In a G.P. of even number of terms, the sum of all terms is five times the sum of the odd terms.

Solution:

(c) 4

Let there be $2 n$ terms in a G.P.

Let $a$ be the first term and $r$ be the common ratio.

$\because S_{2 n}=5\left(S_{\text {odd terms }}\right)$

$\Rightarrow \frac{a\left(r^{2 n}-1\right)}{(r-1)}=5\left(a+a r^{2}+a r^{4}+a r^{6}+\ldots a r^{(2 n-1)}\right)$

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

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

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

$\Rightarrow \frac{\left(r^{n}-1\right)\left(r^{n}+1\right)}{(r-1)}=5 \frac{\left(r^{n}-1\right)\left(r^{n}+1\right)}{(r-1)(r+1)}$

$\Rightarrow\left(r^{n}-1\right)\left(r^{n}+1\right)(r-1)(r+1)-5(r-1)\left(r^{n}-1\right)\left(r^{n}+1\right)=0$

$\Rightarrow\left(r^{n}-1\right)\left(r^{n}+1\right)(r-1)(r+1-5)=0$

But, $r=1$ or $-1$ is not possible.

$\therefore r=4$