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 to Selina Solutions for Class 10 Mathematics prepared by our experienced faculty at BYJU’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.

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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, ………

(ii) 1/8, 1/24, 1/72, 1/216, ………

(iii) 9, 12, 16, 24, ………

Solution:

(i) Given sequence: 8, 24, 72, 216, ………

Since,

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

⇒ 24/8 = 72/24 = 216/72 = ……….. = 3

Therefore 8, 24, 72, 216, ……… is a G.P. with a common ratio 3.

(ii) Given sequence: 1/8, 1/24, 1/72, 1/216, ………

Since,

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

⇒ (1/24)/ (1/8) = (1/72)/ (1/24) = (1/216)/ (1/72) = ……….. = 1/3

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

(iii) Given sequence: 9, 12, 16, 24, ………

Since,

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

12/9 = 16/12 ≠ 24/16

Therefore, 9, 12, 16, 24 …… is not a G.P.

2. Find the 9^th term of the series: 1, 4, 16, 64, …..

Solution:

It’s seen that, the first term is (a) = 1

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

We know that, the general term is

t_n = ar^{n – 1}

Thus,

t₉ = (1)(4)^{9 – 1} = 4⁸ = 65536

3. Find the seventh term of the G.P: 1, √3, 3, 3 √3, …..

Solution:

It’s seen that, the first term is (a) = 1

And, common ratio(r) = √3/1 = √3

We know that, the general term is

t_n = ar^{n – 1}

Thus,

t₇ = (1)(√3)^{7 – 1} = (√3)⁶ = 27

4. Find the 8^th term of the sequence:

Solution:

The given sequence can be rewritten as,

3/4, 3/2, 3, …..

It’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

t_n = ar^{n – 1}

Thus,

t₈ = (3/4)(2)^{8 – 1} = (3/4)(2)⁷ = 3 x 2⁵ = 3 x 32 = 96

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

Solution:

The given sequence can be rewritten as,

12, 4, 4/3, …..

It’s seen that, the first term is (a) = 12

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

We know that, the general term is

t_n = ar^{n – 1}

Thus,

t₁₀ = (12)(1/3)^{10 – 1} = (12)(1/3)⁹ = 12 x 1/(19683) = 4/ 6561

6. Find the nth term of the series:

1, 2, 4, 8, ……..

Solution:

It’s seen that, the first term is (a) = 1

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

We know that, the general term is

t_n = ar^{n – 1}

Thus,

t_n = (1)(2)^{n – 1} = 2^{n – 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/√3)/ (-10) = 1/(-2√3)

We know that, the general term is

t_n = ar^{n – 1}

So,

t_n = (-10)( 1/(-2√3))^{n – 1} = -5/72

Now, equating the exponents we have

n – 1 = 4

n = 5

Thus, the 5^th 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

t_n = ar^{n – 1}

So,

t₅ = ar^{5 – 1} = ar⁴ = 81 …. (1)

And,

t₂ = ar^{2 – 1} = ar¹ = 24 …. (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, ar³……

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

= 16, 24, 36, 54, ……

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

t_n = ar^{n – 1}

So,

t₄ = ar^{4 – 1} = ar³ = 1/18 …. (1)

And,

t₇ = ar^{7 – 1} = ar⁶ = -1/486 …. (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, ar³……

= -3/2, -3/2(-1/3), -3/2(-1/3)², -3/2(-1/3)³, ……

= -3/2, 1/2, -1/6, 1/18, …..

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

t_n = ar^{n – 1}

So,

t₁ = ar^{1 – 1} = a = 2 …. (1)

And,

t₃ = ar^{3 – 1} = ar² = 8 …. (2)

Dividing (2) by (1), we have

ar²/ a = 8/ 2

r² = 4

r = ± 2

Hence, the 2^nd 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 3^rd and 8^th terms of a G.P. is 243. If its 4^th term is 3, find its 7^th term

Solution:

Given,

Product of 3^rd and 8^th 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,

t_n = ar^{n – 1}

So,

t₃ x t₈ = ar^{3 – 1} x ar^{8 – 1} = ar² x ar⁷ = a²r⁹ = 243

Also given,

t₄ = ar^{4 – 1} = ar³ = 3

Now,

a²r⁹ = (ar³) ar⁶ = 243

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

(3) ar⁶ = 243

ar⁶ = 81

ar^{7 – 1} = 81 = t₇

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

Exercise 11(C) Page No: 156

1. Find the seventh term from the end of the series: √2, 2, 2√2, …… , 32

Solution:

Given series: √2, 2, 2√2, …… , 32

Here,

a = √2

r = 2/ √2 = √2

And, the last term (l) = 32

l = t_n = ar^{n – 1} = 32

(√2)( √2)^{n – 1} = 32

(√2)ⁿ = 32

(√2)ⁿ = (2)⁵ = (√2)¹⁰

Equating the exponents, we have

n = 10

So, the 7^th term from the end is (10 – 7 + 1)^th term.

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

Hence,

t₄ = (√2)(√2)^{4 – 1} = (√2)(√2)³ = (√2) x 2√2 = 4

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

2/27, 2/9, 2/3, ……., 162

Solution:

Given series: 2/27, 2/9, 2/3, ……., 162

Here,

a = 2/27

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

r = 3

And, the last term (l) = 162

l = t_n = ar^{n – 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 6^th term of the G.P. = t₆

Hence,

t₆ = ar^6-1

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, ……, 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, ……, 81

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

We know that,

l = t_n = ar^{n – 1} = 81

(1/27)(3)^{n – 1} = 81

3^{n – 1} = 81 x 27 = 2187

3^{n – 1} = 3⁷

n – 1 = 7

n = 8

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

Now,

4^th term from the beginning is t₄ and the 4^th term from the end is (8 – 4 + 1) = 5^th term (t₅)

Thus,

the product of t₄ and t₅ = ar^{4 – 1} x ar^{5 – 1} = ar³ x ar⁴ = a²r⁷ = (1/27)²(3)⁷ = 3

4. If for a G.P., p^th, q^th and r^th terms are a, b and c respectively; prove that:

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

Solution:

Let’s take the first term of the G.P. be A and its common ratio be R.

Then,

p^th term = a ⇒ AR^{p – 1} = a

q^th term = b ⇒ AR^{q – 1} = b

r^th term = c ⇒ AR^{r – 1} = c

Now,

On taking log on both the sides, we get

log( a^q-r x b^r-p x c^p-q ) = log 1

⇒ (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 + ………. to 12 terms

(ii) 0.3 + 0.03 + 0.003 + 0.0003 +….. to 8 terms.

(iii) 1 – 1/2 + 1/4 – 1/8 + …….. to 9 terms

(iv) 1 – 1/3 + 1/3² – 1/3³ + ……… to n terms

(v)

(vi)

Solution:

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

Here,

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

Number of terms, n = 12

Hence,

S_n = a(rⁿ – 1)/ r – 1

⇒ S₁₂ = (1)((3)¹² – 1)/ 3 – 1

= (3¹² – 1)/ 2

= (531441 – 1)/ 2

= 531440/2

= 265720

(ii) Given G.P: 0.3 + 0.03 + 0.003 + 0.0003 +….. to 8 terms

Here,

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

Number of terms, n = 8

Hence,

S_n = a(1 – rⁿ )/ 1 – r

⇒ S₈ = (0.3)(1 – 0.1⁸ )/ (1 – 0.1)

= 0.3(1 – 0.1⁸)/ 0.9

= (1 – 0.1⁸)/ 3

= 1/3(1 – (1/10)⁸)

(iii) Given G.P: 1 – 1/2 + 1/4 – 1/8 + …….. to 9 terms

Here,

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

Number of terms, n = 9

Hence,

S_n = a(1 – rⁿ )/ 1 – r

⇒ S₉ = (1)(1 – (-1/2)⁹ )/ (1 – (-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 – 1/3³ + ……… to n terms

Here,

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

Number of terms is n

Hence,

S_n = a(1 – rⁿ )/ 1 – r

(v) Given G.P:

Here,

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

Number of terms = n

Hence,

S_n = a(1 – rⁿ )/ 1 – r

(vi) Given G.P:

Here,

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

Number of terms = n

Hence,

S_n = a(1 – rⁿ )/ 1 – r

2. How many terms of the geometric progression 1 + 4 + 16 + 64 + …….. must be added to get sum equal to 5461?

Solution:

Given G.P: 1 + 4 + 16 + 64 + ……..

Here,

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

And,

S_n = 5461

We know that,

S_n = a(rⁿ – 1)/ r – 1

⇒ S_n = (1)((4)ⁿ – 1)/ 4 – 1

= (4ⁿ – 1)/3

5461 = (4ⁿ – 1)/3

16383 = 4ⁿ – 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 8^th term is 1/81. Find the sum of its first 10 terms.

Solution:

Given,

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

And, 8^th term = t₈ = ar^{8 – 1} = 1/81

(27)r⁷ = 1/81

r⁷ = 1/(81 x 27)

r⁷ = (1/3)⁷

r = 1/3 (r <1)

S_n = a(1 – rⁿ)/ 1 – r

Now,

Sum of first 10 terms = S₁₀

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

Solution:

Given,

Amount spent on 1^st day = Rs 10

Amount spent on 2^nd day = Rs 20

And amount spent on 3^rd day = Rs 40

It’s seen that,

10, 20, 40, …… 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.

S_n = a(rⁿ – 1)/ r – 1

S₁₂ = (10)(2¹² – 1)/ 2 – 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 4^th term and the 7^th 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,

t_n = ar^{n – 1}

So,

t₄ = ar^{4 – 1} = ar³ = 1/27 …. (1)

t₇ = ar^{7 – 1} = ar⁶ = 1/729 …. (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,

S_n = a(1 – rⁿ)/ 1 – r

S_n = (1 – (1/3)ⁿ )/ 1 – (1/3)

= (1 – (1/3)ⁿ )/ (2/3)

= 3/2 (1 – (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 S_n = 728

1458 – a = 728 x 2 = 1456

Thus, a = 2

7. Find the sum of G.P.: 3, 6, 12, …., 1536.

Solution:

Given G.P: 3, 6, 12, …., 1536

Here,

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

So,

The sum of terms = (lr – a)/ (r – 1)

= (1536 x 2 – 3)/ (2 – 1)

= 3072 – 3

= 3069

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

Solution:

Given G.P: 2 + 6 + 18 + …..

Here,

a = 2 and r = 6/2 = 3

Also given,

S_n = 728

S_n = a(rⁿ – 1)/ r – 1

728 = (2)(3ⁿ – 1)/ 3 – 1 = 3ⁿ – 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