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સમગુણોત્તર શ્રેણીનું પ્રથમ પદ $a$ અને $n$ મું પદ છે. જો $n$ પદોનો ગુણાકાર $P$ હોય, તો સાબિત કરો કે $P^{2}=(a b)^{n}$
Solution
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.
