Question

If the sum of an infinitely decreasing G.P. is $3$, and the sum of the squares of its terms is $\frac{9}{2}$, then the sum of the cubes of the terms is

• A. $\frac{105}{13}$
• B. $\frac{108}{13}$
• C. $\frac{729}{8}$
• D. $\frac{108}{9}$

Answer 1

The correct option is B $\frac{108}{13}$

Let the GP be $a,ar,a{r}^2,a{r}^3,...$

The first term be $a$ and the common ratio be $r$.

Then, it is given that:

$\text{Sum}=\frac{a}{1-r}=3$ ....(1)

Sequence of squares of terms is $a^2,a^2{r}^2,a^2{r}^4,...$

$\text{Sum}=\frac{a^2}{1-r^2}=\frac{9}{2}$ ....(2)

Thus, we have:

$\frac{a^2÷a}{(1-r^2)÷(1-r)}=\frac{9÷3}{2}$

$\Rightarrow \frac{a}{1+r}=\frac{3}{2}$ ....(3)

Now divide equations (1) and (3),

$\frac{1-r}{1+r}=\frac{1}{2}$

$\Rightarrow r=\frac{1}{3}$

Substituting this value of $r$ in equation (1), we get $a=2$

Therefore, $\frac{a^3}{1-r^3}=\frac{8}{(1-\frac{1}{27})}=\frac{108}{13}$

Thus, the answer is option B.