If the first and the `n^(th)` terms of a G.P., are `a` and `b ,` respectively, and if `P` is the product of the first `n` terms prove that `P^2=(a b)^ndot`

Explanation

Let r be the common ratio of the given GP. Then,
`T_(n)=b rArr ar^(n-1)=b rArr r^(n-1)=b/a rArr r=(b/a)^(1/((n-1)))`....(i)
`:.` P = product of first n terms of the GP
`=a.ar.ar^(2) ... ar^(n-1)`
`=a^(n).r^({1+2+3+...+(n-1)})=a^(n).r^(1/2 n(n-1))`
[taking the sum of the AP]
`=a^(n). {(b/a)^(1/((n-1)))}^(1/2 n(n-1))` [using (i)]
`=a^(n).(b/a)^({1/((n-1))xx1/2n (n-1)})=a^(n).(b/a)^(n/2)`
`rArr P^(2)=a^(2n). (b/a)^(n)=(a^(2n). b^(n)/a^(n))=a^(n)b^(n)=(ab)^(n)`.
Hence, `P^(2)=(ab)^(n)`.