Question

The sum of an infinitely decreasing $GP$ is $4$ and the sum of cubes of its term is equal to $\frac{64}{7}$.
The ratio of it's $5^{th}$ to $7^{th}$ term is ...

Answer 1

Moves donot make any sense that must be cubes meaning.

$a+ar+a{r}^2+...\infty =4$

&

$a^3+a^3{r}^3+a^3{r}^6+...\infty =\frac{64}{7}$

Now using formula for sum of infinite GP

$\frac{a}{1-r}=4$ & $\frac{a^3}{1-r^3}=\frac{64}{7}$ ...(a)

$\therefore \frac{A^3}{(1-r{)}^3}=64$ ...(b)

Dividing (a) by (b)

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

$\frac{(1-r{)}^2}{1+r+r^2}=\frac{1}{7}$

$7+7r^2-14r=1+r+r^2$

$6r^2-15r+6=0$

$6r^2-12r-3r+6=0$

$6r(r-2)-3(r-2)=0$

$\therefore r=2$ or $\frac{1}{2}$

$r=2$ (not possible) as $0<r<1$

$r=\frac{1}{2}$

$\therefore \frac{a_7}{95}=\frac{a{r}^6}{a{r}^4}=r^2={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$