Question

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

• A. $\frac{105}{13}$
• B. $\frac{108}{13}$
• C. $\frac{729}{8}$
• D. None of these

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 :

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

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

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

Dividing equation ( 2 ) by ( 1 ) we get,

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

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

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

Dividing equation ( 1 ) by ( 3 ) we get,

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

$\Rightarrow$ $2-2r=1+r$

$\Rightarrow$ $3r=1$

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

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

$\Rightarrow$ $\frac{a}{1-\frac{1}{3}}=3$

$\Rightarrow$ $\frac{a}{\frac{2}{3}}=3$

$\therefore$ $a=2$

$\Rightarrow$ $\frac{a^3}{1-r^3}=\frac{(2{)}^3}{1-{\left(\frac{1}{3}\right)}^3}$

$=\frac{8}{1-\frac{1}{27}}$

$=\frac{8\times 27}{26}$

$=\frac{108}{13}$