# 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, .... is 1/19683?

**Solution:**

(a) The given sequence is 2, 2√2, 4,...

Here, a = 2 and r = 2√2/2 = √2

Let the n^{th} term of the sequence be 128

⇒ a = ar^{n }^{- }^{1} = 128

⇒ (2)(√2)^{n }^{- }^{1} = 128

⇒ (2)(2)^{(n }^{- }^{1)/2} = (2)^{7}

⇒ (2)(2)^{(n }^{- }^{1)/2 + 1} = (2)^{7}

Hence,

⇒ (n - 1)/2 + 1 = 7

⇒ (n - 1)/2 = 6

⇒ n - 1 = 12

⇒ n = 13

Thus, the 13^{th} term of the sequence be 128.

(b) The given sequence is √3,3,3√3,...

Here, a = √3 and r = 3/√3 = √3

Let the nth term of the sequence be 729.

⇒ a = ar^{n }^{- }^{1} = 729

⇒ (√3)(√3)^{n }^{- }^{1} = 729

⇒ a = (3)^{1/2}(√3)^{(n }^{- }^{1)/2} = (3)^{6}

⇒ (3)^{1/2 + (n }^{- }^{1)/2} = (3)^{6}

Hence,

⇒ 1/2 + (n - 1)/2 = 6

⇒ (1 + n - 1)/2 = 6

⇒ n = 12

Thus, the 12^{th} term of the sequence is 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 sequence be 1/19683

⇒ a = ar^{n }^{- }^{1} = 1/19683

⇒ (1/3)(1/3)^{n }^{- }^{1} = 1/19683

⇒ (1/3)^{n} = (1/3)^{9}

⇒ n = 9

Thus, the 9^{th} term of the sequence be 1/19683

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.3 Question 5

## 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, .... is 1/19683 ?

**Summary:**

128 was the 13th term of series a,729 was the 12th term of series b, 1/19683 was the 9th term of series c

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