numberphobia

Sum of first "n" numbers= n(n+1)/2

average of first n numbers= ${(sum of "n" numbers)}/{n}$ = $(n+1)/2$

Therefore, average of first 99 no's = 50

$99+1=100$ , average of $99$ and $1$ is $100/2=50$

$98+2=100$ , average of those two is also $100/2=50$

Continue on until only one number is left: $50$

Average of all those $50$ 's is, of course, $50$ .

Python 2.7:

1

print sum(range(1,100))/99.0