Question 22.

Opening the brackets, we get the series as: $\left(\dfrac{1}{5}\right)^2-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)+\left(\dfrac{1}{5}\right)^4-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^2+\left(\dfrac{1}{5}\right)^6-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^3+...$

These are two infinite GPs when rearranged:

$\left(\dfrac{1}{5}\right)^2+\left(\dfrac{1}{5}\right)^4+\left(\dfrac{1}{5}\right)^6+...-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^2-\left(\dfrac{1}{5}\times\ \dfrac{1}{7}\right)^3-...$

The sum of the first series would be $\dfrac{\dfrac{1}{25}}{1-\dfrac{1}{25}}=\dfrac{1}{24}$

The sum of the second series would be $\frac{\dfrac{1}{35}}{1-\dfrac{1}{35}}=\dfrac{1}{34}$

The answer to the given series would then be $\dfrac{1}{24}-\dfrac{1}{34}=\dfrac{10}{816}=\dfrac{5}{408}$

Therefore, Option B is the correct answer.