Question

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

Answer 1

$\begin{array}{c}{a}_n=a{r}^{n-1}\\ b=a{r}^{n-1}\\ first\_term=a\\ P=a_1\times a_2\times a_3\times ........\times a_n\\ =a\times ar\times a{r}^2\times ........\times a{r}^{n-1}\\ =\underset{n-times}{\underset{}{(a\times a\times a\times ......\times a)}}\times (r\times r^2\times r^3\times ......\times r^n)\\ =a^n{r}^{\frac{n(n-1)}{2}}\\ P^2=a^{2n}.r^{n(n-1)}\\ ={\left[a^2.r^{n-1}\right]}^n={\left[a.a{r}^{n-1}\right]}^n=(ab{)}^n\end{array}$

Hence proved.