Question

Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4^th by 18.

Answer 1

Let the geometric series be a, ar, ar², ar³,...

Third term = ar², first term = a

∴ ar² – a = 9 …........(i)

Second term = ar, fourth term = ar³

ar – ar³ = 18 ….........(ii)

Dividing equation (i) by (ii), we get

`("a"("r"^2 - 1))/("a"("r" - "r"^3))`

= `9/18`

= `1/2`

or 2(r² − 1) = r − r³

∴ r³ + 2r² − r − 2 = 0

or (r − 1) (r + 1) (r + 2) = 0

or r = 1, −1, −2 if r = −2,

From equation (i), a(4 − 1) = 9

∴ a = 3

∴ 4th terms of the geometric progression 3, −6, 12, −24.