Sequences and Series – Exercise 9.3[1-10] – Class XI

  1. Find the 20th and nth terms of the G.P. 5/2, 5/4, 5/8, …

Solution:

The given G.P. is  5/2, 5/4, 5/8, …

Here a = first term = 5/2

r = common ratio = (4/5)/(5/2) = 1/2

a20 = ar20-1 = 5/2(1/2)n-1 = 5/(2)(2)n-1 = 5/2n

an = arn-1 = 5/2(1/2)n-1 = 5/(2)(2)n-1 = 5/2n


  1. Find the 12th term of G.P. whose 8th term is 192 and the common ratio is 2.

Solution:

Common ratio, r = 2

Let a be the first term of the G.P.

∴ a8 = ar8–1 = ar7

⇒ ar7 = 192 a(2)7 = 192 a(2)7 = (2)6(3)

⇒a = [(2)6x3]/(2)73/2

a12 = ar12-1 = 3/2(2)11 = (3)(2) = 3072


  1. The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.

Solution:

Let a be the first term and r be the common ratio of the G.P. According to the given condition,

a5 = ar5–1 = ar4 = p —————–(1)

a8 = a r8–1 = a r7 = q —————-(2)

a11 = a r11–1 = a r10 = s ————– (3)

Dividing equation (2) by (1), we obtain

[ar7]/[ar4] = q/p

r3 = q/p —————(4)

Dividing equation (3) by (2), we obtain

[ar10]/[ar7] = s/q

r3 = s/q —————-(5)

Equating the values of r3 obtained in (4) and (5), we obtain

q/p = s/q

q2 = ps

Thus the given result is proved.


  1. The 4th term of a G.P. is square of its second term, and the first term is –3. Determine its 7th term.

Solution:

Let a be the first term and r be the common ratio of the G.P.

∴ a = –3

It is known that, an = arn–1

∴ a4 = ar3 = (–3) r3

a2 = a r1 = (–3) r

According to the given condition,

(–3) r3 = [(–3) r]2

⇒ –3r3 = 9 r2

⇒ r = –3 a7 = a r7–1 = a

r6 = (–3) (–3)6 = – (3)7 = –2187

Thus, the seventh term of the G.P. is –2187.


  1. Which term of the following sequences:

(a)2, 2√2, 4, …is 128?

(b) √3, 3, 3√3,…is 729?

(c)1/3, 1/9, 1/27,…1/19683 ?

Solution:

(a) The given sequence 2, 2√2, 4, …is 128?

Here a = 2 and r = (2√2)/2 = √2

Let nth term of the given sequence be 128.

an = arn-1

(2)( √2)n-1 = 128

(2)(2)(n-1)/2 = 27

(2)[(n-1)/2]+1 = 27

n-1/2 + 1 = 7

n-1/2 = 6

n – 1 = 12

n = 13

Thus the 13th term of the given sequence is 128.

 

(b) the given sequence is √3, 3, 3 √3

a = √3 and r = 3/√3 = √3

Let the nth term of the given sequence be 729

an = arn-1

arn-1 = 729

(√3)( √3)n-1 = 729

(3)1/2( 3)(n-1)/2 =36

1+n-1/2 = 6

n = 12

Thus the 12th term of the given sequence 729.

 

(c) The given sequence is 1/3, 1/9, 1/27,…

Here, a = 1/3 and r = 1/9 + 1/3 = 1/3

Let the nth term of the given sequence be 1/19683

an = arn-1

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

(1/3)n = (1/3)9

n = 9

Thus, the 9th term of the given sequence is 1/19683


  1. For what values of x, the numbers 2/7, x, –7/2 are in G.P.?

Solution:

The given numbers are –2/7, x, –7/2

Common ratio =

x/(-2/7)=-7x/2

Also, common ratio = (-7/2)/x = –7/2x

-7/2x = –7/2x

x2 = -2×7/-2×7 = 1

x = √1

x = ±1

Thus for x = ±1 the given numbers will be in G.P.


  1. Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015

Solution:

The given G.P. is 0.15, 0.015, 0.0015

Here a = 0.15 and r = 0.015/0.15 = 0.1

1


  1. Find the sum to n terms in the geometric progression √7 , √21, 3√7, …

Solution:

The given G.P. is √7 , √21, 3√7, …

Here, a = √7 and r =√21/7 = √3

1


  1. Find the sum to n terms in the geometric progression 1, -a, a2, -a3…(if a≠-1)

Solution:

The given G.P. is 1, -a, a2, -a3

Here first term = a1 = 1

Common ratio = r = – a

1


  1. Find the sum to n terms in the geometric progression x3, x5, x7…(if x ≠±1)

Solution:

The given G.P. is x3, x5, x7

Here a = x3 and r = x2

1


Sequences and Series – Exercise 9.3[11-20] – Class XI

Sequences and Series – Exercise 9.3[21-32] – Class XI