smc

Note that $a_7-3d=a_4$ and $a_7+3d=a_{10}$ where $d$ is the common difference, so $a_4+a_7+a_{10}=3a_7=17$, or $a_7=\frac{17}{3}$. Likewise, we can write every term in the second equation in terms of $a_9$, giving us $11a_9=77⟹a_9=7$. Then the common difference is $\frac{2}{3}$. Then $a_k-a_9=13-7=6=9\cdot \frac{2}{3}$. This means $a_k$ is $9$ terms after $a_9$, so $k=18⟹\boxed{18}$