The product of two natural numbers is 17. Then, the sum of the reciprocals of their squares is :

A. $$\frac{1}{289}$$

B. $$\frac{289}{290}$$

C. $$\frac{290}{289}$$

D. $$289$$

Answer: Option C

Solution(By Examveda Team)

Let the numbers be a and b
Then,
ab = 17
⇒ a = 1 and b = 17
So,
$$\eqalign{ & = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} \cr & = \frac{{{a^2} + {b^2}}}{{{a^2}{b^2}}} \cr & = \frac{{{1^2} + {{\left( {17} \right)}^2}}}{{{{\left( {1 \times 17} \right)}^2}}} \cr & = \frac{{290}}{{289}} \cr} $$