• Sum of first $n$ numbers is

$\dfrac{n(n+1)}{2}$

• The mean would be $\dfrac{\text{Sum}}{\text{n}}$

$\dfrac{n(n+1)}{2} \times \dfrac{1}{n} \implies {\dfrac{n+1}{2}}$

• Hence, when $n$ is 100, the mean is

${\dfrac{100+1}{2}} \implies \boxed{50.5}$

Generalisation

$\text{Mean of first n positive natural numbers}=\dfrac{\text{Last term}+1}{2}$

first N integer sum = $\cfrac{N (N+1)}{2}$

first N integer mean = $\cfrac{N (N+1)}{2N}$ = $\boxed{\cfrac{(N+1)}{2} }$

$N=100$ , the mean = $\cfrac{(100+1)}{2} = 50.5$

To find the mean, you first need to find the sum.

Use the formula Gauss invented to find the sum over all numbers 1 to n: $n(n+1)/2$

$100(100 + 1)/2$

Then divide that sum by the amount of numbers you have to get the average.

$5050/100 = 50.5$