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Class 10 SELINA Solutions Maths Chapter 11 - Geometric Progression

Geometric Progression Exercise TEST YOURSELF

Solution 1(a)

Correct option: (d) either 1 or –1

When k = 1,

Here, common ratio = –1

When k = –1,

Here, common ratio = 1

Since k cannot take both the values, k can be either 1 or –1.

Solution 1(b)

Correct option: (a) 16

x + 9, 10 and 4 are in G.P.

Solution 1(c)

Correct option: (a)

6th term = ar5 = 48

Common ratio, r = 2

Solution 1(d)

Correct option: (b) 3

Let the three terms in G.P. be a,  and ar, respectively.

a3 = 27

a = 3

Solution 1(e)

Correct option: (b) 3

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

nth term = arn-1

4th term = 27

ar3 = 27

And, 6th term = 243

ar5 = 1458

Solution 1(f)

Correct option: (b) 185

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

3rd term = ar2 = 18

Product of first five terms

= ar0 × ar1 × ar2 × ar3 × ar4

= a5 × r10

= (ar2)5

= 185

Solution 2

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

nth term = arn-1

5th term = 32

ar4 = 32

And, 8th term = 256

ar7 = 256

And, ar4 = 32

a(2)4 = 32

a × 16 = 32

a = 2

Hence, the first term and the common ratio are 2 and 2, respectively.

Solution 3

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

nth term = arn-1

3rd term – 1st term = 9

ar2 – a = 9

a(r2 – 1) = 9

And, 2nd term – 4th term = 18

ar – ar3 = 18

ar(1 – r2) = 18

Substituting in (i),

Then, the G.P. is 3, 3(–2), 3(–2)2,  3(–2)3, …..

That is, 3, –6, 12, –24, …..

Solution 4

x, 2x + 2, 3x + 3, G are four consecutive terms of a G.P.

Then,

And, 2x + 2 = 2(–4) + 2 = –6

Now, 

Solution 5

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

3rd term = ar2 = 2

Product of first five terms

= ar0 × ar1 × ar2 × ar3 × ar4

= a5 × r10

= (ar2)5

= 25

= 32

Solution 6

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

nth term = arn-1

10th term = ar9 = x

16th term = ar15 = y

22nd term = ar21 = z

Now,

And,

From (i) and (ii),

Therefore, x, y and z are in G.P.

Solution 7

Given G.P. is

Here, first term, a = 2 and the common ratio, r =

Let the nth term =

n – 1 = 13

n = 14

Solution 8

Let the first term = a and the common ratio, r = 2

nth term = arn-1

10th term = 768

ar9 = 768

a(2)9 = 768

a × 512 = 768

Therefore, 8th term = ar7

Solution 9

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

nth term = arn-1

4th term = 48

ar3 = 48

And, 7th term = 384

ar6 = 384

a(2)3 = 48 a = 6

Therefore, 6th term = ar5

 = 6(2)5

= 192

Solution 10

 are three consecutive terms of a G.P.

Then,

Hence, the value of x is either 1 or –1.

Geometric Progression Exercise Ex. 11(A)

Solution 8

Solution 18

Solution 9

Solution 19

Solution 10

Solution 20

Solution 11

Solution 2

Solution 12

Solution 3

Solution 13

Solution 4

Solution 14

Solution 15

Solution 21(i)

Solution 5

Solution 16

Solution 21(ii)

Solution 17

Solution 22

Solution 6

Solution 7

Given series: 2, - 6, 18, - 54 ……

Solution 1(a)

Correct option: (c) 32

First term, a = 8

Common ratio, r = –2

nth term = arn-1

Then, 3rd term = 8(–2)2 = 32

Solution 1(b)

Correct option: (a) 2

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

nth term = arn-1

4th term = 16

ar3 = 16

And, 7th term = 128

ar6= 128

Solution 1(c)

Correct option: (c)

Solution 1(d)

Correct option: (c) 35

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

3rd term = ar2 = 3

Product of first five terms

= ar0 × ar1 × ar2 × ar3 × ar4

= a5 × r10

= (ar2)5

= 35

Solution 1(e)

Correct option: (d)

Let the first term = a and the common ratio, r = 2  8th term = 192

ar7 = 192

a(2)7 = 192

a(128) = 192

Geometric Progression Exercise Ex. 11(B)

Solution 2(i)

Solution 2(ii)

Solution 2(iii)

Solution 2(iv)

Solution 2(v)

Solution 2(vi)

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13(i)

 

Solution 13(ii)

Solution 13(iii)

Solution 14

Solution 15

Solution 16

First term (a) =   

  

Solution 17

First term (a) = 125 

  

Solution 18

  

Thus, the given sequence is a G.P. with   

  

Solution 19

  

Solution 20

Let the five terms of the given G.P. be

  

Given, sum of first two terms = -4

  

And, 5th term = 4(3rd term)

ar2 = 4(a)

r2 = 4

r = ±2

When r = +2,

  

When r = -2,

  

Solution 1(a)

Correct option: (b) 80

a = 2, n = 4 and

Solution 1(b)

Correct option: (b) 3

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

nth term = arn-1

4th term = 54

ar3 = 54

And, 7th term = 1458

ar6= 1458

Solution 1(c)

Correct option: (a) 2

Geometric mean between 8 and 32

Solution 1(d)

Correct option: (a)

Solution 1(e)

Correct option: (c) 10(220 – 1)

a = 10, n = 20 and

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