Question

The first term of a G.P. is $a$ and its $n^{t}h$ term is $b$, where n is an even number. If the product of first n numbers of this $G.P$ is $P$. Prove that : $P^2=(ab{)}^n$.

Answer 1

Given,

first term $=a$

last term $b=a{r}^{n-1}$

$P=a_1\times a_2\times a_3\times ..........a_n$

$=a\times ar\times a{r}^2\times a{r}^3\times ......\times a{r}^{n-1}$

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

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

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

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

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

$=(ab{)}^n$

$\therefore P^2=(ab{)}^n$

Hence proved.