if `S` is the sum , `P` the product and `R` the sum of reciprocals of `n` terms in `G.P.` prove that `P^2 R^n=S^n`

Explanation

For a GP, Sum S = `a+ar+ar^2+…+ar^(n-1) = a(1-r^n)/(1-r)`


Product,` P = a(ar)(ar^2)…(ar^(n-1))
`

= `a^nr^(1+2+3+.....+n-1)`


=` a^nr^(n(n-1)/2)`

`P^2 = a^(2n)r^(n(n-1))`

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