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Question
Find five numbers in G.P. such that their product is 243 and sum of second and fourth number is 10.
Solution
Let the five numbers in G.P. be
`"a"/"r"^2, "a"/"r"`, a, ar, ar2.
According to the first condition,
`"a"/"r"^2 xx "a"/"r" xx "a" xx "ar" xx "ar"^2` = 243
∴ a5 = 243
∴ a = 3
According to the second condition,
`"a"/"r" + "ar"` = 10
∴ `1/"r" + "r" = 10/"a"`
∴ `(1 + "r"^2)/"r" = 10/3`
∴ 3r2 – 10r + 3 = 0
∴ 3r2 – 9r – r + 3 = 0
∴ (3r – 1) (r – 3) = 0
∴ r = `1/3, 3`
When a = `3, "r" = 1/3`
`"a"/"r"^2 = 27, "a"/"r" = 9, "a" = 3, "ar" = 1, "ar"^2 = 1/3`
When a = 3, r = 3
`"a"/"r"^2 = 1/3, "a"/"r" = 1, "a" = 3, "ar" = 9, "ar"^2 = 1/3`
∴ the five numbers in G.P. are
27, 9, 3, 1, `1/3 or 1/3`, 1, 3, 9, 27.
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