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RD Chapter 20- Geometric Progressions Ex-20.1 Interview Questions Answers

Question 1 :

Show that each one of the following progressions is a G.P. Also, find thecommon ratio in each case:

(i) 4, -2, 1, -1/2, ….

(ii) -2/3, -6, -54, ….

(iii) a, 3a2/4, 9a3/16, ….

(iv) ½, 1/3, 2/9, 4/27, …

Answer 1 :

(i) 4, -2, 1, -1/2, ….

Let a = 4, b = -2, c =1

In GP,

b=ac

(-2)2 =4(1)

4 = 4

So, the Common ratio =r = -2/4 = -1/2

(ii) -2/3, -6, -54, ….

Let a = -2/3, b = -6,c = -54

In GP,

b=ac

(-6)2 =-2/3 × (-54)

36 = 36

So, the Common ratio =r = -6/(-2/3) = -6 × 3/-2 = 9

(iii) a, 3a2/4,9a3/16, ….

Let a = a, b = 3a2/4,c = 9a3/16

In GP,

b=ac

(3a2/4)2 =9a3/16 × a

9a4/4 = 9a4/16

So, the Common ratio =r = (3a2/4)/a = 3a2/4a = 3a/4

(iv) ½, 1/3, 2/9, 4/27, …

Let a = 1/2, b = 1/3,c = 2/9

In GP,

b=ac

(1/3)2 =1/2 × (2/9)

1/9 = 1/9

So, the Common ratio =r = (1/3)/(1/2) = (1/3) × 2 = 2/3

Question 2 :

Show that the sequence defined by an = 2/3n,n  N is a G.P.

Answer 2 :

Given:

an =2/3n

Let us consider n = 1,2, 3, 4, … since n is a natural number.

So,

a1 =2/3

a2 =2/32 = 2/9

a3 =2/33 = 2/27

a4 =2/34 = 2/81

In GP,

a3/a2 =(2/27) / (2/9)

= 2/27 × 9/2

= 1/3

a2/a1 =(2/9) / (2/3)

= 2/9 × 3/2

= 1/3

Common ratio ofconsecutive term is 1/3. Hence n  N is a G.P.

Question 3 :

Find:

(i) the ninth term of the G.P. 1, 4, 16, 64, ….

(ii) the 10th term of the G.P. -3/4, ½, -1/3, 2/9, ….

(iii) the 8th term of the G.P. 0.3, 0.06, 0.012, ….

(iv) the 12th term of the G.P. 1/a3x3 ,ax, a5x5, ….

(v) nth term of the G.P. √3, 1/√3, 1/3√3, …

(vi) the 10th term of the G.P. √2, 1/√2, 1/2√2, ….

Answer 3 :

(i) the ninth term of theG.P. 1, 4, 16, 64, ….

We know that,

t1 = a= 1, r = t2/t1 = 4/1 = 4

By using the formula,

Tn =arn-1

T9 = 1(4)9-1

= 1 (4)8

= 48

(ii) the 10th termof the G.P. -3/4, ½, -1/3, 2/9, ….

We know that,

t1 = a= -3/4, r = t2/t1 = (1/2) / (-3/4) = ½ × -4/3 = -2/3

By using the formula,

Tn =arn-1

T10 =-3/4 (-2/3)10-1

= -3/4 (-2/3)9

= ½ (2/3)8

(iii) the 8th termof the G.P., 0.3, 0.06, 0.012, ….

We know that,

t1 = a= 0.3, r = t2/t1 = 0.06/0.3 = 0.2

By using the formula,

Tn =arn-1

T8 =0.3 (0.2)8-1

= 0.3 (0.2)7

(iv) the 12th termof the G.P. 1/a3x3 , ax, a5x5,….

We know that,

t1 = a= 1/a3x3, r = t2/t1 = ax/(1/a3x3)= ax (a3x3) = a4x4

By using the formula,

Tn =arn-1

T12 =1/a3x3 (a4x4)12-1

= 1/a3x3 (a4x4)11

= (ax)41

(v) nth term of the G.P.√3, 1/√3, 1/3√3, …

We know that,

t1 = a= √3, r = t2/t1 = (1/√3)/√3 = 1/(√3×√3) = 1/3

By using the formula,

Tn =arn-1

Tn =√3 (1/3)n-1

(vi) the 10th termof the G.P. √2, 1/√2, 1/2√2, ….

We know that,

t1 = a= √2, r = t2/t1 = (1/√2)/√2 = 1/(√2×√2) = 1/2

By using the formula,

Tn =arn-1

T10 =√2 (1/2)10-1

= √2 (1/2)9

= 1/√2 (1/2)8

Question 4 :

Find the 4th term from the end of the G.P. 2/27, 2/9,2/3, …., 162.

Answer 4 :

The nth term from theend is given by:

an = l(1/r)n-1 where, l is the last term, r is the common ratio, n isthe nth term

Given: last term, l =162

r = t2/t1 =(2/9) / (2/27)

= 2/9 × 27/2

= 3

n = 4

So, an =l (1/r)n-1

a4 =162 (1/3)4-1

= 162 (1/3)3

= 162 × 1/27

= 6

 4th termfrom last is 6.

Question 5 :

Which term of the progression 0.004, 0.02, 0.1, …. is 12.5?

Answer 5 :

By using the formula,

Tn =arn-1

Given:

a = 0.004

r = t2/t1 =(0.02/0.004)

= 5

Tn =12.5

n = ?

So, Tn =arn-1

12.5 = (0.004) (5)n-1

12.5/0.004 = 5n-1

3000 = 5n-1

55 = 5n-1

5 = n-1

n = 5 + 1

= 6

 6th termof the progression 0.004, 0.02, 0.1, …. is 12.5.

Question 6 :

Which term of the G.P.:

(i) √2, 1/√2, 1/2√2, 1/4√2, … is 1/512√2 ?

(ii) 2, 2√2, 4, … is 128 ?

(iii) √3, 3, 3√3, … is 729 ?

(iv) 1/3, 1/9, 1/27… is 1/19683 ?

Answer 6 :

(i) √2, 1/√2, 1/2√2,1/4√2, … is 1/512√2 ?

By using the formula,

Tn =arn-1

a = √2

r = t2/t1 =(1/√2) / (√2)

= 1/2

Tn =1/512√2

n = ?

Tn =arn-1

1/512√2 = (√2) (1/2)n-1

1/512√2×√2 = (1/2)n-1

1/512×2 = (1/2)n-1

1/1024 = (1/2)n-1

(1/2)10 =(1/2)n-1

10 = n – 1

n = 10 + 1

= 11

 11th termof the G.P is 1/512√2

(ii) 2, 2√2, 4, … is 128 ?

By using the formula,

Tn =arn-1

a = 2

r = t2/t1 =(2√2/2)

= √2

Tn =128

n = ?

Tn =arn-1

128 = 2 (√2)n-1

128/2 = (√2)n-1

64 = (√2)n-1

26 =(√2)n-1

12 = n – 1

n = 12 + 1

= 13

 13th termof the G.P is 128

(iii) √3, 3, 3√3, … is 729 ?

By using the formula,

Tn =arn-1

a = √3

r = t2/t1 =(3/√3)

= √3

Tn =729

n = ?

Tn =arn-1

729 = √3 (√3)n-1

729 = (√3)n

36 =(√3)n

(√3)12 =(√3)n

n = 12

 12th termof the G.P is 729

(iv) 1/3, 1/9, 1/27… is1/19683 ?

By using the formula,

Tn =arn-1

a = 1/3

r = t2/t1 =(1/9) / (1/3)

= 1/9 × 3/1

= 1/3

Tn =1/19683

n = ?

Tn =arn-1

1/19683 = (1/3) (1/3)n-1

1/19683 = (1/3)n

(1/3)9 =(1/3)n

n = 9

 9th termof the G.P is 1/19683

Question 7 : Which term of the progression 18, -12, 8, … is512/729 ?

Answer 7 :

By using the formula,

Tn =arn-1

a = 18

r = t2/t1 =(-12/18)

= -2/3

Tn =512/729

n = ?

Tn =arn-1

512/729 = 18 (-2/3)n-1

29/(729 ×18) = (-2/3)n-1

29/36 ×1/2×32 = (-2/3)n-1

(2/3)8 =(-1)n-1 (2/3)n-1

8 = n – 1

n = 8 + 1

= 9

 9th termof the Progression is 512/729

Question 8 :

Find the 4th term from the end of the G.P. ½, 1/6, 1/18, 1/54, … , 1/4374

Answer 8 :

The nth term from theend is given by:

an = l(1/r)n-1 where, l is the last term, r is the common ratio, n isthe nth term

Given: last term, l =1/4374

r = t2/t1 =(1/6) / (1/2)

= 1/6 × 2/1

= 1/3

n = 4

So, an =l (1/r)n-1

a4 =1/4374 (1/(1/3))4-1

= 1/4374 (3/1)3

= 1/4374 × 33

= 1/4374 × 27

= 1/162

 4th termfrom last is 1/162.


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RD Chapter 20- Geometric Progressions Ex-20.1 Contributors

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