$32 \cdot 6$ is the sum of all integers between 50 and 350 that end in 1
(A) 5880.
(B) 5539.
(C) 5208.
(D) 4877.
(E) 4566.
(24th American High School Mathematics Examination, 1973)
[Solution] For an arithmetic sequence with the first term $a$ and common difference $d$, the sum of the first $n$ terms $S_{n}$ and the general term $a_{n}$ are given by
$$
S_{n}=\frac{1}{2} n\left(a+a_{n}\right) \text { and } a_{n}=a+(n-1) d \text {, }
$$

When $a=51, \quad d=10$, we get
$$
\begin{array}{cc} 
& a_{n}=341=51+(n-1) \cdot 10, \quad \text { so } \quad n=30, \\
& S_{n}=51+61+\cdots+341=\frac{1}{2} n(51+341)=196 n, \\
\therefore \quad & S=15 \cdot 392=5880 .
\end{array}
$$

Therefore, the answer is $(A)$.