Question

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

Answer 1

Let $a$ be the first term and $r$ be the common ratio of G.P

Given: First term of G.P$=a$

We know that $n^{th}$ term of G.P$=a{r}^{n-1}$

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

Now, $P$ is the product of $n$ terms

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

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

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

$=a^n{r}^{1+2+3+...+(n-1)}$

Consider $1+2+3+..+(n-1)=\frac{(n-1)(n-1+1)}{2}=\frac{n(n-1)}{2}$ since $1+2+3+..n=\frac{n(n+1)}{2}$

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

Hence proved.