Question

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

Answer 1

Let $G.P$ be

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

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

$P=(a\times a{r}^{n-1})(ar\times a{r}^{n-2}).....................\left(ar\frac{n-1}{2}\right)$ for $\frac{n+1}{2}$ terms $n\to$ odd

$(a\times a{r}^{n-1})(ar\times a{r}^{n-2}).....................(a{r}^{\frac{n-1}{2}}\times a{r}^{\frac{n}{2}})$ for $\frac{n}{2}$ terms $n\to$ even

$\Rightarrow P=\left(ab{)}^{\frac{n-1}{2}}\right(\sqrt{ab}{)}^1$ for odd

$(ab{)}^{\frac{n}{2}}$ for even

So,. $p^2=(ab{)}^n$