Thank you for reporting, we will resolve it shortly
Q.
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
Sequences and Series
Solution:
Let the GP be $a, ar, ar ^{2}, ........,$ where $0< \,r <\,1$.
Then $a + ar + ar ^{2}+........ =3$
and $a^{2}+a^{2} r^{2}+a^{2} r^{4}+\ldots=9 / 2$
$\Rightarrow \frac{ a }{1- r }=3$ 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
$a ^{3}+ a ^{3} r ^{3}+ a ^{3} r ^{6}+............$
$=\frac{ a ^{3}}{1- r ^{3}}=\frac{8}{1-(1 / 27)}=\frac{108}{13}$