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`.

Answer 1

Let a be the 1^st term and r be the common ratio of the G.P.
∴ the G.P. is a, ar, ar², ar³, ..., ar^n–1 

∴ S = a + ar + ar² + ... + ar^n–1 = `"a"(("r"^"n" - 1)/("r" - 1))`

P = a(ar) (ar)² ... (ar^n–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, ..., r^n–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)]