Find the 6th term from the end of the G.P. : 16, 8, 4, 2,……, $\dfrac{1}{512}$.

Given,

a = 16

r = $\dfrac{8}{16} = \dfrac{1}{2}$

l = $\dfrac{1}{512}$

We know that,

nth term from end of a G.P. is given by,

T_n = $\dfrac{l}{r^{n - 1}}$

$\Rightarrow T_{\text{6th term from end}}= \dfrac{\dfrac{1}{512}}{\Big(\dfrac{1}{2}\Big)^{6 - 1}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{512}}{\Big(\dfrac{1}{2}\Big)^{5}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{512}}{\Big(\dfrac{1}{32}\Big)} \\[1em] \Rightarrow \dfrac{1}{512} \times 32 \\[1em] \Rightarrow \dfrac{1}{16}.$

Hence, 6th term from end = $\dfrac{1}{16}$.