Question

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

Answer 1

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

According to the given conditions,

$S_2$= - 4 = $\frac{a(1-r^2)}{1-r}$...(1)

$a_5={4\times}{a_3}$

$ar_4=4ar^2$

⇒ $r^2=4$

∴ r = ± 2

From (1), we obtain

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

⇒ - 4 = $\frac{a(1-4)}{-1}$

⇒ - 4 = a (3)

⇒ a = $\frac{-4}{3}$

Also, - 4 = $\frac{a[1-(-2)2]}{1-(-2)}$ For r = - 2

⇒ - 4 = $\frac{a(1-4)}{1+2}$

⇒ - 4 = $\frac{a(-3)}{3}$

⇒ a = 4

Thus, the required G.P. is

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