If the first and the `n^(t h)`term of a GP are a and b, respectively, and if P is the product of n terms, prove that `P^2=(a b)^n`.

Options

A. ab
B. `(ab)^(n)`
C. `(ab)^(n//2)`
D. `(ab)^(2n)`

Answer

B

Explanation

Let r be the common ratio of the G.P. Then,
`b=n^(th)term=ar^(n-1)rArrr^(n-1)=(b)/(a)rArrr=((b)/(a))^((1)/(n-1))`
and,
P = Product of the first n terms
`rArr" "P=a.ar.ar^(2) . . .ar^(n-1)=a^(n)r^(1+2+3+ . . .+(n-1)`
`rArr" "P=a^(n)r^((n(n-1))/(2))=a^(n){((b)/(a))^((1)/(n-1))}^((n(n-1))/(2)`
`rArr" "P=a^(n)((b)/(a))^(n//2)=a^(n//2)b^(n//2)=(ab)^(n//2)`
`rArr" "P^(2)=[(ab)^(n//2)]^(2)=(ab)^(n)`