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Question
Find three numbers in G.P. such that their sum is 21 and sum of their squares is 189.
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Solution
Let the three numbers in G. P. be `"a"/"r"`, a, ar.
According to the given conditions,
`"a"/"r" + "a" + "ar"` = 21
∴ `1/"r" + 1 + "r" = 21/"a"`
∴ `1/"r" + "r" = 21/"a" - 1` ...(i)
Also, `"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.
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