Find a G.P. for which sum of the first two terms is –4 and the fifth term is 4 times the third term.

Let a be the first term and r be the common ratio of the G.P.

According to the given conditions,

$S_{2}=-4=\frac{a\left(1-r^{2}\right)}{1-r}$ $\ldots(1)$

$a_{5}=4 \times a_{3}$

$a r^{4}=4 a r^{2}$

$\Rightarrow r^{2}=4$

$\therefore r=\pm 2$

From (1), we obtain

$-4=\frac{a\left[1-(2)^{2}\right]}{1-2}$ for $r=2$

$\Rightarrow-4=\frac{a(1-4)}{-1}$

$\Rightarrow-4=a(3)$

$\Rightarrow a=\frac{-4}{3}$

Also, $-4=\frac{a\left[1-(-2)^{2}\right]}{1-(-2)}$ for $r=-2$

$\Rightarrow-4=\frac{a(1-4)}{1+2}$

$\Rightarrow-4=\frac{a(-3)}{3}$

$\Rightarrow a=4$

Thus, the required G.P. is

$\frac{-4}{3}, \frac{-8}{3}, \frac{-16}{3}, \ldots$ or $4,-8,16,-32, \ldots$