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Question
Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P2Rn = Sn
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Solution
Let the geometric progression be a + ar + ar2 +….. + arn – 1
The product of these n terms, P = a. ar . ar2….. arn – 1
= `"a"^"n". "r"^(1 + 2 + ...... + ("n" - 1))`
= `"a"^"n""r" ("n" ("n" - 1))/2`
∴ `"P"^2 = "a"^(2"n"). "r"("n"("n" - 1))`
R = `1/"a" + 1/"ar" + 1/"ar"^2 + ....... + 1/"ar"^("n" - 1)`
= `(1/"a" [(1/"r")^"n" - 1])/1/"r" -1`
= `((1 - "r"^"n")"r")/("ar"^"n"(1 - "r"))`
∴ Rn = `((1 - "r"^"n")^"n")/(("a"^"n" "r"^"n"("n" - 1))(1 - "r")^"n")`
Left Side: P2 Rn = `"a"^(2"n") "r"^("n" ("n" -1)) ((1 - "r"^"n")^"n")/(("a"^"n""r"^("n"("n" - 1))(1 - "r")^"n"))`
= `("a"^"n"(1 - "r"^"n")^"n")/((1 - "r")"n") = "S"^"n"`
Whereas S = a + ar + ar2 + .... + arn - 1
= `("a"(1 - "r"^"n"))/(1 - "r")`
Hence, P2Rn = Sn
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