8. Given the sequence $\left\{a_{n}\right\}$, where $a_{n}$ is a real number, and for $n \geqslant 3, n \in$ $\mathbf{N}$, we have $a_{n}=a_{n-1}-a_{n-2}$. If the sum of the first 1985 terms is 1000, and the sum of the first 1995 terms is 4000, then the sum of the first 2002 terms is $\qquad$.
8.3000

Let $a_{1}=a, a_{2}=b$, then the first 6 terms of the sequence are $a, b, b-a$, $-a, -b, a-b$. Also, $a_{7}=a, a_{8}=b$, it is easy to verify that the sum of any consecutive 6 terms of this sequence is 0.
$$
\begin{aligned}
\therefore S_{1995} & =S_{332 \times 6+3}=0+a+b+(b-a) \\
& =2 b=4000 . \\
S_{1 \text { ges }} & =S_{380 \times 6+5} \\
& =0+a+b+(b-a)+(-a)+(-b) \\
& =b-a=1000 .
\end{aligned}
$$

From equations (1) and (2), we solve to get $b=2000, a=1000$.
$$
\text { Then } \begin{aligned}
S_{2002} & =S_{333 \times 6+4}=0+a+b+(b-a)+(-a) \\
& =2 b-a=3000 .
\end{aligned}
$$