2. (1997 National High School Competition Question) 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 the correct conclusion is
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. $x_{100}=-a, S_{100}=b-a$
2. Choose A.

Brief solution: After calculation, the first few terms of the sequence $\left\{x_{n}\right\}$ are $a, b, b-a,-a,-b, a-b, a, b, b-a,-a$, $-b, a-b, \cdots$. From this, we can see that $x_{n+6}=x_{n}$ (in fact, $x_{n+1}=x_{n}-x_{n-1}=\left(x_{n-1}-x_{n-2}\right)-x_{n-1}=x_{n-2}$, so $\left.x_{n+6}=-x_{n+3}=x_{n}\right)$, which means $\left\{x_{n}\right\}$ is a sequence with a period of 6, so $x_{100}=x_{6 \times 16-4}=x_{4}=-a$. Also, $x_{6 k+1}+x_{6 k+2}+x_{6 k+3}+x_{6 k+4}+x_{6 k+5}+x_{6 k+6}=x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=a+b+(b-a)+$ $(-a)+(-b)=0$, so $S_{100}=S_{6 \times 15+4}=S_{4}=a+b+(b-a)+(-a)=2 b-a$.