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

1. 698.67
2. 699.67
3. 697.67
4. 696.67

Correct answer: 697.67

Solution

The sum of squares of the first $n$ natural numbers is $\frac{n(n+1)(2n+1)}{6}$. For $n=45$, the sum is $\frac{45\cdot46\cdot91}{6}=31395$, and the average is $31395/45=697.67$.

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