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. $x_{100}=-a, S_{100}=b-a$
$$
x_{n+2}=x_{n+1}-x_{n}=x_{n}-x_{n-1}-x_{n}=-x_{n-1} \Rightarrow x_{n+3}=-x_{n} \Rightarrow x_{n+6}=x_{n} \text {, }
$$

Then the sequence $\left\{x_{n}\right\}$ is a periodic sequence with a period of 6. Therefore, $x_{100}=x_{4}=-x_{1}=-a$, $S_{100}=-x_{5}-x_{6}=x_{2}+x_{3}=b+b-a=2 b-a$. So the answer is $A$.