This is an arithmetic progression with first term a=1 and common difference d=2. If n is the number of terms, the last term is a+(n-1)d.
The sum Sn of the first n terms is a+a+d+a+2d+...+a+(n-1)d.
There are n/2 pairs of terms:
(a+a+(n-1)d)+(a+d+a+(n-2)d)+(a+2d+a+(n-3)d)+...
(2a+(n-1)d)+(2a+(n-1)d)+(2a+(n-1)d)+...
So each pair has the same sum and there are n/2 pairs, so:
Sn=(n/2)(2a+(n-1)d)=an+nd(n-1)/2.
But Sn=10000, and we know a=1 and d=2, so substitute for these:
n+2n(n-1)/2=n+n2-n=n2=10000, n=100.
The last number is a+(n-1)d=1+198=199, making the series:
1+3+5+7+...+195+197+199.