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) $x_{100}=-a, S_{170}=i-a$
(Note: The problem is provided by a question group)
,- 1. A.
After calculation, the first few terms of the sequence $\left\{x_{n}\right\}$ are
$$
\begin{array}{l}
a, b, b-a,-a,-b, a \sigma-b, a, b, b-a,-a,-b, \\
a-b, \cdots,
\end{array}
$$

From this, we can see that $x_{n+6}=x_{n}$, i.e., $\left\{x_{n}\right\}$ is a sequence with a period of 6.
$$
\begin{array}{l}
\therefore x_{100}=x_{6 \times 16+4}=x_{4}=-a . \\
\text { Also, } x_{6 k+1}+x_{6 k+2}+x_{6 k+3}+x_{6 k+4}+x_{6 k+5}+x_{6 k+6} \\
=x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6} \\
=a+b+(b-a)+(-a)+(-b)+(a-b)=0, \\
\therefore S_{100}=S_{6 \times 16+4}=S_{4}=a+b+(b-a)+(-a) \\
\quad=2 b-a .
\end{array}
$$