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"^2"r"^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.