Question

The sum of an infinite geometric progression is 2 and the sum of the geometric progression made from the cubes of the terms of this infinite series is 24. Then find the first term of the series ?

Answer 1

Let the first term be $a$ and the common ratio be $r.$
The first sum is $\frac{a}{1-r}=2$
Now, the terms have been cubed, so the sum becomes $\frac{a^3}{1-r^3}=24$
Substituting $a$ to be $2(1-r)$ in the second equation, we get $8(1-r{)}^3=24(1-r^3)$
$\Rightarrow 1-3r+3r^2-r^3=3-3r^3$
$\Rightarrow 2r^3+3r^2-3r-2=0$
$\Rightarrow (r-1)(2r^2+5r+2)=0$
So, $r=1,-0.5,-2$
$r$ cannot be $1$,and also $-2$ is not allowed since the terms have to reduce, so $r$ is $-0.5$, we get $a=2(1-r)=3$