Question

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

Answer 1

Let the five terms of the given G.P. be

`a/(r^2), a/r, a, ar, ar^2`

Given, sum of first two terms = – 4

`a/(r^2) + a/r = 4`

`=> (a + ar)/r^2 = -4`

`=> a + ar = -4r^2`

`=> a(1 + r) = -4r^2`

`=> a = -(4r^2)/(1 + r)`

And 5^th term = 4(3^rd term)

`=>` ar² = 4(a)

`=>` r² = 4

`=>` r = ±2

When r = +2,

`a = -(4(2)^2)/(1 + 2) = -16/3`

When r = –2,

`a = -(4(-2)^2)/(1 - 2) = -16`

Thus, the required terms are `a/(r^2), a/r, a, ar, ar^2`

i.e `(-16/3)/4, (-16/3)/2, -16/3, -16/3 xx 2, -16/3 xx 4` OR `16/4, 16/(-2), 16, 16(-2), 16 xx 4`

i.e `-4/3, -8/3, -16/3, -32/3, -64/3` OR `4, -8, 16, -32, 64`