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Question
If S, P, R are the sum, product, and sum of the reciprocals of n terms of a G.P. respectively, then verify that `["S"/"R"]^"n"` = P2
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Solution
Let a be the first term and r be the common, ratio of the G.P.
Then S = `("a"("r"^"n" - 1))/("r" - 1)`
P = a × ar × ar2 × ...... arn–1
= `"a"^"n"*"r"^(1 + 2 + 3 + ... + ("n" - 1))`, where
1 + 2 + 3 + ... + (n – 1) = `(("n" - 1))/2[2(1) + ("n" - 1 - 1)1]`
= `(("n" - 1))/2(2 + "n" - 2) = ("n"("n" - 1))/2`
∴ P = `"a"^"n"*"r"("n"("n" - 1))/2`
and R = `1/"a" + 1/"ar" + 1/"ar"^2 + ... + 1/("ar"^("n" - 1))`
= `("r"^("n" - 1) + "r"^("n" - 2) + "r"^("n" - 3) + ... + 1)/"ar"^("n" - 1)`, where
rn–1 + rn–2 + rn–3 + ... + 1 = 1 + r + r2 + ... + rn–1
= `(1("r"^"n" - 1))/("r" - 1)`
= `("r"^"n" - 1)/("r" - 1)`
∴ R = `("r"^"n" - 1)/(("r" - 1)*"ar"^"n" - 1)`
∴ `("S"/"R")^"n" = [("a"("r"^"n" - 1))/("r" - 1) xx (("r" - 1)"ar"^("n" - 1))/("r"^"n" - 1)]^"n"`
= `("a"^2"r"^("n" - 1))^"n" = "a"^(2"n")*"r"^("n"("n" - 1))`
= `["a"^"n"*"r" ("n"("n" - 1))/2]^2` = P2
Hence, `["S"/"R"]^"n"` = P2
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