# Selina Solutions Concise Maths Class 10 Chapter 11 Geometric Progression

A sequence, in which each of its terms can be obtained by multiplying or dividing its preceding term by a fixed quantity, is called a Geometric Progression. So, this chapter is going to be completely about G.P., its general term, properties and the sum of terms in a G.P. Students who find it difficult to solve problems in this chapter can refer toÂ Selina Solutions for Class 10 Mathematics prepared by our experienced faculty at BYJUâ€™S. The solutions are prepared with an aim to boost confidence among students for taking up their Class 10 final exams fearlessly. This mainly improves problem-solving skills of the students, which are vital from an examination point of view. Selina Solutions for Class 10 Mathematics Chapter 11 Geometric Progression PDF are available right below.

## Access Selina Solutions Concise Maths Class 10 Chapter 11 Geometric Progression

### Exercise 11(A) Page No: 152

1. Find which of the following sequence form a G.P.:

(i) 8, 24, 72, 216, â€¦â€¦â€¦

(ii) 1/8, 1/24, 1/72, 1/216, â€¦â€¦â€¦

(iii) 9, 12, 16, 24, â€¦â€¦â€¦

Solution:

(i) Given sequence: 8, 24, 72, 216, â€¦â€¦â€¦

Since,

24/8 = 3, 72/24 = 3, 216/72 = 3

â‡’ 24/8 = 72/24 = 216/72 = â€¦â€¦â€¦.. = 3

Therefore 8, 24, 72, 216, â€¦â€¦â€¦ is a G.P. with a common ratio 3.

(ii) Given sequence: 1/8, 1/24, 1/72, 1/216, â€¦â€¦â€¦

Since,

(1/24)/ (1/8) = 1/3, (1/72)/ (1/24) = 1/3, (1/216)/ (1/72) = 1/3

â‡’ (1/24)/ (1/8) = (1/72)/ (1/24) = (1/216)/ (1/72) = â€¦â€¦â€¦.. = 1/3

Therefore 1/8, 1/24, 1/72, 1/216, â€¦â€¦â€¦ is a G.P. with a common ratio 1/3.

(iii) Given sequence: 9, 12, 16, 24, â€¦â€¦â€¦

Since,

12/9 = 4/3; 16/12 = 4/3; 24/16 = 3/2

12/9 = 16/12 â‰  24/16

Therefore, 9, 12, 16, 24 â€¦â€¦ is not a G.P.

2. Find the 9th term of the series: 1, 4, 16, 64, â€¦..

Solution:

Itâ€™s seen that, the first term is (a) = 1

And, common ratio(r) = 4/1 = 4

We know that, the general term is

tn = arn â€“ 1

Thus,

t9 = (1)(4)9 â€“ 1 = 48 = 65536

3. Find the seventh term of the G.P: 1, âˆš3, 3, 3 âˆš3, â€¦..

Solution:

Itâ€™s seen that, the first term is (a) = 1

And, common ratio(r) = âˆš3/1 = âˆš3

We know that, the general term is

tn = arn â€“ 1

Thus,

t7 = (1)(âˆš3)7 â€“ 1 = (âˆš3)6 = 27

4. Find the 8th term of the sequence:

Solution:

The given sequence can be rewritten as,

3/4, 3/2, 3, â€¦..

Itâ€™s seen that, the first term is (a) = 3/4

And, common ratio(r) = (3/2)/ (3/4) = 2

We know that, the general term is

tn = arn â€“ 1

Thus,

t8 = (3/4)(2)8 â€“ 1 = (3/4)(2)7 = 3 x 25 = 3 x 32 = 96

5. Find the 10th term of the G.P. :

Solution:

The given sequence can be rewritten as,

12, 4, 4/3, â€¦..

Itâ€™s seen that, the first term is (a) = 12

And, common ratio(r) = (4)/ (12) = 1/3

We know that, the general term is

tn = arn â€“ 1

Thus,

t10 = (12)(1/3)10 â€“ 1 = (12)(1/3)9 = 12 x 1/(19683) = 4/ 6561

6. Find the nth term of the series:

1, 2, 4,Â 8, â€¦â€¦..

Solution:

Itâ€™s seen that, the first term is (a) = 1

And, common ratio(r) = 2/ 1 = 2

We know that, the general term is

tn = arn â€“ 1

Thus,

tn = (1)(2)n â€“ 1 = 2n – 1

### Exercise 11(B) Page No: 154

1. Which term of the G.P. :

Solution:

In the given G.P.

First term, a = -10

Common ratio, r = (5/âˆš3)/ (-10) = 1/(-2âˆš3)

We know that, the general term is

tn = arn â€“ 1

So,

tn = (-10)( 1/(-2âˆš3))n â€“ 1 = -5/72

Now, equating the exponents we have

n â€“ 1 = 4

n = 5

Thus, the 5th of the given G.P. is -5/72

2. The fifth term of a G.P. is 81 and its second term is 24. Find the geometric progression.

Solution:

Given,

t5 = 81 and t2 = 24

We know that, the general term is

tn = arn â€“ 1

So,

t5 = ar5 â€“ 1 = ar4 = 81 â€¦. (1)

And,

t2 = ar2 â€“ 1 = ar1 = 24 â€¦. (2)

Dividing (1) by (2), we have

ar4/ ar = 81/ 24

r3 = 27/ 8

r = 3/2

Using r in (2), we get

a(3/2) = 24

a = 16

Hence, the G.P. is

G.P. = a, ar, ar2, ar3â€¦â€¦

= 16, 16 x (3/2), 16 x (3/2)2, 16 x (3/2)3

= 16, 24, 36, 54, â€¦â€¦

3. Fourth and seventh terms of a G.P. are 1/18 and -1/486 respectively. Find the G.P.

Solution:

Given,

t4 = 1/18 and t7 = -1/486

We know that, the general term is

tn = arn â€“ 1

So,

t4 = ar4 â€“ 1 = ar3 = 1/18 â€¦. (1)

And,

t7 = ar7 â€“ 1 = ar6 = -1/486 â€¦. (2)

Dividing (2) by (1), we have

ar6/ ar3 = (-1/486)/ (1/18)

r3 = -1/27

r = -1/3

Using r in (1), we get

a(-1/3)3 = 1/18

a = -27/ 18 = -3/2

Hence, the G.P. is

G.P. = a, ar, ar2, ar3â€¦â€¦

= -3/2, -3/2(-1/3), -3/2(-1/3)2, -3/2(-1/3)3, â€¦â€¦

= -3/2, 1/2, -1/6, 1/18, â€¦..

4. If the first and the third terms of a G.P are 2 and 8 respectively, find its second term.

Solution:

Given,

t1 = 2 and t3 = 8

We know that, the general term is

tn = arn â€“ 1

So,

t1 = ar1 â€“ 1 = a = 2 â€¦. (1)

And,

t3 = ar3 â€“ 1 = ar2 = 8 â€¦. (2)

Dividing (2) by (1), we have

ar2/ a = 8/ 2

r2 = 4

r = Â± 2

Hence, the 2nd term of the G.P. is

When a = 2 and r = 2 is 2(2) = 4

Or when a = 2 and r = -2 is 2(-2) = -4

5. The product of 3rdÂ and 8thÂ terms of a G.P. is 243. If its 4thÂ term is 3, find its 7thÂ term

Solution:

Given,

Product of 3rdÂ and 8thÂ terms of a G.P. is 243

The general term of a G.P. with first term a and common ratio r is given by,

tn = arn â€“ 1

So,

t3 x t8 = ar3 â€“ 1 x ar8 â€“ 1 = ar2 x ar7 = a2r9 = 243

Also given,

t4 = ar4 â€“ 1 = ar3 = 3

Now,

a2r9 = (ar3) ar6 = 243

Substituting the value of ar3in the above equation, we get,

(3) ar6 = 243

ar6 = 81

ar7 â€“ 1 = 81 = t7

Thus, the 7th term of the G.P is 81.

### Exercise 11(C) Page No: 156

1. Find the seventh term from the end of the series: âˆš2, 2, 2âˆš2, â€¦â€¦ , 32

Solution:

Given series: âˆš2, 2, 2âˆš2, â€¦â€¦ , 32

Here,

a = âˆš2

r = 2/ âˆš2 = âˆš2

And, the last term (l) = 32

l = tn = arn â€“ 1 = 32

(âˆš2)( âˆš2)n â€“ 1 = 32

(âˆš2)n = 32

(âˆš2)n = (2)5 = (âˆš2)10

Equating the exponents, we have

n = 10

So, the 7th term from the end is (10 â€“ 7 + 1)th term.

i.e. 4th term of the G.P

Hence,

t4 = (âˆš2)(âˆš2)4 â€“ 1 = (âˆš2)(âˆš2)3 = (âˆš2) x 2âˆš2 = 4

2. Find the third term from the end of the G.P.

2/27, 2/9, 2/3, â€¦â€¦., 162

Solution:

Given series: 2/27, 2/9, 2/3, â€¦â€¦., 162

Here,

a = 2/27

r = (2/9) / (2/27)

r = 3

And, the last term (l) = 162

l = tn = arn â€“ 1 = 162

(2/27) (3)n – 1 = 162

(3)n – 1 = 162 x (27/2)

(3)n – 1 = 2187

(3)n – 1 = (3)7

n – 1 = 7

n = 7+1

n = 8

So, the third term from the end is (8 – 3 + 1)th term

i.e 6th term of the G.P. = t6

Hence,

t6 = ar6-1

t6 = (2/27) (3)6-1

t6 = (2/27) (3)5

t6 = 2 x 32

t6 = 18

3. Find the G.P. 1/27, 1/9, 1/3, â€¦â€¦, 81; find the product of fourth term from the beginning and the fourth term from the end.

Solution:

Given G.P. 1/27, 1/9, 1/3, â€¦â€¦, 81

Here, a = 1/27, common ratio (r) = (1/9)/ (1/27) = 3 and l = 81

We know that,

l = tn = arn â€“ 1 = 81

(1/27)(3)n â€“ 1 = 81

3n â€“ 1 = 81 x 27 = 2187

3n â€“ 1 = 37

n â€“ 1 = 7

n = 8

Hence, there are 8 terms in the given G.P.

Now,

4th term from the beginning is t4 and the 4th term from the end is (8 â€“ 4 + 1) = 5th term (t5)

Thus,

the product of t4 and t5 = ar4 â€“ 1 x ar5 â€“ 1 = ar3 x ar4 = a2r7 = (1/27)2(3)7 = 3

4. If for a G.P.,Â pth,Â qthÂ andÂ rthÂ terms are a, b and c respectively; prove that:

(q – r) log a + (r – p) log b + (p – q) log c = 0

Solution:

Letâ€™s take the first term of the G.P. be A and its common ratio be R.

Then,

pth term = a â‡’ ARp â€“ 1 = a

qth term = b â‡’ ARq â€“ 1 = b

rth term = c â‡’ ARr â€“ 1 = c

Now,

On taking log on both the sides, we get

log( aq-r x br-p x cp-q ) = log 1

â‡’ (q – r)log a + (r – p)log b + (p – q)log c = 0

– Hence Proved

### Exercise 11(D) Page No: 156

1. Find the sum of G.P.:

(i) 1 + 3 + 9 + 27 + â€¦â€¦â€¦. to 12 terms

(ii) 0.3 + 0.03 + 0.003 + 0.0003 +â€¦..Â toÂ 8 terms.

(iii) 1 â€“ 1/2 + 1/4 â€“ 1/8 + â€¦â€¦.. to 9 terms

(iv) 1 – 1/3 + 1/32 â€“ 1/33 + â€¦â€¦â€¦ to n terms

(v)

(vi)

Solution:

(i) Given G.P: 1 + 3 + 9 + 27 + â€¦â€¦â€¦. to 12 terms

Here,

a = 1 and r = 3/1 = 3 (r > 1)

Number of terms, n = 12

Hence,

Sn = a(rn – 1)/ r â€“ 1

â‡’ S12 = (1)((3)12 – 1)/ 3 â€“ 1

= (312 â€“ 1)/ 2

= (531441 â€“ 1)/ 2

= 531440/2

= 265720

(ii) Given G.P: 0.3 + 0.03 + 0.003 + 0.0003 +â€¦..Â to 8 terms

Here,

a = 0.3 and r = 0.03/0.3 = 0.1 (r < 1)

Number of terms, n = 8

Hence,

Sn = a(1 – rn )/ 1 – r

â‡’ S8 = (0.3)(1 – 0.18 )/ (1 â€“ 0.1)

= 0.3(1 â€“ 0.18)/ 0.9

= (1 â€“ 0.18)/ 3

= 1/3(1 â€“ (1/10)8)

(iii) Given G.P: 1 â€“ 1/2 + 1/4 â€“ 1/8 + â€¦â€¦.. to 9 terms

Here,

a = 1 and r = (-1/2)/ 1 = -1/2 (| r | < 1)

Number of terms, n = 9

Hence,

Sn = a(1 – rn )/ 1 – r

â‡’ S9 = (1)(1 – (-1/2)9 )/ (1 â€“ (-1/2))

= (1 + (1/2)9)/ (3/2)

= 2/3 x ( 1 + 1/512 )

= 2/3 x (513/512)

= 171/ 256

(iv) Given G.P: 1 – 1/3 + 1/32 â€“ 1/33 + â€¦â€¦â€¦ to n terms

Here,

a = 1 and r = (-1/3)/ 1 = -1/3 (| r | < 1)

Number of terms is n

Hence,

Sn = a(1 – rn )/ 1 – r

(v) Given G.P:

Here,

a = (x + y)/ (x â€“ y) and r = 1/[(x + y)/ (x â€“ y)] = (x – y)/ (x + y) (| r | < 1)

Number of terms = n

Hence,

Sn = a(1 – rn )/ 1 – r

(vi) Given G.P:

Here,

a = âˆš3 and r = 1/âˆš3/ âˆš3 = 1/3 (| r | < 1)

Number of terms = n

Hence,

Sn = a(1 – rn )/ 1 – r

2. How many terms of the geometric progression 1 + 4 + 16 + 64 + â€¦â€¦..Â mustÂ be added to get sum equal to 5461?

Solution:

Given G.P: 1 + 4 + 16 + 64 + â€¦â€¦..

Here,

a = 1 and r = 4/1 = 4 (r > 1)

And,

Sn = 5461

We know that,

Sn = a(rn – 1)/ r â€“ 1

â‡’ Sn = (1)((4)n – 1)/ 4 â€“ 1

= (4n â€“ 1)/3

5461 = (4n â€“ 1)/3

16383 = 4n â€“ 1

4n = 16384

4n = 47

n = 7

Therefore, 7 terms of the G.P must be added to get a sum of 5461.

3. The first term of a G.P. is 27 and its 8th term is 1/81. Find the sum of its first 10 terms.

Solution:

Given,

First term (a) of a G.P = 27

And, 8th term = t8 = ar8 – 1 = 1/81

(27)r7 = 1/81

r7 = 1/(81 x 27)

r7 = (1/3)7

r = 1/3 (r <1)

Sn = a(1 – rn)/ 1 – r

Now,

Sum of first 10 terms = S10

4. A boy spendsÂ Rs.10 on first day,Â Rs.20 on second day,Â Rs.40 on third day and so on. Find how much, in all, will he spend in 12 days?

Solution:

Given,

Amount spent on 1st day = Rs 10

Amount spent on 2nd day = Rs 20

And amount spent on 3rd day = Rs 40

Itâ€™s seen that,

10, 20, 40, â€¦â€¦ forms a G.P with first term, a = 10 and common ratio, r = 20/10 = 2 (r > 1)

The number of days, n = 12

Hence, the sum of money spend in 12 days is the sum of 12 terms of the G.P.

Sn = a(rn – 1)/ r â€“ 1

S12 = (10)(212 – 1)/ 2 â€“ 1 = 10 (212 – 1) = 10 (4096 – 1) = 10 x 4095 = 40950

Therefore, the amount spent by him in 12 days is Rs 40950

5. The 4th term and the 7th term of a G.P. are 1/27 and 1/729 respectively. Find the sum of n terms of the G.P.

Solution:

Given,

t4 = 1/27 and t7 = 1/729

We know that,

tn = arn â€“ 1

So,

t4 = ar4 â€“ 1 = ar3 = 1/27 â€¦. (1)

t7 = ar7 â€“ 1 = ar6 = 1/729 â€¦. (2)

Dividing (2) by (1) we get,

ar6/ ar3 = (1/729)/ (1/27)

r3 = (1/3)3

r = 1/3 (r < 1)

In (1)

a x 1/27 = 1/27

a = 1

Hence,

Sn = a(1 – rn)/ 1 – r

Sn = (1 â€“ (1/3)n )/ 1 â€“ (1/3)

= (1 â€“ (1/3)n )/ (2/3)

= 3/2 (1 â€“ (1/3)n )

6. A geometric progression has common ratio = 3 and last term = 486. If the sum of its terms is 728; find its first term.

Solution:

Given,

For a G.P.,

r = 3, l = 486 and Sn = 728

1458 â€“ a = 728 x 2 = 1456

Thus, a = 2

7. Find the sum of G.P.: 3, 6, 12, â€¦., 1536.

Solution:

Given G.P: 3, 6, 12, â€¦., 1536

Here,

a = 3, l = 1536 and r = 6/3 = 2

So,

The sum of terms = (lr â€“ a)/ (r – 1)

= (1536 x 2 â€“ 3)/ (2 – 1)

= 3072 â€“ 3

= 3069

8. How many terms of the series 2 + 6 + 18 + â€¦.. must be taken to make the sum equal to 728?

Solution:

Given G.P: 2 + 6 + 18 + â€¦..

Here,

a = 2 and r = 6/2 = 3

Also given,

Sn = 728

Sn = a(rn – 1)/ r â€“ 1

728 = (2)(3n – 1)/ 3 â€“ 1 = 3n â€“ 1

729 = 3n

36 = 3n

n = 6

Therefore, 6 terms must be taken to make the sum equal to 728.

9. In a G.P., the ratio between the sum of first three terms and that of the first six terms is 125: 152.

Find its common ratio.

Solution:

Given,

Therefore, the common ratio is 3/5.

### Exercises of Selina Solutions Concise Maths Class 10 Chapter 11 Geometric Progression

Exercise 11(A) Solutions

Exercise 11(B) Solutions

Exercise 11(C) Solutions

Exercise 11(D) Solutions