The average of the squares of the first 45 natural numbers is

to find the average of the squares of the first 45 natural numbers.

​$$\text{Sum of squares} = \frac{n(n+1)(2n+1)}{6}$$​​

​$$\text{Average} = \frac{\text{Sum of squares of first } n \text{ natural numbers}}{n}$$​​

sum of squares of the first 45 natural numbers:

​$$= \frac{45(45+1)(2(45)+1)}{6}\\ \ \\ = \frac{45 \times 46 \times 91}{6} = 31,395$$​​

Now,

​$$\text{Average} = \frac{31,395}{45} = 697.67$$​​

Thus, the average of the squares of the first 45 natural numbers is 697.67.