10-28 Let $a_{1}, a_{2}, \cdots a_{k}$ be a finite arithmetic sequence, and
$$
\begin{array}{c}
a_{4}+a_{7}+a_{10}=17, \\
a_{4}+a_{5}+a_{6}+a_{7}+a_{8}+a_{9}+a_{10}+a_{11}+a_{12}+a_{13}+a_{14}=77 .
\end{array}
$$

If $a_{k}=13$, then $k$ equals
(A) 16 .
(B) 18 .
(C) 20 .
(D) 22 .
(E) 24 .
(44th American High School Mathematics Examination, 1993)
[Solution 1] In an arithmetic sequence with an odd number of terms, the middle term is their average. Since $a_{4}, a_{7}, a_{10}$ itself is an arithmetic sequence and their sum is 17, then $a_{7}=\frac{17}{3}$.
Since $a_{4}, a_{5}, \cdots, a_{14}$ is an arithmetic sequence of 11 terms with a sum of 77, the middle term $a_{9}=\frac{77}{11}=7$.
Let $d$ be the common difference of the arithmetic sequence. Since $a_{7}$ and $a_{9}$ differ by $2d$, then $d=\frac{2}{3}$. Since $a_{7}=a_{1}+6d$, we get $a_{1}=\frac{5}{3}$.
From $a_{k}=a_{1}+(k-1)d=\frac{5}{3}+(k-1)\frac{2}{3}=13$, we get $k=18$. Therefore, the answer is $(B)$.
[Solution 2] Given $a_{4}+a_{7}+a_{10}=17$, we have $3a_{7}=17$. If the common difference is $d$, and given $a_{4}+a_{5}+\cdots+a_{14}=77$, we have $11a_{7}+22d=77$, i.e., $a_{7}+2d=7$, then $a_{9}=7$, and $d=\frac{2}{3}$. Let $a_{9}+xd=13$, i.e., $7+x \cdot \frac{2}{3}=13$, solving for $x$ gives $x=9$, then $a_{18}=13$, i.e., $k=18$. Therefore, the answer is $(B)$.