​$$\text{Find the value of } \left( \frac{1}{6} \right)^{-2} + \left( \frac{1}{5} \right)^{-2} + \left( \frac{1}{4} \right)^{-2}.$$​​

Correct option is A

Given:

​$$\left(\frac{1}{6}\right)^{-2} + \left(\frac{1}{5}\right)^{-2} + \left(\frac{1}{4}\right)^{-2}$$

Formula Used:

$$\dfrac{1}{a^{-n}} = a^{n}$$ for a ≠ 0​

Solution:

$$\left(\frac{1}{6}\right)^{-2} + \left(\frac{1}{5}\right)^{-2} + \left(\frac{1}{4}\right)^{-2}$$

=$$\frac1{6^{-2}}+\frac 1{5^{-2}}+\frac 1{4^{-2}}$$​​

​$$=6^2 + 5^2 + 4^2$$​​
=36 + 25 + 16 = 77