The average of the squares of the first 48 natural numbers is

to find the average of the squares of the first 48 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:

​$$= \frac{48(48+1)(2(48)+1)}{6}\\ \ \\ = \frac{48 \times 49 \times 97}{6} = 38,024$$​​

Now,

​$$\text{Average} = \frac{38,024}{48} = 792.17$$​​

Thus, the average of the squares of the first 45 natural numbers is 792.17

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