NCERT Solutions
Class 11 Maths
Sequences and Series

Ex.9.3 Q.6
Which term of the following sequences:
(a) 2, 2√2, 4, ... is 128?
(b) √3, 3, 3√3, ... is 729?
(c) ,
,
, ... is
?
(a) The given sequence is 2, 2√2, 4, ...
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 = 128
=> 2(n-1) ÷ (2 + 1) = 128
=> 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) Given sequence is: √3, 3, 3√3, ...
Here, a = √3 and r = (3 ÷ √3) = √3
Let nth term of the given sequence be 729
an = arn-1 => √3 × (√3) n-1 = 729
=> 31÷2 × 3(n-1) ÷ 2 = 729
=> (n - 1) ÷ 2 = 6 –
=> (n – 1) ÷ 2 =
=> n – 1 = 11
=> n = 12
Thus, the 12th term of the given sequence is 729.
(c) Given sequence is: ,
,
, ...
Here, a = and r = (
) ÷ (
) =
Let nth term of the given sequence be
=> an = arn-1
=> ( ) × (
)n-1 =
=> ( )n = (
)9
=> n = 9
Thus, the 9th term of the given sequence is