If the G.P.'s 5, 10, 20, ... and 1280, 640, 320, ... have their nth terms equal,

Given :

First term, $a=5$

Common ratio, $r=2$

$a_{n}=(5)(2)^{n-1}$   .. (1)

Similarly, $a_{n}=(1280)\left(\frac{1}{2}\right)^{n-1} \ldots(2)$

From $(1)$ and $(2)$

$(5)(2)^{n-1}=(1280)\left(\frac{1}{2}\right)^{n-1}$

$\Rightarrow \frac{1}{256}=\left(\frac{1}{4}\right)^{n-1}$

$\Rightarrow\left(\frac{1}{4}\right)^{4}=\left(\frac{1}{4}\right)^{n-1}$

$\Rightarrow n-1=4$

$\Rightarrow n=5$