$10 \cdot 33$ For $p=1,2, \cdots, 10$, let $S_{p}$ be the sum of the first 40 terms of an arithmetic sequence with the first term $p$ and common difference $2 p-1$. Then $S_{1}+S_{2}+\cdots+S_{10}$ is
(A) 80000.
(B) 80200.
(C) 80400.
(D) 80600.
(E) 80800.
(25th American High School Mathematics Examination, 1974)
[Solution] For each $p=1,2, \cdots 10$, we have
$$
\begin{aligned}
S_{p}= & p+[p+(2 p-1)]+[p+2(2 p-1)]+\cdots \\
& +[p+39(2 p-1)] \\
= & 40 p+\frac{40 \cdot 39}{2}(2 p-1) \\
= & (40+40 \cdot 39) p-20 \cdot 39 \\
= & 40^{2} p-20 \cdot 39 .
\end{aligned}
$$

Then
$$
\begin{aligned}
\sum_{p=1}^{10} S_{p} & =40^{2} \sum_{p=1}^{10} p-(10 \cdot 20 \cdot 39) \\
& =40^{2} \frac{10 \cdot 11}{2}-(10 \cdot 20 \cdot 39) \\
& =80200 .
\end{aligned}
$$

Therefore, the answer is $(B)$.