 12

d = -12/3

d = -4

Now, substitute value of d in equation (i),

a + 2d = 1

a + 2(-4) = 1

a – 8 = 1

a = 8 + 1

a = 9

S32 = (n/2) (2a + (n – 1)d)

= (32/2) (2(9) + (32 – 1)(-4))

= 16 (18 + (31)(-4))

= 16 (18 – 124)

= 16 (-106)

= – 1696

Therefore, the sum of its first 32 terms is – 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.

Solution:-

From the question it is given that,

S6 = 36

S16 = 256

We know that,

Sn = (n/2) (2a + (n – 1)d)

S6 = (6/2) (2a + (6 – 1)d) = 36

3 (2a + 5d) = 36

Divide both the side by 3,

2a + 5d = 12 … [equation (i)]

Now, S16 = (16/2) (2a + (16 – 1)d) = 256

8 (2a + 15d) = 256

Divide both the side by 8,

2a + 15d = 32 … [equation (ii)]

Then, subtract equation (ii) from equation (i) we get,

(2a + 5d) – (2a + 15d) = 12 – 32

2a + 5d – 2a – 15d = -20

-10d = -20

d = -20/-10

d = 2

substitute the value of d in equation (i) to find a,

2a + 5d = 12

2a + 5(2) = 12

2a + 10 = 12

2a = 12 – 10

2a = 2

a = 2/2

a = 1

So, S10 = (n/2) (2a + (n – 1)d)

= (10/2) ((2 × 1) + (10 – 1)2)

= 5 (2 + 18)

= 5 (20)

= 100

Therefore, the sum of first 10 terms is 100.

12. Show that a1, a2, a3, … form an A.P. where an is defined as an = 3 + 4n. Also find the sum of first 15 terms.

Solution:-

From the question it is given that,

nth term is 3 + 4n

So, an = 3 + 4n

Now, we start giving values, 1, 2, 3, … in the place of n, we get,

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

So, The numbers are 7, 11, 15, 19, ….

Then, first term a = 7, common difference d = 11 – 7 = 4

We know that,

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

Therefore, the sum of first 15 terms is 525.

13.

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

Solution:-

From the question it is given that,

nth term is 3 + 4n

So, an = 3 – 4n

Now, we start giving values, 1, 2, 3, … in the place of n, we get,

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

So, The numbers are -1, -5, -9, -13, ….

Then, first term a = -1, common difference d = -5 – (-1) = -5 + 1 = -4

We know that,

S20 = (n/2) (2a + (n – 1)d)

= (20/2) ((2 × (-1)) + (20 – 1) × (-4))

= 10 (-2 + (19 × (-4)))

= 10(-2 – 76)

= 10 (-78)

= -780

Therefore, the S20 is -780.


(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.

Solution:-

From the question it is given that,

First term a = 5

And also it is given that, the sum of first four terms is half the sum of next four terms,

a1 + a2 + a3 + a4 = ½ (a5 + a6 + a7 + a8)

then,

a + (a + d) + (a + 2d) + (a + 3d) = ½ ((a + 4d) + (a + 5d) + (a + 6d) + (a + 7d))

a + a + d + a + 2d + a + 3d = ½ (a + 4d + a + 5d + a + 6d + a + 7d)

4a + 6d = ½ (4a + 22d)

By cross multiplication,

2(4a + 6d) = (4a + 22d)

2 ((4 × 5) + 6d) = ((4 × 5) + 22d) … [given a = 5]

2(20 + 6d) = (20 + 22d)

40 + 12d = 20 + 22d

40 – 20 = 22d – 12d

20 = 10d

d = 20/10

d = 2

Therefore, the common difference d is 2.

Exercise 9.4

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

Solution:-

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, …

Solution:-

From the question,

First term a = 1/6

Then, r = (1/3) ÷ (1/6)

r = (1/3) × (6/1)

r = 6/3

r = 2

Therefore, next term = 2/3 × 2 = 4/3

(ii) Find the next term of the list of numbers 3/16, -3/8, ¾, -3/2,…

Solution:-

From the question,

First term a = 3/16

Then, r = (-3/8) ÷ (3/16)

r = (-3/8) × (16/3)

r = (-3 × 16)/(8 × 3)

r = (-1 × 2)/ (1 × 1)

r = -2

Therefore, next term = -3/2 × (-2) = 6/2 = 3

(iii) Find the 15th term of the series √3 + 1/√3 + 1/3√3 + …

Solution:-

From the question,

First term a = √3

Then, r = (1/√3) ÷ (√3)

r = (1/√3) × (1/√3)

r = (1 × 1)/( √3 × √3)

r = 1/(√3)2

r = 1/3

So, a15 = arn – 1

= √3(1/3)15 – 1

= √3(1/3)14

= √3 × (1/314)

Therefore, a15 = √3 × (1/314)

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

Solution:-

From the question it is given that,

First term a = 1/√2

Then, r = -2 ÷ (1/√2)

r = (-2/1) × (√2/1)

r = (-2 × √2)/(1 × 1)

r = -2√2

So, 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

Therefore, an = (-1)n – 1 × 2(3n 4)/2

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

Solution:-

From the question it is given that,

First term a = 5,

Then, r = (25) ÷ (5)

r = (25) × (1/5)

r = 5

So, a10 = arn – 1

= 5 × (5)10 – 1

= 5 × 5 9

= 59 + 1 … [by am × an = am + n]

= 510

Therefore, an = arn – 1

= 5 × 5n – 1

= 5n – 1 + 1

= 5n

(vi) Find the 6th and the nth terms of the list of numbers 3/2, ¾, 3/8,…

Solution:-

From the question it is given that,

First term a = 3/2,

Then, r = (3/4) ÷ (3/2)

r = (3/4) × (2/3)

r = (3 × 2)/(4 × 3)

r = (1 × 1)/(2 × 1)

r = ½

So, an = arn – 1

= (3/2) × (1/2)n – 1

= 3 × ½ × (½)n – 1

= 3 × (½)n – 1 + 1

= 3 × (½)n

= 3/2n

Therefore, a6 = 3/2n

= 3/26

= 3/64

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

Solution:-

From the question it is given that,

Last term = 12288

First term a = 3,

Then, r = (-6) ÷ (3)

r = (-6) × (1/3)

r = (-6 × 1)/(1 × 3)

r = (-2 × 1)/(1 × 1)

r = -2

Then, 6th term from the end,

a6 = l × (1/r)n – 1

= 12288 × (1/-2)6 – 1

= 12288 × (1/-25)

= 12288/-32

= – 384

3. Which term of the G.P.

(i) 2, 2√2, 4, … is 128?

Solution:-

From the question it is given that,

Last term = 128

First term a = 2,

Then, r = (2√2) ÷ (2)

r = (2√2)/2

r = √2

Then, an = arn – 1

So, 128 = 2(√2)n – 1

27 = 2(√2)n – 1

27/2 = (√2)n – 1

27 – 1 = (√2)n – 1

26 = (√2)n – 1

(√2)n -1 = (√2)12

Now, comparing the powers

n – 1 = 12

n = 12 + 1

n = 13

Therefore, 128 is the 13th term.

(ii) 1, 1/3, 1/9, … is 1/243

Solution:-

From the question it is given that,

Last term (an) = 1/243

First term a = 1,

Then, r = (1/3) ÷ (1)

r = (1/3) × (1/1)

r = 1/3

Then, an = arn – 1

1/243 = 1 × (1/3)n – 1

(1/3)5 = (1/3)n – 1

By comparing both left hand side and right hand side,

5 = n – 1

n = 5 + 1

n = 6

Therefore, 1/243 is 6th term.

(iii) 1/3, 1/9, 1/27, … is 1/19683?

Solution:-

From the question it is given that,

Last term (an) = 1/19683

First term a = 1/3

Then, r = (1/9) ÷ (1/3)

r = (1/9) × (3/1)

r = 1/3

Then, an = arn – 1

1/19683 = (1/3) × (1/3)n – 1

(1/3)9 = (1/3)n – 1 + 1

(1/3)9 = (1/3)n

By comparing both left hand side and right hand side,

9 = n

n = 9

Therefore, 1/19683 is 9th term.

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

Solution:-

From the question it is given that,

Last term (an) = 729

First term a = 3

Then, r = (-3√3) ÷ 3

r = (-3√3/3)

r = -√3

Then, an = arn – 1

729 = (3) × (-√3)n – 1

729/3 = (-√3)n – 1

243 = (-√3)n – 1

(-√3)10 = (-√3) n – 1

By comparing both left hand side and right hand side,

10 = n – 1

n = 10 + 1

n = 11

Therefore, 729 is 11th term.

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

Solution:-

From the question it is given that,

a8 = 192 and r = 2

Then, by the formula an = arn – 1

a8 = ar8 – 1

192 = a(2)8 – 1

192 = a(2)7

a = 192/27

a = 192/128

a = 3/2

Now, a12 = (3/2)(2)12 – 1

= (3/2) × (2)11

= (3/2) × 2048

= 3072

a8 = 3072

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

Solution:-

From the question it is given that,

a3 = 24

a6 = 192

Then, by the formula an = arn – 1

a6 = ar6 – 1

192 = ar6 – 1

192 = ar5 … [equation (i)]

Now, a3 = arn – 1

24 = ar3 – 1

24 = ar2 … [equation (ii)]

By dividing equation (i) by equation (ii)

ar5/ar2 = 192/24

r5 – 2 = 8

r3 = 8

r3 = 23

r = 2

Now, substitute the value r in equation (i),

192 = ar5

192 = a (2)5

a = 192/32

a = 6

So, a10 = ar10 – 1

= ar9

= 6(2)9

= 6 (512)

= 3072

7. Find the number of terms of a G.P. whose first term is ¾, common ratio is 2 and the last term is 384.

Solution:-

From the question it is given that,

First term of G.P. a = ¾

Common ratio (r) = 2

Last term = 384

Then, by the formula an = arn – 1

384 = (3/4) (2)n – 1

(384 × 4)/3 = (2)n – 1

(1536)/3 = (2)n – 1

  512 = 2n – 1

29 = 2n – 1

By comparing both left hand side and right hand side,

9 = n – 1

n = 9 + 1

n = 10

The number of terms of a G.P. is 10.

8. Find the value of x such that,

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

Solution:-

From the question,

x2 = -2/7 × -7/2

x2 = 1

x = ± 1

Therefore, x = 1 or x = – 1

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

Solution:-

From the question,

(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 let us take x – 16 = 16

Or x = 0

So, x = 0, 16

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

Solution:-

From the question,

(x + 3)2 = x(x + 9)

x2 + 6x + 9 = x2 + 9x

9 = 9x – 6x

9 = 3x

X = 9/3

X = 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.

Solution:-

From the question it is given that,

a4 = x

a7 = y

a10 = z

Now we have to prove that, x, y, z are in G.P.

Then, by the formula an = arn – 1

a4 = ar4 – 1

a4 = a3

a4 = x

So, a7 = a7 – 1

a7 = a6

a7 = y

a10 = a10 -1

a10 = a9

a10 = z

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

y2 = xz

Then, xz = ar3 × ar9

= a2r3 + 9

= a2r12

y2 = (ar6)2

y2 = a2r12

By comparing left hand side and right hand side

LHS = RHS

Therefore, 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.

Solution:-

From the question it is given that,

a5 = p

a8 = q

a11 = s

Now we have to prove that, q² = ps

Then, by the formula an = arn – 1

a5 = ar5 – 1

a5 = a4

a5 = p

So, a8 = a8 – 1

a8 = a7

a8 = q

a11 = a11 -1

a11 = a10

a11 = s

p, q, s are in G.P. then,

q2 = (ar7)2

= ar14

Then, px = ar4 × ar10

= a2r4 + 10

= a2r14

Therefore, q2 = ps

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

Solution:-

From the question it is given that,

a, b, c are in G.P.

We have to show that a², b², c² are also in G.P

Then,

b² = ac … equation (i)

Therefore, a², b², c² will be in G.P.

if (b²)² = a² x c²

(ac)² = a²c² … [from the equation (i)]

a²c² = a²c²

Therefore, it is proved that a2, b2, c2 are also in G.P.

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

Solution:-

From the question it is given that,

a, b and c are in A.P.

So, 2b = a + c

We have to show that 3a, 3b, 3c are also in G.P.

If (3b)2 = 3a x 3c

32b = 3a + c

Now, comparing the results we get,

2b = a + c

Therefore, 3a, 3b, 3c are in G.P

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

Solution:-

From the question it is given that,

a, b and c are in A.P.

So, 2b = a + c

We have to show that 10ax + 10, 10bx + 10, 10cx + 10, x ≠ 0, are also in G.P.

2b = a + c

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

(102bx + 20) = 10ax + 10 + cx + 10

(102bx + 20) = 10ax + cx + 20

By comparing left hand side and right hand side we get,

2bx + 20 = ax + cx + 20

2bx = ax + cx

2b = a + c

Therefore, 10ax + 10, 10bx + 10, 10cx + 10 are in G.P.

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

Solution:-

From the question,

(a2 + 2)2 = a(a3 + 10)

a4 + 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

Then, 2a – 1 = 0

a = ½

a – 2 = 0

a = 2

Therefore, a = 2 or a= ½

15. The first and the second terms of a GP. are x-4 and xm . If its 8th term is x52, then find the value of m.

Solution:-

From the question it is given that,

First term of G.P. a1 = x-4

Second term of G.P. a2 = xm

Eighth term of G.P a8 = x52

Then, r = a2/a1

= xm/x-4

= xm – (-4) … [by am/an = am – n]

= xm + 4

a8 = ar8 – 1

a8 = ar7

x52 = x-4 × r7

= x-4 × x7(m + 4)

= x-4 + 7m + 28

X52 = x7m + 24

By comparing left hand side and right hand side we get,

52 = 7m + 24

7m = 52 – 24

7m = 28

m = 28/7

m = 4

Therefore, the value of m is 4.

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

Solution:-

From the question it is given that,

4th term a4 = 54

7th term a7 = 1458

ar3 = 54

ar6 = 1458

Now dividing we get,

ar6/ar3 = (1458/54)

r6 – 3 = 27

r3 = 33

r = 3

Then, ar3 = 54

a × 27 = 54

a = 54/27

a = 2

Therefore G.P. is 2, 6, 18, 54, …

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

Solution:-

From the question it is given that,

a1 = – 3

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

an = arn – 1

a4 = ar4 – 1

= ar3

a3 = ar3 – 1

a3 = ar2

So, ar3 = ar2

ar3 = a2r2

r3/r2 = a2/a

r3 – 2 = a2 – 1

r = a

Therefore, a7 = ar7 – 1

a7 = ar6

= (-3) (-3)6

= -31 + 6

= – 37

a7 = – 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.

Solution:-

From the question it is give that,

The sum of first three terms of a G.P. is 39/10

The product of first three terms of a G.P. is 1

Let us assume that a be the first term and ‘r’ be the common ratio,

And also assume that, three terms of the G.P. is a/r, a, ar,

The sum of three terms = (a/r) + a + ar = 39/10

Take out ‘a’ as common then, we get

a(1/r + 1 + r) = 39/10 … [equation (i)]

Now, product of three terms = (a/r) × a × ar = 1

a3r/r = 1

a3= 1

a3 = 13

a = 1

Substitute the value of ‘a’ in equation (i),

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

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

By cross multiplication we get,

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

10 + 10r + 10r2 = 39r

Transposing 39r from right hand side to left hand side it becomes – 39r,

10 + 10r + 10r2 – 39r = 0

10r2 – 29r + 10 = 0

10r2 – 25r – 4r + 10 = 0

5r(2r – 5) – 2(2r – 5) = 0

(2r – 5) (5r – 2) = 0

So, 2r – 5 = 0

r = 5/2

5r – 2 = 0

r = 2/5

Therefore, r = 5/2 or 2/5

Then the terms if r = 5/2 are, 1, 5/2, 25/4, …

The terms if r = 2/5 are, 1, 2/5, 4/25, …

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.

Solution:-

From the question it is give that,

The sum of first three terms of a A.P. is 15

Let us assume three numbers are a – d, a, a + d.

The sum of three terms = a – d + a + a + d = 15

a = 15/3

a = 5

Then, adding 1, 4, 19 in the terms

The numbers become, a – d + 1, a + 4, a + d + 19

Therefore, b2 = ac

(a + 4)2 = (a – d + 1) (a + d + 19)

Simplify the above terms,

a2 + 8a + 16 = a2 + ad + 19a – ad – a2 – 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

d2 + 21d – 3d – 63 = 0

d(d + 21) – 3(d + 21) = 0

(d + 21) (d – 3) = 0

So, d + 21 = 0

d = – 21

d – 3 = 0

d = 3

Then the terms if d = 3 and a = 5,

Then G.P. 5 – 3 = 2, 5, 5 + 3 = 8

The terms if d = – 21 are 5 – (-21) = 5 + 21 = 26, 5, 5 – 21 = – 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 ratio of the G.P.

Solution:-

From the question it is given that,

Three numbers form an increasing G.P.

Let us assume the three numbers a/r, a, are

Then, double the middle term we get,

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

So, 2(2a) = a/r + ar

4a = a(1/r + r)

4 = 1/r + r

By cross multiplication,

4r = 1 + r2

r2 – 4r + 1 = 0

r = – b ± √(b2 – 4ac)/2a

= – (-4) ± √((-4)2 – 4 × 1 × 1)/(2 × 1)

= 4 ± √(16 – 4)/2

= 4 ± √12/2

= 4 ± 2√3/2

= 2 ± √3

Therefore, the common ratio of the G.P. is 2 ± √3.

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

Solution:-

From the question it is given that,

Three numbers are in G.P. whose sum is 70.

Let us assume the three number be a/r, a, ar

Then, sum = (a/r) + a + ar = 70

Take out a as common,

a((1/r) + 1 + r) = 70 … [equation (i)]

Now, multiplying the extremes by 4 and mean by 5,

Then, (a/r) × 4 = 4a/r

(a × 5) = 5a

(ar × 4) = 4ar

4a/r, 5a, 4ar

Therefore, these are in A.P.

So, 2(5a) = (4a/r) + 4ar

10a = 4a(1/r) + r

Divide both the side by 2 we get,

(10/2)a = (4/2)a (1/r) + r

5a = 2a((1/r) + r)

5r = 2 + 2r2

2r2 – 5r + 2 = 0

2r2 – r – 4r + 2 = 0

r(2r – 1) – 2(2r – 1) = 0

(2r – 1) (r – 2) = 0

So, 2r – 1 = 0

2r = 1

r = ½

r – 2 = 0

r = 2

Now substitute the value r in equation (i),

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

a ((½) + 3) = 70

a ((1 + 6)/2) = 70

a (7/2) = 70

a = 70 × (2/7)

a = 10 × 2

a = 20

Then,

r = 2, a = 20

= (a/r), a, ar

= (20/2), 20, (20 × 2)

= 10, 20, 40

Then,

r = ½, a = 20

= (a/r), a, ar

= (20/½), 20, (20 × ½)

= (20 × 2), 20, 10

= 40, 20, 10

22.

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

Solution:-

From the question it is given that,

a, b, c are in A.P. as well in G.P.

We have to prove that, a = b = c.

a, b, c are in A.P.

2b = a + c

b = (a + c)/2 … [equation (i)]

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

b2 = ac … [equation (ii)]

Now substitute the value of ‘b’ in equation (ii),

((a + c)/2)2 = ac

(a + c)2/4 = ac

(a + c)2 = 4ac

(a + c)2 – 4ac = 0

Then, (a – c)2 = 0

(a – c) = 0

a = c … [equation (iii)]

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

Substitute the value of a

Then, 2b = a + a

2b = 2a

Therefore, b = a … [equation (iv)]

By comparing equation (iii) and equation (iv),

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

Solution:-

From the question it is given that,

a, b, c are in A.P.

So, 2b = a + c

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

b2 = ac

from question 22(i) a = b = c,

Given, ab – c + bc – a + ca – b

Therefore, b – c = 0, c – a = 0 and a – b = 0

So, a0 + b0+ c0

We know that, x0 = 1

1 + 1 + 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.

Solution:-

From the question it is given that,

First term of G.P = a

Common ratio = r

Then the terms of G.P. is a, ar, ar2.

By squaring the terms of G.P. we get,

a2, a2r2, a2r4

We know that, b2 = 4ac

(a2r2)2 = a2 × a2r4

a4r4 = a4r4

Therefore, the first term is a2

Common ratio is r2

Then, nth term will be

an = arn – 1

an = a2(rn – 1)2

an = 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, …, ARn-1 form a G.P. and find the common ratio.

Solution:-

From the question it is given that,

The corresponding terms of two G.P.’s a, ar, ar², …, arn-1 and A, AR, AR2, …, ARn-1

We have to show that, the products of the corresponding terms of two G.P.’s form a G.P.

Consider first and second term,

So, ratio = second term/third term = arAR/aA

= rR

Then, Consider second and third term,

So, ratio = third term/second term = ar2AR2/arAR

= rR

By comparing the both the results 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.

Solution:-

From the question it is given that,

a, b, c are in G.P.

We know that, b2 = ac

We have to show that, 1/a, 1/b, 1/c are also in G.P.

(1/b)2 = (1/a) × (1/c)

(1/b2) = = (1/ac)

By cross multiplication we get,

ac = b2

Hence it is proved that, 1/a, 1/b, 1/c are 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.

Solution:-

From the question it is given that,

K is any positive real number

Ka, Kb, Kc are three consecutive terms of a G.P.

We know that, b2 = ac

(Kb)2 = ka × Kc

K2b = k a + c … [ from a× an = am + n]

By comparing left hand side and right hand side we get,

2b = a + c

Therefore, 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 GP.

Solution:-

From the question it is given that,

p, q, r are in A.P.

So, 2p = p + r

We have to show that pth, qth and rth terms of any G.P.

Pth term in G.P. = ARp – 1

Qth term in G.P. = ARq – 1

Rth term in G.P. = ARr – 1

So, if (ARq – 1)2 = ARp – 1 × ARr – 1

A2R2q – 2 = A2Rp – 1 + r – 1

A2R2q – 2 = A2Rp + r – 2

R2q – 2 = Rp + r – 2

By comparing left hand side and right hand side we get,

2p – 2 = p + r – 2

2p = P + r

Therefore, p, q, r are in A.P

26.

If a, b, c are in GP., prove that the following are also in G.P.
(i) a3, b3, c3

Solution:-

From the question it is given that,

a, b, c are in GP.

So, b2 = ac

We have to prove that, a3, b3, c3 are in G.P.

Then, (b3)2 = a3 × c3

It can be written as, (b2)3 = (a × c)3

b2 = ac

Therefore, it is proved that a3, b3, c3 are in G.P.


(ii) a2 + b2, ab + bc, b2 + c2.

Solution:-

From the question it is given that,

a, b, c are in GP.

So, b2 = ac

We have to prove that, a2 + b2, ab + bc, b2 + c2. are in G.P.

Then, (ab + bc)2 = (a2 + b2) (b2 + c2)

a2b2 + b2c2 + 2ab2c = a2b2 + a2c2 + b2c2 + b4

By transposing and simplification, we get,

b4 + a2c2 – 2ab2c = 0

(b2 – ac)2 = 0

b2 = ac

Therefore, a2 + b2, ab + bc, b2 + c2 are in GP.

27. If a, b, c, d are in G.P., show that
(i) a2 + b2, b2 + c2, c2 + d2 are in G.P.

Solution:-

From the question it is given that

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

So, bc = ad … [equation (i)]

b2 = ac … [equation (ii)]

c2 = bd … [equation (iii)]

We have to show that, a2 + b2, b2 + c2, c2 + d2 are in G.P.

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

Consider the LHS = (b2 + c2)2

= b4 + c4 + 2b2c2

From the equation (ii) and equation (iii),

= a2c2 + b2d2 + a2d2 + b2c2 

= c2(a2 + b2) + d2(a2 + b2)

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

Now consider the RHS = (a2 + b2) (c2 + d2)

By comparing the LHS and RHS

LHS = RHS

Hence it is proved that, a2 + b2, b2 + c2, c2 + d2 are in G.P.


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

Solution:-

From the question it is given that

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

We have to prove that, (b – c)2 + (c – a)2 + (d – b)2 = (a – d)2.

Consider the LHS = (b – c)2 + (c – a)2 + (d – b)2

We know that, the first, second and third terms of G.P. generally a, ar, ar2

So, LHS = (ar – ar2)2 + (ar2 – a)2 + (ar3 – ar)2

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

By taking out a2 as common we get,

= 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)

Now, consider the RHS = (a – d)2

= (a – ar3)2

= a2 (1 – r3)2

= a2(1 + r6 – 2r3)

= a2 (r6 – 2r3 + 1)

By comparing the LHS and RHS

LHS = RHS

Hence it is proved that, (b – c)2 + (c – a)2 + (d – b)2 = (a – d)2.


Exercise 9.5

1. Find the sum of:

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

Solution:-

From the question,

First term a = 2,

Common ratio r = 6/2 = 3

Number of terms n = 20

So, S20 = a(rn – 1)/r – 1

= 2(320 – 1)/3 – 1

= 2(320 – 1)/2

= 320 – 1

Therefore, S20 = 320 – 1


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

Solution:-

From the question,

First term a = 1,

Common ratio r = √3/1 = √3

Number of terms n = 10

So, S10 = a(rn – 1)/r – 1

= 1((√3)10 – 1)/ √3 – 1

Multiplying (√3 + 1) for both numerator and denominator we get,

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

= (35 – 1) (√3 + 1))/3 – 1 … [by rationalizing the denominator]

= ((243 – 1)( √3 + 1))/2

= 242(√3 + 1)/2

= 121(√3 + 1)

Therefore, S10 = 121(√3 + 1

(iii) 6 terms of the GP 1, -2/3, 4/9, …

Solution:-

From the question,

First term a = 1,

Common ratio r = -2/3 × 1= -2/3

Number of terms n = 6

So, S6 = a(rn – 1)/r – 1

= 1[1 – (-2/3)6]/(1 + (2/3))

= (3/5) (1 – (-26/36))

= (3/5) (1 – (64/729))

= (3/5) ((729 – 64)/729)

= 3/5 × (665/729)

= 133/243

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

Solution:-

From the question,

First term a = 1,

Common ratio r = 2/3 × 1= -/3

Number of terms n = 5

So, Sn = a(1 – rn)/1 – r

= 1[1 – (2/3)n]/(1 – 2/3)

Sn = 3[1 – (2/3)n]

Then, S5 = 3[1 – (2/3)5]

= 3[1 – (32/243)]

= 3((243 – 32)/243)

= 211/81

(v) n terms of the G.P. √7, √21, 3√7, …

Solution:-

From the question,

First term a = √7,

Common ratio r = √21/√7 = √3

Number of terms n = 10

So, Sn= a(rn – 1)/r – 1

= √7((√3)n – 1)/ √3 – 1

Multiplying (√3 + 1) for both numerator and denominator we get,

= √7((√3n – 1) (√3 + 1))/ ((√3 – 1) (√3 + 1)

= [√7((√3)n – 1) (√3 + 1)]/((√3)2 – 12) … [by rationalizing the denominator]

= (√7[(√3)n – 1] (√3 + 1))/3 – 1

Therefore, Sn = √7/2 [(√3)n – 1] (√3 + 1)

(vi) n terms of the G.P. 1, -a, a2, -a3, … (a ≠ -1)

Solution:-

From the question,

First term a = 1,

Common ratio r = -a/1 = -a

So, Sn= a(1 – rn)/1 – r

= 1[1- (-a)n]/(1 – (-a))

= (1 – (-a)n)/( 1 + a)

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

Solution:-

From the question,

First term a = x3,

Common ratio r = x5/x3 = x5 – 3 = x2

So, Sn= a(1 – rn)/1 – r

= x3[(1 – (x2)n]/(1 – x2) if r < 1

= x3(1 – x2n)/1 – x2

And also 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 + ….

Solution:-

From the question it is given that,

a = √2

r = √3

We know that, Sn = a(rn – 1)/(r – 1)

S10 = √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)]

Rationalizing the denominator, we get,

= 242(√6 + √2)/(3 – 1)

= 242(√6 + √2)/2

= 121(√6 + √2)

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

Solution:-

From the question it is given that,

First term a = 81

r = -27/81

= -1/3

Last term l = -1/27

Sn = (a – lr)/(l – 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.

Solution:-

From the question it is given that,

The nth term of a G.P. Tn = 128

The sum of its n terms Sn = 255

Common ratio r = 2

We know that, Tn = arn – 1

128 = a2n – 1

a = 128/2n – 1 … [equation (i)]

Also we know that, Sn = a(rn – 1)/(r – 1)

255 = a(2n – 1)/(2 – 1)

By cross multiplication we get,

255 = a(2n – 1)

a = 255/(2n – 1) … [equation (ii)]

Now, consider both the equation(i) and equation (ii)

255/(2n – 1) = 128/(2n – 1)

By cross multiplication we get,

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

By simplification,

2n = 256

2n = 28

By comparing both LHS and RHS, we get,

Then, 128 = a27

128 = a × 128

a = 128/128

a = 1

Therefore, the value of a is 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.

Solution:-

From the question it is given that,

the sum of first six terms of any G.P. is equal to 9 times the sum of the first three terms,

S6 = 9S3

We know that,

Sn = a(rn – 1)/(r – 1)

S6 = a(r6 – 1)/(r – 1)

S3 = a(r3 – 1)/(r – 1)

Now,

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

By simplification we get,

r6 – 1 = 9(r3 – 1)

(r6 – 1)/(r3 – 1) = 9

[(r3 + 1) (r3 – 1)]/(r3 – 1) = 9

r3 + 1 = 9

r3 = 9 – 1

r3 = 8

r3 = 23

r = 2

Therefore, common ratio r = 2

6.

(i) How many terms of the G.P. 3, 32, 33, … are needed to give the sum 120?

Solution:-

From the question it is given that,

Terms of the G.P. 3, 32, 33, …

Sum of the terms = 120

The first term a = 3

r = 32/3

= 9/3

= 3

We know that, Sn = a(rn – 1)/r – 1 = 120

3(3n – 1)/(3 – 1) = 120

3(3n – 1)/2 = 120

By cross multiplication we get,

3n – 1 = (120 ×2)/3

3n – 1 = 240/3

3n – 1 = 80

3n = 80 +1

3n = 81

3n = 34

Therefore, n = 4

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

Solution:-

From the question it is given that,

Terms of the G.P. 1, 4, 16, …

Sum of the terms = 341

The first term a = 1

r = 4/1

= 4

We know that, Sn = a(rn – 1)/r – 1 = 341

1(4n – 1)/(4 – 1) = 341

1(4n – 1)/3 = 341

By cross multiplication we get,

4n – 1 = (341 × 3)

4n – 1 = 1023

4n = 1023 + 1

4n = 1024

ML Aggarwal Solutions for Class 10 Maths Chapter 9 Image 5

4n = 45

Therefore, n = 5

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

Solution:-

From the question it is given that,

Terms of the G.P. 1, √2 > 2, 2 √2, …

Sum of the terms = 1023(√2 + 1)

The first term a = 1

r = √2/1 = √2

We know that, Sn = a(rn – 1)/r – 1 = 1023( √2 + 1)

1[(√2n – 1)]/(√2 – 1) = 1023(√2 + 1)

(√2n – 1) = 1023(√2 + 1)( √2 – 1)

(√2)n – 1 = 1023[(√2)2 – 12]

(√2)n – 1 = 1023(2 – 1)

(√2)n – 1 = 1023(1)

(√2)n – 1 = 1023

(√2)n = 1023 + 1

(√2)n = 1024

ML Aggarwal Solutions for Class 10 Maths Chapter 9 Image 6

(√2)n = 210

(√2)n = (√2)20

n = 20

8. How many terms of the 2/9 – 1/3 + ½ + … will make the sum 55/72?

Solution:-

From the question it is given that,

Terms of G.P. is 2/9 – 1/3 + ½ + …

Sum of the terms = 55/72

The first term a = 2/9

r = -1/3 ÷ 2/9 = (-1/3) × (9/2) = – 3/2

We know that, Sn = a(rn – 1)/r – 1 = 55/72

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

1 – (-3/2)n = (55/72) × (5/2) × (9/2)

(1 – (-1))n (3/2)n = 275/32

1 + 1(3/2)n = 275/32

(3/2)n = 275/32 – 1

(3/2)n = (275 – 32)/32

(3/2)n = 243/32

(3/2)n = (3/2)5

Therefore, n = 5

9. The 2nd and 5th terms of a geometric series are -½ and sum 1/16 respectively. Find the sum of the series up to 8 terms.

Solution:-

From the question it is given that,

a2 = -½

a5 = 1/16

We know that, a2 = arn – 1

= ar2 – 1

a2 = ar = -½ … [equation (i)]

a5 = ar5 – 1

= ar4

a5 = ar4 = 1/16 … [equation (ii)]

Now, dividing equation (ii) by (i) we get,

r3 = 1/16 ÷ (-½)

= (1/16) × (-2)

= -1/8

r3 = (-1/2)3

So, r = -½

ar = -½

a × (-½) = -½

a = – ½ × (-2/1)

a = 1

Therefore, a = 1 and r = -½

Then, S8 = a(1 – rn)/(1 – r)

= 1[1 – (-1/2)8]/(1 + ½)

= [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.

Solution:-

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

a8 = ar7 = 1/81

ar7 = 1/81

27r7 = 1/81

r7 = 1/(81 × 27)

r7 = 1/2187

r7 = 1/(37)

r = 1/3

So, 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.

Solution:-

From the question it is given that,

Common ratio r = 3

Last term = 486

Sum of the terms = 728

We know that, Sn = a(rn – 1)/(r – 1)

= a(3n – 1)/(3 – 1) = 728

a(3n – 1)/2 = 728

a(3n – 1) = 728 × 2

a(3n – 1) = 1456 … [equation (i)]

Then, last term = arn – 1

486 = a × 3n – 1

486 = a(3n/3)

486 × 3 = a3n

1458 = a3n … [equation (ii)]

Consider equation (i), a(3n – 1) = 1456

a3n – a = 1456

Substitute the value of a3n in equation (i),

1458 – a = 1456

a = 1458 – 1456

a = 2

Therefore, the first term a is 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.

Solution:-

From the question it is given that,

First term a is = 7

Then, last term is = 448

Sum = 889

We know that, last term = arn – 1

7rn -1 = 448

rn – 1 = 448/7

rn – 1 = 64 … [equation (i)]

So, sum = a(rn – 1)/(r – 1) = 889

7(rn – 1)/(r – 1) = 889

(rn – 1)/(r – 1) = 889/7

(rn – 1)/(r – 1) = 127 … [equation (ii)]

Consider the equation (i),

rn/r = 64

rn = 64r

Now substitute the value of rn in equation (ii),

(64r – 1)/(r – 1) = 127

64r – 1 = 127r – 127

127r – 64r = -1 + 127

63r = 126

r = 126/63

r = 2

Therefore, 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.

Solution:-

From the question it is given that,

Common ratio r = 3

S7 = 2186

We know that, Sn = a(rn – 1)/(r – 1)

S7 = a(3n – 1)/(3 – 1)

2186 = a(3n – 1)/2

By cross multiplication,

(2186 × 2) = a(37 – 1)

(4372) = a(2187 – 1)

4372 = a2186

a = 4372/2186

a = 2

Then, a3 = ar3 – 1

= ar2

= 2 × 32

= 2 × 9

a3 = 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.

Solution:-

From the question it is given that,

First term of a G.P. is a = 5

The sum of first three terms is S3 = 31/5

We know that, Sn = a(rn – 1)/(r – 1)

S3 = a(r3 – 1)/(r – 1)

31/5 = 5(r3 – 1)/(r – 1)

31/(5 × 5) = (r3 – 1)/(r – 1)

31/25 = (r3 – 1)/(r – 1)

(r – 1) (r2 + r + 1)/(r – 1) = 31/25

r2 + r + 1 = 31/25

By cross multiplication we get,

25(r2 + r + 1) = 31

25r2 + 25r + 25 = 31

Transposing 31 from right hand side to left hand side it becomes – 31,

25r2 + 25r + 25 – 31 =

25r2 + 25r – 6 = 0

25r2 + 30r – 5r – 6 = 0

5r(5r + 6) – 1(5r + 6) = 0

(5r – 1) (5r + 6) = 0

Take 5r – 1 = 0

r = 1/5

or 5r + 6 = 0

r = -6/5

Therefore, common ratio r = 1/5 or -6/5.

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

Solution:-

Froom the question it is given that,

Ratio of the sum of first three terms to the sum of first six terms S3 ÷ S6 = 125 : 152

We know that, Sn = a(r3 – 1)/(r – 1)

S3 : S6 = 125 : 152

[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

(r3 – 1)/[(r3 + 1) (r3 – 1)] = 125/152

1/(r3 + 1) = 125/152

By cross multiplication,

(1 × 152) = (r3 + 1) × 125

152 = 125r3 + 125

125r3 = 152 – 125

125r3 = 27

r3 = 27/125

r3 = (3/5)3

r = 3/5

Therefore, common ratio r = 3/5

16.
ML Aggarwal Solutions for Class 10 Maths Chapter 9 Image 7

Solution:-

From the question it is given that,

n = 1, 2, 3, 4, …., 50

Then, 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

We know that, Sn = [a(an – 1)/(r – 1)] – 50

= [2(250 – 1)/(2 – 1)] – 50

= (2 × 250) – 2 – 50

= 251 – 52

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

Solution:-

From the question it is given that, x(x + y) + x2(x2 + y2) + x3 (x3 + y3) …

Then, Sn = x2 + xy + x4 + x2y2 + x6 + x3y3 + … n terms

By separating the terms,

Sn = x2 + x4 + x6 + … n terms G.P. … (1)

Sn = xy + x2y2 + x3y3 + … … (2)

In G.P. (1) first term a = x2, r = x4/x2 = x4 – 2 = x2

In G.P. (2) first term a = xy, r = x2y2/xy = x2 – 1y2-1 = xy

Sn = a(rn – 1)/(r – 1)

Sn = [(x2((x2)n – 1))/(x2 – 1)] + [(xy((xy)n – 1))/(xy – 1)]

Sn = [x2(x2n – 1)/(x2 – 1)] + [xy((xy)n – 1)/(xy – 1)]

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

Solution:-

From the question it is given that,

1 + (1 + x) + (1 + x + x2) + … to n terms

Now, multiply and divide by (1 – x) we get,

= [(1 – x)/(1 – x)] + [((1 – x)(1 + x))/(1 – x)] + [(1 – x) (1 + x + x2)/(1 – x)] + …

By taking common we get,

= 1/(1 – x) [(1 – x) + (1 + x)2 + (1 + x3) + ….]

= 1/(1 – x) [1 – x + 1 + x2 + 1 + x3 + …]

= 1/(1 – x) [(1 + 1 + 1 + … n terms) – (x + x2 + x3+ … n terms)]

We know that, Sn = a(1 – rn)/(1 – r)

= (1/(1 – x)) [n – (x(1 – xn)/(1 – x))]

= (1/(1 – x)) [(n(1 – x) – x(1- xn))/(1- x)]

= (1/(1 – x2)) [n(1 – x) – x(1 – xn)]

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

(i) 7 + 77 + 777 + … 

Solution:-

Consider the given numbers 7 + 77 + 777 + … n terms

Take out 7 as common we get,

= 7 (1 + 11 + 111 + … n terms)

= 7/9 (9 + 99 + 999 + … n terms)

= 7/9 ((10 – 1) + (100 – 1) + (1000 – 1) + … n terms)

= 7/9 (10 + 100 + 100 + … n terms – (1 + 1 + 1 + …. n terms))

We know that, Sn = a(rn – 1)/(r – 1)

First term a = 10

Common ratio r = 10

= 7/9 [(10(10n – 1)/(10 – 1)) – n]

= 7/9 [(((10 × 10n) – 10)/9) – n]

= 7/81 [10n + 1 – 10 – 9n]

= 7/81 [10n + 1 – 9n – 10]

(ii) 8 + 88 + 888 + …

Solution:-

Consider the given numbers 8 + 88 + 888 + … n terms

Take out 8 as common we get,

= 8(1 + 11 + 111 + … n terms)

= 8/9 (9 + 99 + 999 + … n terms)

= 8/9 ((10 – 1) + (100 – 1) + (1000 – 1) + … n terms)

= 8/9 (10 + 100 + 100 + … n terms – (1 + 1 + 1 + …. n terms))

We know that, Sn = a(rn – 1)/(r – 1)

First term a = 10

Common ratio r = 10

= 8/9 [(10(10n – 1)/(10 – 1)) – n]

= 8/9 [(((10 × 10n) – 10)/9) – n]

= 8/81 [10n + 1 – 10 – 9n]

= 8/81 [10n + 1 – 9n – 10]

 

(iii) 0.5 + 0.55 + 0.555 + …

Solution:-

Consider the given numbers 0.5 + 0.55 + 0.555 + … n terms

Take out 5 as common we get,

= 5(0.1 + 0.11 + 0.111 + … n terms)

= 5/9 (0.9 + 0.99 + 0.999 + … n terms)

= 5/9 ((1 – 0.1) + (1 – 0.01) + (1 – 0.001) + … n terms)

= 5/9 (1 + 1 + 1 + … n terms – (0.1 + 0.01 + 0.001 + …. n terms))

We know that, Sn = a(1 – rn)/(1 – r)

= 5/9 [n – (0.1(1 – (-0.1)n)/(1 – 0.1))]

= 5/9 [n – ((1/9) (1 – (1/10n)))]

= 5/81 [9n – 1 + (1/10n)]

Chapter – test

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

Solution:-

From the question it is given that,

First term a = – 5

Common difference d = -3

Then the first four terms are = – 5 + (-3) = -5 – 3 = – 8

-8 + (-3) = – 8 – 3 = -11

– 11 + (-3) = – 11 – 3 = – 14

Therefore, first four terms are -5, -8, -11 and -14.

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

(i) 0, ¼, ½, ¾, …

Solution:-

From the question it is given that,

First term a = 0

Common difference = ¼ – 0 = ¼

So, next three numbers are ¾ + ¼ = 4/4 = 1

1 + ¼ = (4 + 1)/4 = 5/4

5/4 + ¼ = 6/4 = 3/2

Therefore, the next three term are 1, 5/4 and 3/2.

(ii) 5, 14/3, 13/3, 4, …

Solution:-

From the question it is given that,

First term a = 5

Common difference = 14/3 – 5 = (14 – 15)/3 = -1/3

So, next three numbers are 4 + (-1/3) = (12 – 1)/3 = 11/3

11/3 + (-1/3) = (11 – 1)/3 = 10/3

10/3 + (-1/3) = (10 – 1)/3 = 9/3 = 3

Therefore, the next three term are 11/3, 10/3 and 3.

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

Solution:-

From the question it is given that,

nth term is 6n + 2

So, Tn = 6n + 2

Now, we start giving values, 1, 2, 3, … in the place of n, we get,

T1 = (6 × 1) + 2 = 6 + 2 = 8

T2 = (6 × 2) + 2 = 12 + 2 = 14

T3 = (6 × 3) + 2 = 18 + 2 = 20

T4 = (6 × 4) + 2 = 24 + 2 = 26

Therefore, A.P. is 8, 14, 20, 26, …

So, common difference 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.

Solution:-

From the question,

The first term a = 9

Then, difference d = 12 – 9 = 3

15 – 12 = 3

18 – 15 = 3

Therefore, common difference d = 3

From the formula, an = a + (n – 1)d

Tn = a + (n – 1)d

= 9 + (n – 1)3

= 9 + 3n – 3

= 6 + 3n

So, T16 = a + (n – 1)d

= 9 + (16 – 1)3

= 9 + (15)(3)

= 9 + 45

= 54

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

Solution:-

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 + (n – 1)(-3)

-40 – 17 = -3n + 3

– 57 – 3 = -3n

n = -60/-3

n = 20

Therefore, 6th term form 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.

Solution:-

From the question it is given that,

a8 = 31

a15 = the 15th term is 16 more than its 11th term = a11 + 16

we know that, an = a + (n – 1)d

So, a8 = a + 7d = 31 … [equation (i)]

a15 = a + 14d = a + 10d + 16

14d – 10d = 16

4d = 16

d = 16/4

d = 4

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

a + (7 × 4) = 31

a + 28 = 31

a = 31 – 28

a = 3

So, 3 + 4 = 7, 7 + 4 = 11, 11 + 4 = 15

Therefore, A.P. is 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.

Solution:-

From the question it is given that,

a17 = 5 more than twice its 8th term = 2a8 + 5

a11 = 43

an = ?

We know that, a11 = a + 10d = 43 … [equation (i)]

a17 = 2a8 + 5

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

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

2a – a = 16d – 14d – 5

a = 2d – 5 … [equation (ii)]

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

2d – 5 + 10d = 43

12d = 43 + 5

12d = 48

d = 48/12

d = 4

To find out the value of a substitute the value of d in equation (i)

a + (10 × 4) = 43

a + 40 = 43

a = 43 – 40

a = 3

Then, 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.

Solution:-

From the question it is given that,

a19 = 19th term of an A.P. is equal to three times its 6th term = 3a6

a9 = 19

As we know, an = a + (n – 1)d

a9 = a + 8d = 19 … [equation (i)]

Then, a19 = 3(a + 5d)

a + 18d = 3a + 15d

3a – a = 18d – 15d

2a = 3d

a = (3/2)d

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

(3/2)d + 8d = 19

(3d + 16d)/2 = 19

(19/2)d = 19

d = (19 × 2)/19

d = 2

To find out the value of a substitute the value of d in equation (i)

a + 8d = 19

a + (8 × 2) = 19

a + 16 = 19

a = 19 – 16

a = 3

Therefore, 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?

Solution:-

From the question it is given that,

a3 = 4

a9 = – 8

We know that, a3 = a + 2d = 4 … [equation (i)]

a9 = a + 8d = -8 … [equation (ii)]

Now, subtracting equation (i) from equation (ii)

(a + 8d) – (a + 2d) = -8 – 4

a + 8d – a – 2d = -12

6d = -12

d = -12/6

d = -2

To find out the value of a substitute the value of d in equation (i)

a + 2d = 4

a + (2 × (-2)) = 4

a – 4 = 4

a = 4 + 4

a = 8

let us assume nth term be zero, then

a + (n – 1)d = 0

8 + (n – 1)(-2) = 0

-2n + 2 = -8

-2n = -8 – 2

-2n = -10

n = -10/-2

n = 5

Therefore, 0 will be the fifth term.

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

Solution:-

From the question it is given that,

First term a = 5

nth term = -55

Common difference d = 2 – 5 = – 3

We know that, an = a + (n – 1)d

– 55 = 5 + (n – 1)(-3)

-55 – 5 = – 3n + 3

-60 – 3 = -3n

-63 = -3n

n = -63/-3

n = 21

Therefore, -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.

Solution:-

From the question it is given that,

The 24th term of an A.P. is twice its 10th term = a24 = 2a10

We have to show that, 72nd term is four times its 15th term = a72 = 4a15

We know that, a24 = a + 23d = 2a10

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

a + 23d = 2a + 18d

2a – a = 23d – 18d

a = 5d … [equation (i)]

a72 = 4a15

a + 71d = 4(a + 14d)

Substitute the value of a we get,

5d + 71d = 4(5d + 14d)

76d = 4(19d)

Therefore, it is proved that 72nd term is four times its 15th term.

12. Which term of the list of numbers 20, 19¼, 18½, 17¾, … is the first negative term?

Solution:-

From the question it is given that,

First term a = 20

Common difference d = 19¼ – 20 = 77/4 – 20 = (77 – 80)/4 = -¾

We know that, an = a + (n – 1)d

an = 20 + (n – 1) (-¾)

an = 20 – ¾n + ¾

an = 20 + ¾ – ¾n

an = (80 + 3)/4 – ¾n

an = 83/4 – ¾n < 0

83/4 < ¾n

83 < 3n

83/3 < n

28 < n

Therefore, 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)

Solution:-

From the question it is given that,

pth term = q

qth term = p

We have to show that, nth term is (p + q – n)

We know that, an = a + (n – 1)d

So, pth term = a + (p -1)d = q … [equation (i)]

qth term = a + (q – 1)d = p … [equation (ii)]

Now subtracting equation (ii) from equation (i), we get

q – p = (a + (p – 1)d) – (a + (q – 1)d)

q – p = (a + pd – d) – (a +qd – d)

q – p = a + pd – d – a – qd + d

q – p = pd – qd

q – p = d (p – q)

d = (q – p)/(p – q)

d = -(p – q)/(p – q)

d = – 1

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

a + (p -1)(-1) = q

a – p + 1= q

a = q + p – 1

Then, nth term = a + (n – 1)d

= (p + q – 1) + (n – 1) (- 1)

= (p + q – 1) – n + 1

= p + q – 1 – n + 1

= p + q – n

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

Solution:-

The three digits numbers which are divisible by 9 are 108, 117, 126, …, 999

Then, first term a = 108

Common difference = 9

Last term = 999

We know that, l = an = a + (n – 1)d

999 = 108 + (n – 1)9

999 – 108 = 9n – 9

891 + 9 = 9n

900 = 9n

n = 900/9

n = 100

Therefore, there are 100 three digits numbers.

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

Solution:-

From the question it is given that,

The sum of three numbers in A.P. = – 3

The product of three numbers in A.P. = 8

Let us assume the 3 numbers which are in A.P. are, a – d, a, a + d

Now adding 3 numbers = a – d + a + a + d = – 3

3a = -3

a = -3/3

a = -1

From the question, product of 3 numbers is – 35

So, (a – d) × (a) × (a + d) = 8

a(a2 – d2) = 8

-1 ((-1)2 – d2 = 8

1 – d2 = 8/-1

1 – d2 = -8

d2 = 8 + 1

d2= 9

d = √9

d = ±3

Therefore, the numbers are if d = 3 (a – d) = – 1 – 3 = – 4

a = -1

(a + d) = – 1 + 3 = 2

If d = – 6

The numbers are (a – d) = – 1 – (-3) = – 1 + 3 = 2

a = -1

(a + d) = -1 + (-3) = -1 – 3 = -4

Therefore, the numbers -4, -1, 2,… and 2, -1, -4,… are in A.P.

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.

Solution:-

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 360o.

a – 3d + a – d + a + d + a+ 3d = 360o

4a = 360o

a = 360/4

a = 90o

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 = 120o

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

a – d = 90o – 10o = 80o

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

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

17. The nth term of an A.P. cannot be n² + n + 1. Justify your answer.

Solution:-

From the question it is given that,

The nth term of an A.P. cannot be n² + n + 1.

Now, we start giving values, 1, 2, 3, … in the place of n, we get,

a1 = 12 + 1 + 1 = 1 + 2 = 3

a2 = 22 + 2 + 1 = 4 + 3 = 7

a3 = 32 + 3 + 1 = 9 + 4 = 13

a4 = 42 + 4 + 1 = 16 + 5 = 21

Then, difference d = a2 – a1 = 7 – 3 = 4

d = a3 – a2 = 13 – 7 = 6

d = a4 – a3 = 21 – 13 = 8

Therefore, common difference d is not same in the numbers.

Hence, the numbers are not form A.P.

So, an ≠ n2 + n + 1

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

Solution:-

From the question it is given that,

nth term is 15 – 4n

So, an = 15 – 4n

Now, we start giving values, 1, 2, 3, … in the place of n, 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

S20 = (n/2) [2a + (n – 1)d]

= (20/2) [(2 × 11) + (20 – 1)(-4)]

= 10 [22 – (19) (-4)]

= 10 [22 – 76]

= 10(-54)

= -540

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

19. Find the sum : 18 + 15½ + 13 + … + (-49½)

Solution:-

From the question it is given that,

First term a = 18

Common difference d = 15½ – 18

= 31/2 – 18

= (31 – 36)/2

= -5/2

Last term = -49½ = -99/2

We know that, an = a + (n – 1)d

-99/2 = 18 + (n – 1)(-5/2)

(-99/2) – (18/1) = (n – 1)(-5/2)

(-99 – 36)/2 = (-5/2)(n – 1)

(-135/2) = (-5/2) (n – 1)

(-135/2) × (-2/5) = n – 1

-135/-5 = n – 1

27 = n – 1

n = 27 + 1

n = 28

Then, Sn = (n/2) [2a + (n – 1)d]

S28 = (28/2) [(2 × 18) + (28 – 1)(-5/2)]

S28 = 14[36 + (27 × (-5/2))]

S28 = 14[36 – (135/2)]

S28 = 14 [(72 – 135)/2]

S28 = 14 (-63/2)

S28 = – 441

20.

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

Solution:-

From the question it is given that,

Terms of the A.P. is -6, (-11/2) – 5, …

The first term a = -6

Common difference d = (-11/2) – (-6)

= (-11/2) + 6

= (-11 + 12)/2

= ½

The terms are make the sum – 25

Then, Sn = (n/2)(2a + (n – 1)d)

-25 = (n/2) [(2 × (-6)) + (n – 1) (½)]

(-25 × 2) = n [-12 + ½n – ½]

-50 = n [(-25/2) + (½n)]

½n2 – (25/2)n + 50 = 0

n2 – 25n + 100 = 0

n2 – 5n – 20n + 100 = 0

n(n – 5) – 20(n – 5) = 0

(n – 5) (n – 20) = 0

So, n – 5 = 0

n = 5

or n – 20 = 0

n = 20

Therefore, number of terms are 5 or 20.

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

Solution:-

From the question it is given that,

First term a = 2

Last term = x

Common difference d = 5 – 2 = 3

Then, sum of the terms = 155

L = a + (n – 1)d

x = 2 + (n – 1)3

x = 2 + 3n – 3

x = 3n – 1 … [equation (i)]

We know that, 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)

310 = 3n2 + n

3n2 + n – 310 = 0

3n2 – 30n + 31n – 310 = 0

3n(n – 10) + 31(n – 10) = 0

(n – 10) (3n + 31) = 0

So, n – 10 = 0

n = 10

or 3n + 31 = 0

n = -31/3

negative is not possible.

Therefore, n = 10

Now, substitute the value of n in equation (i),

x = 3n – 1

= (3 × 10) – 1

= 30 – 1

= 29

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.

Solution:-

From the question it is given that,

The third term of an A.P. a3 = 5

The ratio of its 6th term to the 10th term a6 : a10 = 7 : 13

We know that, an = a + (n – 1)d

a3 = a + (3 – 1)d = 5

= a + 2d = 5 … [equation (i)]

Then, a6/a10 = 7/13

(a + 5d)/(a + 9d) = 7/13

By cross multiplication we get,

13(a + 5d) = 7(a + 9d)

13a + 75d = 7a + 63d

13a – 7a + 65d – 63d = 0

6a + 2d = 0

Divide by 2 on both side we get,

3a + d = 0

d = -3a … [equation (ii)]

Substitute the value of d in equation (i),

a + 2(-3a) = 5

a – 6a = 5

-5a = 5

a = -5/5

a = -1

Now substitute the value of a in equation (ii),

d = -3(-1)

d = 3

Then, sum of first 20 terms,

= (n/2) [2a + (n – 1)d]

= (20/2)[(2 × (-3)) + (2- – 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.

Solution:-

From the question it is given that,

First term a = 2

Last term = 29

The sum of terms = 155

We know that, last term = an = a + (n – 1)d

29 = 2 + (n – 1)d

29 – 2 = d(n – 1)

27 = d(n – 1) … (i)

Then, Sn = (n/2)[2a + (n – 1)d]

155 = (n/2)[(2 ×2) + 27]

155 = (n/2)[4 + 27]

155 = (31/2)n

n = (155 × 2)/31

n = 10

d(n – 1) = 27

d(10 – 1) = 27

d(9) = 27

d = 27/9

d = 3

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

Solution:-

From the question it is given that,

First term a = 10

The sum of first 14 terms of an A.P. = 1505

25th term = ?

We know that, Sn = (n/2) [2a + (n – 1)d]

S14 = (n/2) [2a + (n – 1)d]

1505 = (14/2) [(2 × 10) + (14 – 1)d]

1505 = 7[20 + 13d]

1505/7 = 20 + 13d

215 = 20 + 13d

13d = 215 – 20

13d = 195

d = 195/13

d = 15

Then, an = a + (n – 1)d

a25 = 10 + (25 – 1)(15)

= 10 + (24)15

= 10 + 360

= 370

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

Solution:-

From the question it is given that,

The sum of first n term of an A.P. is 3n² + 4n

Sn = 3n2 + 4n

So, 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

Therefore, 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.

Solution:-

From the question it is given that,

The sum of first 10 terms = – 150

The sum of next 10 terms = – 550

A.P = ?

We know that, Sn = (n/2) [2a + (n – 1)d]

S10 = (n/2) [2a + (10 – 1)d]

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

-150 = 5[2a + 9d]

– 150 = 10a + 45d … [equation (i)]

Then, S20 = S10 + S10

= – 150 – 550

= -700

S20 = (20/2) [2a + 19d]

-700 = 10(2a + 19d)

-700 = 20a + 190d … [equation (ii)]

Now, multiplying equation (i) by 2 we get,

20a + 90d = – 300 … [equation (iii)]

Subtract equation (iii) from equation (ii),

(20a + 190d) – (20a + 90d) = -700 – (-300)

20a + 190d – 20a – 90d = -700 + 300

100d = -400

d = -400/100

d = -4

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

10a + 45(-4) = -150

10a – 180 = – 150

10a = -150 + 180

10a = 30

a = 30/10

a = 3

a2 = 3 + (-4) = 3 – 4 = -1

a3 = -1 + (-4) = -1 – 4 = -5

a3 = -5 + (-4) = -5 – 4 = -9

Therefore, 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.

Solution:-

From the question it is given that,

an = 107

The sum of first m terms of an A.P. Sm = 4m2 – m

nth term Sn is = 4n2 – n

Then, Sn – 1 = 4(n – 1)2 – (n – 1)

= 4(n2 – 2n + 1) – n + 1

= 4n2 – 8n + 4 – n + 1

= 4n2 – 9n + 5

Therefore, an = Sn – Sn – 1

107 = 4n2 – n – 4n2 + 9n – 5

107 = 8n – 5

107 + 5 = 8n

112 = 8n

n = 112/8

n = 14

an = 8n – 5

a21 = (8 × 14) – 5

a21 = 168 – 5

a21 = 163

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

Solution:-

From the question it is given that,

The geometric progression whose 4th term a4 = 54

The geometric progression whose 7th term a7 = 1458

We know that, an = arn – 1

a4 = ar4 – 1

a4 = ar3 = 54

a7 = ar6 = 1458

By dividing both we get,

ar6/ar3 = 1458/54

r6 – 3 = 27

r3 = 33

r = 3

To find out a, consider ar3 = 54

a(3)3 = 54

a = 54/27

a = 2

Therefore, a = 2, r = 3

So, G.P. is 2, 6, 18, 54,…

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.

Solution:-

From the question it is given that,

The fourth term of a G.P. is the square of its second term = a4 = (a2)2

The first term a1 = – 3

We know that, an = arn – 1

a4 = ar4 – 1

a4 = ar3

a2 = ar

Now, ar3 = (ar)2

ar3 = a2r2

r3/r2 = a2/a

r3 – 2 = a2 – 1 … [from am/an = am – n]

r = a

a1 = -3

a7 = ar7 – 1

a7 = ar6

= -3 × (-3)6

= -3 × 729

= -2187

Therefore, the 7th term a7 = -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.

From the question it is given that,

a4 = x

a10 = y

a16 = z

Now, we have to show that x, y and z are in G.P.

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 = a2r18

Then, xz = ar3 × ar15

= a1 + 1 r3 + 15 … [from am × an = am + n]

= a2r18

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, ¾ are needed to give the sum 3069/512?

Solution:-

From the question it is given that,

Sum of the terms Sn = 3069/512

First term a = 3

Common ratio r = (3/2)/3

= (3/2) × (1/3)

= ½

We know that, Sn = a(1 – rn)/(1 – r)

(3069/512) = 3[1 – (½)n]/ (1 – ½)

(3069/512) = (2 × 3) [1 – (½)n]

1 – (½)n = 3069/(512 × 6)

1 – (½)n = 1023/1024

(½)n = 1 – (1023/1024)

(½)n = (1024 – 1023)/1024

(½)n = 1/1024

ML Aggarwal Solutions for Class 10 Maths Chapter 9 Image 8

(½)n = (½)10

By comparing both LHS and RHS,

n = 10

Therefore, there are 10 terms are in the G.P.

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

Solution:-

Consider the given numbers 3 + 33 + 333 + … n terms

Take out 3 as common we get,

= 3 (1 + 11 + 111 + … n terms)

= 3/9 (9 + 99 + 999 + … n terms)

= 3/9 ((10 – 1) + (100 – 1) + (1000 – 1) + … n terms)

= 3/9 (10 + 100 + 100 + … n terms – (1 + 1 + 1 + …. n terms))

We know that, Sn = a(rn – 1)/(r – 1)

First term a = 10

Common ratio r = 10

= 3/9 [(10(10n – 1)/(10 – 1)) – n]

= 3/9 [(((10 × 10n) – 10)/9) – n]

= 3/81 [10n + 1 – 10 – 9n]

= 1/27 [10n + 1 – 9n – 10]

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

Solution:-

Consider the given numbers 7 + 7.7 + 7.77 + 7.777 + … to 50 terms

Take out 7 as common we get,

= 7(1 + 1.1 + 1.11 + 1.111 + … 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)

= 7/9 [500 – (1(1 – (0.1)50)/(1 – 0.1))]

= 7/9 [500 – ((10/9) (1 – (1/1050)))]

= 7/81 [4500 – 10 + 10-49]

= 7/81[4490 + 10-49]

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