The 6th term from the end of the G.P. 8, 4, 2, ……, $\dfrac{1}{1024}$ is :

1. $\dfrac{1}{32}$

2. $\dfrac{1}{16}$

3. $\dfrac{1}{64}$

4. $\dfrac{1}{128}$

G.P. : 8, 4, 2, ……, $\dfrac{1}{1024}$.

a = 8

r = $\dfrac{4}{8} = \dfrac{1}{2}$

l = $\dfrac{1}{1024}$.

We know that,

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

Substitute values we get:

$\Rightarrow {\text{6th term from the end}} = \dfrac{l}{r^{6 - 1}} \\[1em] = \dfrac{\dfrac{1}{1024}}{\Big(\dfrac{1}{2}\Big)^{5}} \\[1em] = \dfrac{1}{1024} \times \dfrac{32}{1}\\[1em] = \dfrac{1}{32}.$

Hence, the 6th term from the end is $\dfrac{1}{32}$.

Hence, option 1 is the correct option.