Which term of the G.P. $2, 2\sqrt{2}, 4, …….. \text{ is } 128\sqrt{2} ?$

Given,

G.P. : $2, 2\sqrt{2}, 4, ……..$

In above G.P.,

First term (a) = 2

Common ratio (r) = $\dfrac{2\sqrt{2}}{2} = \sqrt{2}$

Let n^th term of G.P. be $128\sqrt{2}$.

$\Rightarrow ar^{n-1} = 128\sqrt{2} \\[1em] \Rightarrow 2 \times (\sqrt{2})^{n - 1} = 128\sqrt{2} \\[1em] \Rightarrow 2 \times (2^{\dfrac{1}{2}})^{n - 1} = (2)^7.\sqrt{2} \\[1em] \Rightarrow 2 \times (2)^{\dfrac{n - 1}{2}} = 2^7.2^{\dfrac{1}{2}} \\[1em] \Rightarrow (2)^{1 + \dfrac{n - 1}{2}} = (2)^{7 + \dfrac{1}{2}} \\[1em] \Rightarrow (2)^{\dfrac{n - 1 + 2}{2}} = (2)^{\dfrac{14 + 1}{2}} \\[1em] \Rightarrow (2)^{\dfrac{n + 1}{2}} = (2)^{\dfrac{15}{2}} \\[1em] \Rightarrow \dfrac{n + 1}{2} = \dfrac{15}{2} \\[1em] \Rightarrow n + 1 = 15 \\[1em] \Rightarrow n = 15 - 1 = 14.$

Hence, 14^th term of G.P. = $128\sqrt{2}$.