A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.

Solution: Let $a$ is the first term of GP and $r$ is the common ratio and there are $2n$ terms in GP.
Then, GP will be,
$a,ar,a{r}^2,a{r}^3\dots a{r}^{2n-1}$
Sum of all $2n$ terms of this GP will be,
$S_{2n}=\frac{a\left(r^{2n}-1\right)}{r-1}$
All odd terms in this GP will be,
$a,a{r}^2,a{r}^4,\dots a{r}^{2n-2}$
It is another GP with first term $a$ and common ratio $r^2$ with $n$ terms.
$\therefore S_n=\frac{a\left({\left(r^2\right)}^n-1\right)}{r^2-1}$
Now, we are given,
$S_{2n}=5S_n$
$\therefore \frac{a\left(r^{2n}-1\right)}{r-1}=5\cdot \frac{a\left({\left(r^2\right)}^n-1\right)}{r^2-1}$
$\Rightarrow \left(r+1\right)=5$
$\Rightarrow r=4$
So, common ratio of given GP is $4$.