Question

A geometric progression with common ratio $r,$ consists of an even number of terms. If the sum of all terms is $5$ times the sum of the terms occupying the odd places, then $\sum _{i=1}^4(ir{)}^2$ is

• A. $1456$
• B. $120$
• C. $1172$
• D. $480$

Answer 1

The correct option is D $480$

Let the G.P. be $a,ar,a{r}^2,\dots ,a{r}^{2n-1}.$
Then, $S_{2n}=5(a+a{r}^2+\cdots +n\text{terms})$
$\Rightarrow \frac{a}{1-r}(1-r^{2n})=5\cdot \frac{a}{1-r^2}(1-(r^2{)}^n)$
$\Rightarrow r+1=5$
$\Rightarrow r=4$

Now, $\sum _{i=1}^4(ir{)}^2$
$=16\sum _{i=1}^4{i}^2$
$=16\times \frac{4\times 5\times 9}{6}$
$=480$