The average of first 99 positive integers is:

The correct answer is Option (2) → 50

The sum of the first \(n\) positive integers is \(\frac{n(n+1)}{2}\), so their average is \(\frac{n+1}{2}\), the middle number.

• Sum of 1 to 99 \(\displaystyle =\frac{99\times100}{2}=4950\)

• Average \(\displaystyle =\frac{4950}{99}=50\)