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Question
Answer the following:
Find three numbers in G.P. such that their sum is 35 and their product is 1000
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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"` = 35
∴ `"a"(1/"r" + 1 + "r")` = 35 ....(i)
Also, `("a"/"r")("a")("ar")` = 1000
∴ a3 = 1000
∴ a = 10
Substituting the value of a in (i), we get
`10(1/"r" + 1 + "r")` = 35
∴ `1/"r" + "r" + 1 = 35/10`
∴ `1/"r" + "r" = 35/10 - 1`
∴ `1/"r" + "r" = 25/10`
∴ `1/"r" + "r" = 5/2`
∴ 2r2 – 5r + 2 = 0
∴ (2r – 1) (r – 2) = 0
∴ r = `1/2` or r = 2
When r = `1/2`, a = 10
`"a"/"r" = 10/((1/2))` = 20, a = 10 and ar = `10(1/2)` = 5
When r = 2, a = 10
`"a"/"r" = 10/2` = 5, a = 10 and ar = 10 (2) = 20
Hence, the three numbers in G.P. are 20, 10, 5 or 5, 10, 20.
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