Question

Which term of the following sequences:

(a) 2 ,$2\sqrt{2}$ , 4 ,.... is 128 ?

(b) $\sqrt{3}$, 3, $3\sqrt{3}$,... is 729 ?

(c) $\frac{1}{3},\frac{1}{9},\frac{1}{27}$ ,.... is $\frac{1}{19683}$ ?

Answer 1

(a) The given sequence is 2 , $2\sqrt{2}$ , 4,....

Here, a = 2 and r = $\frac{2\sqrt 2}{2}$= $\sqrt{2}$

Let the n th term of the given sequence be 128

an = ar n - 1

⇒ (2)$(\sqrt{2})$ n - 1 = 128

⇒ (2)(2)n - $\frac{1}{2}$ = (2$)^7$

⇒ (2)n - $\frac{1}{2}$ + 1 = (2$)^7$

⇒ ∴ n - $\frac{1}{2}$ + 1 = 7

⇒ n - $\frac{1}{2}$ = 6

⇒ n - 1 = 12

⇒ n = 13

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

(b) The given sequence is $\sqrt{3}$ , 3 , $3\sqrt{3}$ ,...

Here, a = $\sqrt{3}$ and r = $\frac{3}{\sqrt{3}}$= $\sqrt{3}$

Let the n th term of the given sequence be 729.

an = arn - 1

∴ arn - 1 = 729

⇒ ( $\sqrt{3}$)( $\sqrt{3}$n - 1) = 729

⇒ (3) $\frac{1}{2}$** (3)n - $\frac{1}{2}$ = (3)**$^6$

⇒ (3) $\frac{1}{2}$** + n - $\frac{1}{2}$ = (3**$)^6$

∴ $\frac{1}{2}$ + n - $\frac{1}{2}$= 6

⇒ n = 12

Thus, the 12th term of the given sequence is 729.

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(c) The given sequence is $\frac{1}{3},\frac{1}{9},\frac{1}{12}$,…..

**Here, a = $\frac{1}{3}$ and r = $\frac{1}{9}$ ÷ **$\frac{1}{3}=\frac{1}{3}$

**Let the n th term of the given sequence be **$\frac{1}{19683}$

**an = arn **$^-1$

**∴ arn $^-1$ = **$\frac{1}{19683}$

**⇒ **$(\frac{1}{3})(\frac{1}{3})$**n $^6$ = **$\frac{1}{19683}$

⇒ $(\frac{1}{3})$**n = **$(\frac{1}{3})^9$

⇒ n = 9

**Thus, the 9th term of the given sequence is **$\frac{1}{19683}$