Question

If $\frac{25}{k}=1^2-\frac{2^2}{5^2}+\frac{2^3}{5^3}-\frac{4^2}{5^4}+\frac{2^5}{5^5}-\frac{2^6}{5^6}+.....\infty$ then find the value of $k$

Answer 1

$\frac{25}{k}=1^2-\frac{2}{5}+{\left(\frac{2}{5}\right)}^2-{\left(\frac{2}{5}\right)}^3+{\left(\frac{2}{5}\right)}^4-{\left(\frac{2}{5}\right)}^5+..........+\infty$

$\Rightarrow \frac{25}{k}=\frac{1}{1-r},$ here $r=\frac{-2}{5}$

$\therefore \frac{25}{k}=\frac{1}{1-\left(\frac{-2}{5}\right)}=\frac{1}{\frac{7}{5}}=\frac{5}{7}$

$\Rightarrow \frac{25}{k}=\frac{5}{7}$

$\Rightarrow k=\frac{7}{5}.$

Hence, the answer is $\frac{7}{5}.$