The 4th term from the end of the G.P. $\dfrac{2}{27}, \dfrac{2}{9}, \dfrac{2}{3}, ……, 162$ is :

1. 18

2. 2

3. 6

4. $\dfrac{2}{3}$

G.P. : $\dfrac{2}{27}, \dfrac{2}{9}, \dfrac{2}{3}, ……, 162$.

a = $\dfrac{2}{27}$

r = $\dfrac{\dfrac{2}{9}}{\dfrac{2}{27}} = \dfrac{2}{9} \times \dfrac{27}{2}$ = 3

l = 162.

We know that,

${\text{nth term from the end}} = \dfrac{l}{r^{n - 1}}$

Substitute values we get:

$\Rightarrow {\text{4th term from the end}} = \dfrac{l}{r^{4 - 1}} \\[1em] = \dfrac{162}{3^{3}} \\[1em] = \dfrac{162}{27} \\[1em] = 6.$

Hence, the 6th term from the end is 6.

Hence, option 3 is the correct option.