Question

The sum of an infinite geometric progression is $2$ and the sum of the geometric prgression made from the cubes of this infinite series is $24$. Then find the series.

Answer 1

$\frac{a}{1-r}=2$ and $\frac{a^3}{1-r^3}=24$ where $r<1$
Eliminating $a$, we get $\frac{1-r^3}{{(1-r)}^3}=\frac{8}{24}$
or $3(1+r+r^2)={(1-r)}^2=1-2r+r^2$
or $2r^2+5r+2=0$
$(2r+1)(r+2)=0$ $\therefore r=-\frac{1}{2}$
Putting the value of $r$, we get $a=3$
$\therefore 3-\frac{3}{2}+\frac{3}{4}-\frac{3}{8}+.......$ is the series.