jmc

The $200$ numbers sum up to $10,000$, so their average is $\frac{10,000}{200}=50$.

Then we can represent the sequence as $$50-199d,50-197d,\dots ,50-d,50+d,50+3d,\dots ,50+197d,50+199d.$$Since all the terms are at least 10, in particular the first and last term of the sequence, we know $50-199d\ge 10$ and $50+199d\ge 10$. This means $50-199|d|\ge 10$ so $|d|\le \frac{40}{199}$ which means $d$ is at most $\frac{40}{199}$ and at least $-\frac{40}{199}$.

The 50th term is $50-101d$.

$$L=50-101\times \frac{40}{199}=50-\frac{4040}{199}$$$$G=50-101\times \left(-\frac{40}{199}\right)=50+\frac{4040}{199}$$We can check that both of these sequences meet all the conditions of the problem (the lower bound, upper bound, and total sum).

Hence, $G-L=2\times \frac{4040}{199}=\boxed{\frac{8080}{199}}$.

Note: The condition that each term is at most 100 is unnecessary in solving the problem! We can see this when we apply the condition to the first and last term (similar to when we applied the condition that all terms are at least 10), $50-199d\le 100$ and $50+199d\le 100$ which means $50+199|d|\le 100$ so $|d|\le \frac{50}{199}$ which is a higher bound than we already have.