9. Let the sum of the first four terms of an arithmetic sequence be 26, and the sum of the last four terms be 110, and the sum of all terms of this sequence be 187. Then, the number of terms in this sequence is ( ).
(A) 11
(B) 22
(C) 8
(D) 16
9.A.

Let the sequence have $n$ terms: $a_{1}, a_{2}, \cdots, a_{n}$. According to the problem, we have
$$
\begin{array}{l}
a_{1}+a_{2}+a_{3}+a_{4}=26, \\
a_{n-3}+a_{n-2}+a_{n-1}+a_{n}=110 .
\end{array}
$$

Since the arithmetic sequence satisfies
$$
a_{1}+a_{n}=a_{2}+a_{n-1}=a_{3}+a_{n-2}=a_{4}+a_{n-3} \text {, }
$$

we have $4\left(a_{1}+a_{n}\right)=26+110=136, a_{1}+a_{n}=34$.
Also, $\frac{n\left(a_{1}+a_{n}\right)}{2}=187$, which means $\frac{n \times 34}{2}=187$, thus $n=11$.