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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" = "P"^2`.
Solution
Let a be the 1st term and r be the common ratio of the G.P.
∴ the G.P. is a, ar, ar2, ar3, ..., arn–1
∴ S = a + ar + ar2 + ... + arn–1 = `"a"(("r"^"n" - 1)/("r" - 1))`
P = a(ar) (ar)2 ... (arn–1)
= `"a"^"n"*"r"^(1 + 2 + 3 + ... + ("n" - 1))`
= `"a"^"n"*"r"^(("n"("n" - 1))/2)`
∴ P = `"a"^(2"n")*"r"^("n"("n" - 1)` ...(i)
R = `1/"a" + 1/"ar" + 1/"ar"^2 + ... + 1/"ar"^("n" - 1)`
= `("r"^("n" - 1) + "r"^("n" - 2) + "r"^("n" - 3) + ... + "r"^2 + "r" + 1)/("a"*"r"^("n" - 1)`
= `(1 + "r" + "r"^2 + ... + "r"^("n" - 2) + "r"^("n" - 1))/("a"*"r"^("n" - 1)`
1, r, r2, ..., rn–1 are in G.P., with a = 1, r = r
∴ R = `1/"ar"^("n" - 1)(("r"^"n" - 1)/("r" - 1)) = 1/("a"^2*"r"^("n" - 1)) xx "a" xx (("r"^"n" - 1)/("r" - 1))`
∴ R = `1/("a"^2*"r"^("n" - 1))"S"`
∴ `"a"^2*"r"^("n" - 1) = "S"/"R"`
∴ `("a"^2*"r"^("n" - 1))^"n" = ("S"/"R")^"n"`
∴ `"a"^(2"n")*"r"^("n"("n" - 1)) = ("S"/"R")^"n"`
∴ P2 = `("S"/"R")^"n"`. ...[From (i)]
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