Example 4 (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 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. $x_{100}=-a, S_{100}=b-a$
Solution: Choose A. Reason: 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, $x_{6 k+1}+x_{6 k+2}+x_{6 k+3}+x_{6 k+4}+x_{6 k+5}+x_{6 k+6}=x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=a+b+$ $(b-a)+(-a)+(-b)+(a-b)=0$, so
$$
S_{100}=S_{6 \times 16+1}=S_{4}=a+b+(b-a)+(-a)=2 b-a .
$$