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Question
Find four numbers in G.P. such that sum of the middle two numbers is `10/3` and their product is 1
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Solution
Let the four numbers in G.P. be `"a"/"r"^3, "a"/"r", "ar", "ar"^3`
Since their product is 1, `"a"/"r"^3*"a"/"r"*"ar"*"ar"^3` = 1
∴ a4 = 1
∴ a = 1
Also the sum of middle two numbers is `10/3`
∴ `"a"/"r" + "ar" = 10/3`
∴ `"a"(1/"r" + "r") = 10/3`
∴ `1/"r" + "r" = 10/3` as a = 1
∴ `(1 + "r"^2)/"r" = 10/3`
∴ 3 + 3r2 = 10r
∴ 3r2 – 10r + 3 = 0
∴ (r – 3)(3r – 1) = 0
∴ r = 3 or r = `1/3`
Taking r = 3, `"a"/"r"^3 = 1/27, "a"/"r" = 1/3`, ar3 = 27 and the four numbers are `1/27, 1/3, 3, 27`.
Taking r = `1/3`, `"a"/"r"^3 = 1/((1/27))` = 27, `"a"/"r" = 1/((1/3))` = 3, `"ar" = 1/3`, ar3 = `1/27` and the our numbers are 27, 3, `1/3`, `1/27`.
Hence, the required numbers in G.P. are `1/27, 1/3, 3, 27` or `27, 3, 1/3, 1/27`.
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