10.77 For a finite sequence $A=\left(a_{1}, a_{2}, \cdots, a_{n}\right)$, we define
$$
\frac{S_{1}+S_{2}+\cdots+S_{n}}{n}
$$

as the Cesaro sum of $A$, where $S_{k}=a_{1}+a_{2}+\cdots+a_{k}(1 \leqslant k \leqslant n)$. If a 99-term sequence $\left(a_{1}, a_{2}, \cdots, a_{99}\right)$ has a Cesaro sum of 1000, then the Cesaro sum of the 100-term sequence $\left(1, a_{1}, a_{2}, \cdots, a_{99}\right)$ is
(A) 991 .
(B) 999 .
(C) 1000 .
(D) 1001 .
(E) 1009 .
(43rd American High School Mathematics Examination, 1992)
[Solution] From $\frac{S_{1}+S_{2}+\cdots+S_{n}}{n}=\frac{n a_{1}+(n-1) a_{2}+\cdots+a_{n}}{n}$
Assuming $\frac{99 a_{1}+98 a_{2}+\cdots+a_{99}}{99}=1000$, then
$$
99 a_{1}+98 a_{2}+\cdots+a_{99}=99 \cdot 1000 \text {, }
$$

Therefore,
$$
\begin{aligned}
& \frac{100 \cdot 1+99 a_{1}+98 a_{2}+\cdots+a_{99}}{100} \\
= & \frac{100+99 \cdot 1000}{100}=991 .
\end{aligned}
$$

Hence, the answer is $(A)$.