10.35 Two people start walking towards each other from points $M$ and $N$, which are 72 miles apart. The first person walks at a speed of 4 miles per hour, while the second person walks 2 miles in the first hour, $2 \frac{1}{2}$ miles in the second hour, 3 miles in the third hour, and so on (forming an arithmetic sequence). They will meet
(A) within 7 hours.
(B) within $8 \frac{1}{4}$ hours.
(C) closer to $M$.
(D) closer to $N$.
(E) at the midpoint of $M N$.
(8th American High School Mathematics Examination, 1957)
［Solution］Let the two people meet after $t$ hours. According to the problem, we have:
$$
4 t+\left\{2+2 \frac{1}{2}+\cdots+\left[2+\frac{1}{2}(t-1)\right]\right\}=72 \text {, }
$$

or
$$
\begin{array}{c}
4 t+\frac{t}{2}\left[2 \cdot 2+\frac{1}{2}(t-1)\right]=72, \\
t^{2}+23 t-288=0,
\end{array}
$$

Solving this, we get $t=9$ (the negative value has been discarded).
Thus, when they meet, the first person has walked $4 \cdot 9=36$ miles.
Therefore, the answer is $(E)$.