We claim the answer is $6$. Suppose the first term of the AP is $a$ and that the progression has $k$ terms. Note that $a,k\in\mathbb Z^+$. Then, the sum of the terms of the AP is $$a+(a+2)+\ldots+(a+2k-2)=ak+2(1+\ldots+k-1)=ak+(k-1)k=k(a+k-1).$$ Thus, we must have $k(a+k-1)=200$. Now, since $a\ge1$, we have $a+k-1\ge k$, so $k$ must be the smallest factor. Now, note that $200=2^3\cdot5^2$ has $12$ positive divisors, giving $6$ possible values for $k$. Each value of $k$ corresponds to a single value of $a$, so each of these $6$ values gives one of the progressions in Omar's list. Since no other progression can be on the list by the work above, the answer is $6$. $\blacksquare$ Extra: The 6 progressions are $$\begin{align*} 200\\ 99,101\\ 47,49,51,53\\ 36,38,40,42,44\\ 18,20,22,24,26,28,30,32\\ 11,13,15,17,19,21,23,25,27,29 \end{align*}$$