smc

Let the first term of the arithmetic sequence be $a$ and the common difference be $d$. The $50^{\text{th}}$ term of the sequence is $a+49d$, so the sum of the first $50$ terms is $\frac{50(a+a+49d)}{2}$. The $51^{\text{th}}$ term of the sequence is $a+50d$ and the $100^{\text{th}}$ term of the sequence is $a+99d$, so the sum of the next $50$ terms is $\frac{50(a+50d+a+99d)}{2}$. Substituting in values results in this system of equations. $$2a+49d=8$$ $$2a+149d=108$$ Solving for $a$ yields $a=\frac{-41}{2}$. The first term is $-20.5$, which is answer choice $\boxed{-20.5}$.