If the first and the nth term of a G.P. are a ad b, respectively,


If the first and the $n^{\text {th }}$ term of a G.P. are $a$ ad $b$, 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.

$b=a r^{n-1} \ldots(1)$

$P=$ Product of $n$ terms

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

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

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

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

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

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

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

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

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

$=(a b)^{n} \quad[U \sin g(1)]$

Thus, the given result is proved.

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