smc

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: $\frac{10}{2}(2a+9d)=100⟺2a+9d=20$ Sum of the first 100 terms: $\frac{100}{2}(2a+99d)=10⟺2a+99d=\frac{1}{5}$ 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, $\boxed{-110}$.