Question

Find three numbers in G. P. such that their sum is 21 and sum of their squares is 189.

Answer 1

Let the three numbers in G. P. be `a/r`, a, ar.

According to the first condition,

`a/r + a + ar` = 21

∴ `1/r + 1 + r = 21/a`

∴ `1/r + r = 21/a - 1`        ...(i)

According to the second condition,

`a^2/r^2 + a^2 + a^2r^2` = 189

∴ `1/r^2 + 1 + r^2 = 189/a^2`

∴ `1/r^2 + r^2 = 189/a^2 - 1`     ...(ii)

On squaring equation (i), we get

`1/r^2 + r^2 + 2 = 441/a^2 - 42/a + 1`

∴ `(189/a^2 - 1) + 2 = 441/a^2 - 42/a + 1`   ...[From (ii)]

∴ `189/a^2 + 1 = 441/a^2 - 42/a + 1`

∴ `441/a^2 - 189/a^2 - 42/a ` = 0

∴ `252/a^2 = 42/a`

∴ 252 = 42a

∴ a = 6

Substituting the value of a in (i), we get

`1/r + r= 21/6 - 1`

∴ `(1 + r^2)/r = 15/6`

∴ `(1 + r^2)/r = 5/2`

∴ 2r² – 5r + 2 = 0

∴ 2r² –  4r –  r + 2 = 0

∴ (2r – 1) (r – 2) = 0

∴  r = `1/2 or 2`

When a = 6, r = `1/2.`

`a/r` = 12, a = 6, ar = 3

When a = 6, r = 2

`a/r` = 3, a = 6, ar = 12

∴ the three numbers are 12, 6, 3 or 3, 6, 12.