If the first and the $n^{\text {th }}$ term of a $G.P.$ are $a$ and $b$, respectively, and if $P$ is the product of $n$ terms, prove that $P^{2}=(a b)^{n}$

The first term of the $G.P.$ is $a$ and the last term is $b$

Therefore, the $G.P.$ is $a, a r, a r^{2}, a r^{3} \ldots a r^{n-1},$ where $r$ is the common ratio.

$b=a r^{n-1}$        .........$(1)$

$P=$ Product of $n$ terms

$=(a)(a r)\left(a r^{2}\right) \ldots \ldots\left(a r^{n-1}\right)$

$=(a \times a \times \ldots a)\left(r \times r^{2} \times \ldots . r^{n-1}\right)$

$ = {a^n}{r^{1 + 2 + .....(n - 1)}}$         ........$(2)$

Here, $1,2, \ldots \ldots(n-1)$ is an $A.P.$

$\therefore 1+2+\ldots \ldots \ldots+(n-1)$

$=\frac{n-1}{2}[2+(n-1-1) \times 1]=\frac{n-1}{2}[2+n-2]=\frac{n(n-1)}{2}$

$P=a^{n} r^{\frac{n(n-1)}{2}}$

$\therefore P^{2}=a^{2 n} r^{n(n-1)}$

$=\left[a^{2} r^{(n-1)}\right]^{n}$

$=\left[a \times a r^{n-1}\right]^{n}$

$=(a b)^{n}$       [ Using $(1)$ ]

Thus, the given result is proved.