If the sum of the first $10$ terms and the sum of the first $100$ terms of a given arithmetic progression are $100$ and $10$, 
respectively, then the sum of first $110$ terms is:
$\text{(A)} \ 90 \qquad \text{(B)} \ -90 \qquad \text{(C)} \ 110 \qquad \text{(D)} \ -110 \qquad \text{(E)} \ -100$
Let $a$ be the first term of the sequence and let $d$ be the common difference of the sequence.
Sum of the first 10 terms: $\frac{10}{2}(2a+9d)=100 \Longleftrightarrow 2a+9d=20$
Sum of the first 100 terms: $\frac{100}{2}(2a+99d)=10 \Longleftrightarrow 2a+99d=\frac{1}{5}$
Solving the system, we get $d=-\frac{11}{50}$, $a=\frac{1099}{100}$. The sum of the first 110 terms is $\frac{110}{2}(2a+109d)=55(-2)=-110$
Therefore, $\boxed{D}$.