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

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

Given, S₂ = -4

Then, from the question we have

\( s_2=-4=\dfrac{a(1-r^n)}{1-r}\)......(1)

And,

a₅ = 4 x a₃

ar⁴ = 4ar²

r² = 4

r = ± 2

Using the value of r in (1), we have

\( s_2=-4=\dfrac{a(1-(2)^2}{1-2}\)

\( -4=\dfrac{a(1-4)}{-1}\)

-4 = a(3)

\( a=\dfrac{-4}{3}\)

for r = -2

\( s_2=-4=\dfrac{a(1-(2)^2}{1-(-2)}\)

\( -4=\dfrac{a(1-4)}{1+2}\)

\( -4=\dfrac{a(-3)}{3}\)

= 4

Therefore, the required G.P is -4/3, -8/3, -16/3, …. Or 4, -8, 16, -32, ……