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`

Answer

N/a

Explanation

Let 'a' be the first term and 'r' be the common ratio of G.P.
`therefore` S = sum of n terms
` = a + ar + ar^(2) + … + a * r^(n - 1)`
`S = (a(1 - r^(n)))/( 1 - r)" "…(1)" "(Let r lt 1)`
`R = (1)/(a) +_ (1)/(ar) + (1)/(ar^(2)) +… +(1)/(ar^(n - 1))`
`= ((1)/(a) * ((1)/(r^(n)) - 1))/((1)/(r) - 1)" "(because (1)/(r) gt 1)`
`= ((1 - r^(n)))/(ar^(n - 1)(1 - r))` ...(2)
Dividing equation (1) by (2),
`(S)/(R) = (a(1 - r^(n)) * a * r^(n - 1) * (1 - r))/((1 - r)(1 - r^(n))) = a^(2) * r^(n- 1)`
`rArr" "((S)/(R))^(n) = (a^(2) * r^(n - 1))^(n)`
`rArr" "(S^(n))/(R^(n)) = a^(2n) * r^(n(n-1))` ...(3)
Now p = product of n terms
`= (a) (ar) (ar^(2)) .... (a * r^(n - 1))`
`= a^(n) * r^(1) + 2 + 3 + ... + (n - 1)`
`= a^(n) * r^((n - 1)/(2) (1 + n -1)) = a^(n) * r^((n(n-))/(2))`
`rArr" "P^(2) = a^(2n) * r^(n(n - 1))`
`rArr" "P^(2) = (S^(n))/(R^(n))` [From equation (3)]
`rArr P^(2) * R^(n) = S^(n)` Hence Proved.