1. Given the sequence $\left\{x_{n}\right\}$ satisfies $x_{n+1}=x_{n}-x_{n-1}(n \geqslant 2), x_{1}=a, x_{2}=b$, and let $s_{n}=x_{1}+x_{2}+\cdots+x_{n}$, then which of the following conclusions is correct?
A. $x_{100}=-a, s_{100}=2 b-a$
B. $x_{100}=-b, s_{100}=2 b-a$
C. $x_{100}=-b, s_{100}=b-a$
D. $x_{100}=-a, s_{100}=b-a$
1. From $x_{n+1}=x_{n}-x_{n-1}=\left(x_{n-1}-x_{n-2}\right)-x_{n-1}=-x_{n-2}$, we have $x_{n+6}=-x_{n+3}=x_{n}$, so $\left\{x_{n}\right\}$ is a sequence with a period of 6.
$$
\begin{array}{l}
\text { Therefore, } x_{100}=x_{4}=-x_{1}=-a \quad \text { and since } x_{6 k+1}+x_{6 k+2}+x_{6 k+3}+x_{6 k+4}+x_{6 k+5}+x_{6 k+6}=x_{1}+x_{2}+ \\
x_{3}+x_{4}+x_{5}+x_{6}=x_{1}+x_{2}+x_{3}-x_{1}-x_{2}-x_{3}=0 . \\
\text { Therefore, } s_{100}=s_{4}=x_{1}+x_{2}+x_{3}+x_{4}=x_{2}+x_{3}=x_{2}+x_{2}-x_{1}=2 b-a \text {. Hence, the answer is A. }
\end{array}
$$