3. Let $a_{1}, a_{2}, \cdots, a_{k}$ be a finite arithmetic sequence, and $a_{4}+a_{7}+a_{10}=17, a_{4}+a_{5}+a_{6}+\cdots+a_{14}=$ 77, and $a_{k}=13$, then $k=$
A. 16
B. 18
C. 19
D. 20
3. B In the arithmetic sequence $\left\{a_{n}\right\}$ with common difference $d$, $S_{2 n-1}=(2 n-1) a_{n}$, so $3 a_{7}=17$, which gives $a_{7}=\frac{17}{3}$; similarly, $a_{9}=\frac{77}{11}=7$. From $a_{9}-a_{7}=2 d$, we get $d=\frac{2}{3}$, thus $a_{1}=\frac{5}{3}, a_{k}=13=\frac{5}{3}+\frac{2}{3}(k-1)$, solving for $k$ gives $k=18$.