Which term of the G.P.

(i) 2, $2\sqrt{2}$, 4, …. is 128?

(ii) $1, \dfrac{1}{3}, \dfrac{1}{9}, …. \text{ is } \dfrac{1}{243}$ ?

(i) Given a = 2, r = $\sqrt{2}$.

Let nth term be 128.

By formula, a_n = ar^{n - 1}.

$\Rightarrow 128 = 2(\sqrt{2})^{n - 1} \\[1em] \Rightarrow \dfrac{128}{2} = (2)^{(n - 1)/2} \\[1em] \Rightarrow 64 = (2)^{(n - 1)/2} \\[1em] \Rightarrow 2^6 = (2)^{(n - 1)/2} \\[1em] \Rightarrow \dfrac{n - 1}{2} = 6 \\[1em] \Rightarrow n - 1 = 12 \\[1em] \Rightarrow n = 13.$

Hence, 128 is 13th term of the G.P.

(ii) Given a = 1, r = $\dfrac{1}{3}$.

Let nth term be $\dfrac{1}{243}$.

By formula, a_n = ar^{n - 1}.

$\Rightarrow \dfrac{1}{243} = 1\Big(\dfrac{1}{3}\Big)^{n - 1} \\[1em] \Rightarrow \dfrac{1}{3^5} = \dfrac{1}{3^{(n - 1)}} \\[1em] \Rightarrow n - 1 = 5 \\[1em] \Rightarrow n = 6.$

Hence, $\dfrac{1}{243}$ is 6th term of the G.P.