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

According to the given conditions,

$A_{2}=-4=\frac{a\left(1-r^{2}\right)}{1-r}$       …….$(1)$

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

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

$\therefore r=\pm 2$

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 $4,-8,-16,-32 \ldots$