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The product of two natural numbers is 17. Then, the sum of the reciprocals of their squares is :
Answer & Solution
Correct Answer:
Option
C
Let the numbers be a and b
Then,
ab = 17
⇒ a = 1 and b = 17
So,
$$\eqalign{ & = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} \cr & = \frac{{{a^2} + {b^2}}}{{{a^2}{b^2}}} \cr & = \frac{{{1^2} + {{\left( {17} \right)}^2}}}{{{{\left( {1 \times 17} \right)}^2}}} \cr & = \frac{{290}}{{289}} \cr} $$
Then,
ab = 17
⇒ a = 1 and b = 17
So,
$$\eqalign{ & = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} \cr & = \frac{{{a^2} + {b^2}}}{{{a^2}{b^2}}} \cr & = \frac{{{1^2} + {{\left( {17} \right)}^2}}}{{{{\left( {1 \times 17} \right)}^2}}} \cr & = \frac{{290}}{{289}} \cr} $$
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