Question

If the sum of an infinitely decreasing G.P is 3,and the sum of the squares of its term is 9/2, the sum of the cubes of the term is

• A. $105/13$
• B. $108/13$
• C. $729/8$
• D. None of these

Answer 1

Answer: (B)

$108/13$

Let the GP be $a,ar,a{r}^2,...$ where $0<r<1$.

Then $a+ar+a{r}^2+...=3$ and $a^2+a^2{r}^2+a^2{r}^4+...=\frac{9}{2}$

$\Rightarrow \frac{a}{1-r}=3$ and $\frac{a^2}{1-r^2}=\frac{9}{2}$

We have $\frac{a}{1-r}=3\Rightarrow a=3(1-r)$

$\therefore \frac{9(1-r{)}^2}{1-r^2}=\frac{9}{2}$

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

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

Putting $r=\frac{1}{3}$ in $\frac{a}{1-r}=3\Rightarrow a=2$

Now, the required sum of cubes is given by $a^3+a^3{r}^3+a^3{r}^5+...=\frac{a^3}{1-r^3}=\frac{2^3}{1-{\left(\frac{1}{3}\right)}^3}=\frac{8}{1-\frac{1}{27}}=\frac{108}{13}$