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 $\frac{64}{7}$. Find the sixth term of the progression.

Answer 1

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

From the given conditions

$\frac{a}{1-r}=4⟹a=4(1-r)$

$a^3+a^3{r}^3+a^3{r}^6+\cdots +\infty =\frac{64}{7}⟹\frac{a^3}{1-r^3}=\frac{64}{7}$

$⟹\frac{64(1-r{)}^3}{1-r^3}=\frac{64}{7}$

$7(1+3r^2-3r-r^3)=1-r^3⟹6r^3-21r^2+21r-6=0⟹r=1,2,\frac{1}{2}$

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

$⟹a=2$

Sixth term is $2\times (\frac{1}{2}{)}^6=\frac{1}{32}$