Question

The sum of an infinitely decreasing geometric progression is equal to $4$ and the sum of the cubes of its terms is equal to $192$. Find the first term and the common ratio of the progression.

Answer 1

Let the first term and the common ratio be $a,r$ respectively

From the given conditions

$\frac{a}{1-r}=4⟹a=4(1-r)$

$a^3+a^3{r}^3+a^3{r}^6+\cdots +\infty =192⟹\frac{a^3}{1-r^3}=192$

$⟹\frac{64(1-r{)}^3}{1-r^3}=192$

$(1+3r^2-3r-r^3)=3(1-r^3)⟹2r^3+3r^2-3r-2=0⟹r=1,-2,-\frac{1}{2}$

The $GP$ is a decreasing $GP$ so $r=-\frac{1}{2}$

$⟹a=6$

The first term and common ratio are $6,-\frac{1}{2}$ respectively