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.