Question

The sum of an infinite G.P. with terms $\left\{t_n\right\}$ is 10, and the sum of another G.P with terms $\left\{t_n^2\right\}$ is 25. Then the sum of G.P. with terms $\left\{t_n^3\right\}$ is

• A. 40
• B. $\frac{4000}{49}$
• C. 48
• D. $\frac{8000}{81}$

Answer 1

The correct option is B $\frac{4000}{49}$

Let the original G.P. be $a,ar,a{r}^2\dots \dots \infty .Then\frac{a}{1-r}=10or\frac{a^2}{(1-r{)}^2}=100\dots \dots \left(i\right)$
Also, second G.P. is $a^2,a^2{r}^2,a^2{r}^4,\dots \dots Then\frac{a^2}{1-r^2}=25\dots \dots \left(ii\right)$
$\left(i\right)÷\left(ii\right)\Rightarrow \frac{1+r}{1-r}=4\Rightarrow r=\frac{3}{5}$
Hence, a = 4
Third G.P. is $a^3,a^3{r}^3,a^3{r}^6,\dots \dots$
$S_{\infty }=\frac{a^2}{1-r^3}=\frac{64\times 125}{98}=\frac{4000}{49}$