# If S be the sum, P be the product and R be the sum of the reciprocals of n terms of a geometric progression. Prove that P^2=(S/R)^n.

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### 1 Answer

We are given that for n terms of a geometric progression, s is their sum, p is their product and r is the sum of their reciprocals.

Let the first term of the GP be a and the common ratio be r.

The n terms are a, ar, ar^2, ...ar^(n-1)

S = a + ar + ar^2 +...ar^(n - 1) = a*(r^n - 1)/ (r - 1)

P = a*ar*ar^2...ar^(n-1) = a^n*r^(1 + 2 + ...(n-1)

=> a^n*r^((n-1)(n)/2)

R = 1/a + 1/ar + 1/ar^2 +...1/r^(n - 1) = (1/a)[ 1 + 1/r + 1/r^2... + 1/r^(n-1)

=> (1/a)[((1/r)^n - 1)/(1/r - 1)

=> (1/a)[r*(1 - r^n)/(1 - r)*r^n]

S/R = [a*(r^n - 1)/ (r - 1)] / (1/a)[r*(1 - r^n)/(1 - r)*r^n]

=>[a^2] / [1/r^(n - 1)]

=> a^2*r^(n - 1)

(S/R)^n = a^2*n*r^[(n - 1)(n)]

P^2 = [a^n*r^((n-1)(n)/2)]^2

=> a^2*n*r^[(n - 1)*n]

**Therefore we get that P^2 = (S/R)^n**