Find the geometric progression (GP) for which the sum of the first two terms is $-4$ and the fifth term is $4$ times the third term.
Let the first term of GP $= t_1 = a$; $\;$ common ratio $= r$
$n^{th}$ term of GP $= t_n = ar^{n-1}$
$2^{nd}$ term of GP $= t_2 = ar^{2 - 1} = ar$
$3^{rd}$ term of GP $= t_3 = ar^{3 - 1} = ar^2$
$5^{th}$ term of GP $= t_5 = ar^{5 - 1} = ar^4$
Now, $\;$ $t_1 + t_2 = -4$ $\;\;\;$ [given]
i.e. $\;$ $a + ar = -4$ $\implies$ $a = \dfrac{-4}{1 + r}$ $\;\;\; \cdots \; (1)$
$t_5 = 4 \times t_3$ $\;\;\;$ [given]
i.e. $\;$ $a r^4 = 4 \times ar^2$
i.e. $\;$ $r^2 = 4$ $\implies$ $r = \pm 2$
Substituting $r = +2$ in equation $(1)$ gives
$a = \dfrac{-4}{1 + 2} = \dfrac{-4}{3}$
Then the GP is $\;$ $\dfrac{-4}{3}, \; \dfrac{-8}{3}, \; \dfrac{-16}{3}, \; \dfrac{-32}{3}, \; \dfrac{-64}{3}, \cdots$
Substituting $r = -2$ in equation $(1)$ gives
$a = \dfrac{-4}{1 - 2} = 4$
Then the GP is $\;$ $4, \; -8, \; 16, \; -32, \; 64, \cdots$