Question

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

Answer 1

Geometric progression

$1^{st}$ term $=a$

$n^{th}$ term $=b$

$P=$ product of n terms

$G.P.=a,ar,a{r}^2.....a{r}^{n-1}$

$\to b=a{r}^{n-1}$

$\to P=\left(a\right)\left(ar\right)\left(a{r}^2\right)...\left(a{r}^{n-1}\right)$

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

$\to P=a^n{r}^{(1+2+3+...+n+1)}$

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

Now,

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

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

$\to P^2=(a\cdot b{)}^n$

Hence

$P^2=(a\cdot b{)}^2$