Question

If the sum of an infinitely decreasing $GP$ is $3$ , and the sum of the squares of its terms is $9/2$, then sum of the cubes of the terms is

• A. $\frac{107}{12}$
• B. $\frac{105}{17}$
• C. $\frac{108}{13}$
• D. $\frac{97}{12}$

Answer 1

Answer: (C)

Let the GP be $a,ar,a{r}^2,........,$ where $0<r<1$.
Then $a+ar+a{r}^2+........=3$
and $a^2+a^2{r}^2+a^2{r}^4+\dots =9/2$
$\Rightarrow \frac{a}{1-r}=3$ and $\frac{a^2}{1-r^2}=\frac{9}{2}$
$\Rightarrow \frac{9(1-r{)}^2}{1-r^2}=\frac{9}{2}$
$\Rightarrow \frac{1-r}{1+r}=\frac{1}{2}$
$\Rightarrow r=\frac{1}{3}$
Putting $r=\frac{1}{3}$ in $\frac{a}{1-r}=3$, we get $a=2$
Now, the required sum of the cubes is
$a^3+a^3{r}^3+a^3{r}^6+............$
$=\frac{a^3}{1-r^3}=\frac{8}{1-(1/27)}=\frac{108}{13}$