Selina Solutions Concise Maths Class 10 Chapter 11 Geometric Progression

 

Selina Solutions Concise Maths Class 10 Chapter 11 Geometric Progression

Selina Class 10 ICSE solution Chapter 11 Geometric Progression will give strength to you and the board will also clear the exam, Mathematics Subject demands time and practice, then keep practicing and you will definitely get success.

Selina Publishers Solutions for Class 10 is designed by experienced teachers keeping in mind the syllabus of the examinations and the requirement of the students. The ICSE solution for class 10 presents a firm platform for students to prepare more and more questions and refer to the correct answers. That can provide complete solutions for students.

Exercise 11(A) Solutions

Exercise 11(C) Solutions

Exercise 11(B) Solutions

Exercise 11(D) Solutions

Exercise 11(A) 

Question 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.

Question 2. Find the 9th 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

tn = arn – 1

Thus,

t9 = (1)(4)9 – 1 = 48 = 65536

Question 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

tn = arn – 1

Thus,

t7 = (1)(√3)7 – 1 = (√3)6 = 27

Question 4. Find the 8th 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

tn = arn – 1

Thus,

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

Question 5. Find the 10th 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

tn = arn – 1

Thus,

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

Question 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

tn = arn – 1

Thus,

tn = (1)(2)n – 1 = 2n – 1

Exercise 11(B) Page No: 154

Question 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

tn = arn – 1

So,

tn = (-10)( 1/(-2√3))n – 1 = -5/72

Now, equating the exponents we have

n – 1 = 4

n = 5

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

Question 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 – 1

So,

t5 = ar5 – 1 = ar4 = 81 …. (1)

And,

t2 = ar2 – 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, ……

Question 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 – 1

So,

t4 = ar4 – 1 = ar3 = 1/18 …. (1)

And,

t7 = ar7 – 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, …..

Question 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 – 1

So,

t1 = ar1 – 1 = a = 2 …. (1)

And,

t3 = ar3 – 1 = ar2 = 8 …. (2)

Dividing (2) by (1), we have

ar2/ a = 8/ 2

r2 = 4

r = ± 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

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

Solution:

Given,

Product of 3rd and 8th 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 – 1

So,

t3 x t8 = ar3 – 1 x ar8 – 1 = ar2 x ar7 = a2r9 = 243

Also given,

t4 = ar4 – 1 = ar3 = 3

Now,

a2r9 = (ar3) ar6 = 243

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

(3) ar6 = 243

ar6 = 81

ar7 – 1 = 81 = t7

Thus, the 7th term of the G.P is 81.

Exercise 11(C) Page No: 156

Question 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 = tn = arn – 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 7th term from the end is (10 – 7 + 1)th term.

i.e. 4th term of the G.P

Hence,

t4 = (√2)(√2)4 – 1 = (√2)(√2)3 = (√2) x 2√2 = 4

Question 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 = tn = arn – 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

Question 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 = tn = arn – 1 = 81

(1/27)(3)n – 1 = 81

3n – 1 = 81 x 27 = 2187

3n – 1 = 37

n – 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 – 4 + 1) = 5th term (t5)

Thus,

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

Question 4. If for a G.P., pth, qth and rth 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,

pth term = a ⇒ ARp – 1 = a

qth term = b ⇒ ARq – 1 = b

rth term = c ⇒ ARr – 1 = c

Now,

On taking log on both the sides, we get

log( aq-r x br-p x cp-q ) = log 1

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

– Hence Proved

Exercise 11(D) Page No: 156

Question 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,

Sn = 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,

Sn = 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,

Sn = 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,

Sn = 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,

Sn = 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,

Sn = a(1 – rn )/ 1 – r

Question 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,

Sn = 5461

We know that,

Sn = a(rn – 1)/ r – 1

⇒ Sn = (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.

Question 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

Question 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 1st day = Rs 10

Amount spent on 2nd day = Rs 20

And amount spent on 3rd 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.

Sn = 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

Question 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 – 1

So,

t4 = ar4 – 1 = ar3 = 1/27 …. (1)

t7 = ar7 – 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,

Sn = a(1 – rn)/ 1 – r

Sn = (1 – (1/3)n )/ 1 – (1/3)

= (1 – (1/3)n )/ (2/3)

= 3/2 (1 – (1/3)n )

Question 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 – a = 728 x 2 = 1456

Thus, a = 2

Question 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

Question 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,

Sn = 728

Sn = 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.

Question 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(C) Solutions

Exercise 11(B) Solutions

Exercise 11(D) Solutions

More Selina Concise Solutions Class 10

Selina Concise Class 10 Physics Solutions 

Selina Concise Class 10 Maths Solutions 

Selina Concise Class 10 Chemistry Solutions 

Selina Concise Class 10 Biology Solutions