Question

Find three numbers in G.P. whose sum is 19 and product is 216.

Answer 1

Let the numbers are `a/r`, a, ar

Sum = 19

`a/r + a + ar` = 19

Product = 216

`a/r xx a xx ar` = 216

a³ = 216

= 6³

a = 6

⇒ `a(1/r + 1 + r) = 19`

⇒ `6((1 + r + r^2)/r) = 19`

⇒ 6 + 6r + 6r² = 19r

⇒ 6r² − 13r + 6 = 0

⇒ 6r² − 9r − 3r + 6 = 0

⇒ 3r(2r − 3) − 2(2r − 3) = 0

⇒ (2r − 3)(3r − 2) = 0

2r − 3 = 0

2r = 3

r = `3/2`

3r − 2 = 0

3r = 2

r = `2/3`

If r = `3/2` Numbers are `a/r`, a, ar

= `6/(3/2), 6, 6 xx 3/2`

= 4, 6, 9

If r = `2/3` Numbers are `a/r`, a, ar

= `6/(2/3), 6, 6 xx 2/3`

= 9, 6, 4