Find the

The given GP is $8,4,2 \ldots \frac{1}{1024} \cdot \rightarrow(1)$

First term in the GP, $a_{1}=a=8$

Second term in the GP, $a_{2}=a r=4$

The common ratio, $r=\frac{4}{8}=\frac{1}{2}$

The last term in the given GP is $\frac{1}{1024}$.

Second last term in the GP $=a_{n-1}=a r^{n-2}$

Starting from the end, the series forms another GP in the form,

$a r^{n-1}, a r^{n-2}, a r^{n-3} \ldots a r^{3}, a r^{2}, a r, a \rightarrow(2)$

Common ratio of this GP is $\frac{1}{r}$.

So, common ratio = 2

$a=\frac{1}{1024}$

So, $6^{\text {th }}$ term of the GP $(2)$,

$a 6=a r^{5}$

$=\frac{1}{1024} \times 2^{5}=\frac{1}{32}$

Hence, the $6^{\text {th }}$ term from the end of the given GP is $\frac{1}{32}$.