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

Let the G.P. be `a,ar,ar^(2),ar^(3),ar^(n-1)`…
According to the given information, we have
`S=(a(r^(n)-1))/(r-1)`
`P=(a)(ar)(ar^(2))(ar^(3))….(ar^(n-1))`
`=a^(n)xxr^(1+2+….+n-1)`
`=a^(n)r^((n(n-1))/2)`
`R=1/a+1/(ar)+…+1/(ar^(n-1))`
`=(1/a(1-(1/r)^(n)))/(1-1/r)`
`=((r^(n)-1))/(ar^(n-1)(r-1))`
`thereforeP^(2)R^(n)=a^(2n)r^(n(n-1))((r^(n)-1)^(n))/(a^(n)r^(n(n-1))(r-1)^(n))`
`=(a^(n)(r^(n)-1)^(n))/((r-1)^(n))`
`=[(a(r^(n)-1))/((r-1))]^(n)`
`=S^(n)`