$10 \cdot 25$ is the sum of an arithmetic sequence with an even number of terms. The sums of the odd-numbered terms and the even-numbered terms are 24 and 30, respectively. If the last term exceeds the first term by 10.5, then the number of terms in the arithmetic sequence is
(A) 20.
(B) 18.
(C) 12.
(D) 10.
(E) 8.
(24th American High School Mathematics Examination, 1973)
[Solution] Let $a, d$ and $2 n$ represent the first term, common difference, and number of terms of the sequence, respectively. If $S_{1}$ and $S_{2}$ represent the sums of all odd-numbered terms and all even-numbered terms, respectively, then
$$
\begin{array}{l}
S_{1}=\frac{n}{2}[2 a+(n-1) \cdot 2 d]=24, \\
S_{2}=\frac{n}{2}[2(a+d)+(n-1) \cdot 2 d]=30 .
\end{array}
$$

The difference $S_{2}-S_{1}=\frac{n}{2} \cdot 2 d=6$, i.e., $n d=6$.
If $l$ represents the last term, then $\quad l-a=a+(2 n-1) d-a=10.5$.
$\therefore 2 n d-d=10.5$, i.e., $12-d=10.5$, yielding $d=1.5$.
$\therefore n=\frac{6}{d}=\frac{6}{1.5}=4$, or $2 n=8$.
Therefore, the answer is $(E)$.