Find a GP for which the sum of the first two terms is -4 and the fifth term is 4 times the third term.

Let first term = $a$ and common ratio = $r$.

First two terms:

$$a + ar = -4$$

$$a(1 + r) = -4 \quad \text{...(i)}$$

Also, $T_5 = 4T_3$

$$ar^4 = 4(ar^2)$$

$$r^2 = 4$$

$$r = 2 \text{ or } -2$$

From (i):

Case 1: $r = 2$

$$a(1 + 2) = -4$$

$$3a = -4$$

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

$$\text{GP: } -\frac{4}{3}, -\frac{8}{3}, -\frac{16}{3}, \dots$$

Case 2: $r = -2$

$$a(1 - 2) = -4$$

$$-a = -4$$

$$a = 4$$

GP: 4, -8, 16, -32, ...