2. Given an arithmetic sequence $\left\{a_{n}\right\}$ with a non-zero common difference, and $a_{2} 、 a_{3} 、 a_{9}$ form a geometric sequence. Then the value of $\frac{a_{4}+a_{5}+a_{6}}{a_{2}+a_{3}+a_{4}}$ is
(A) $\frac{8}{3}$
(B) $\frac{7}{3}$
(C) 3
(D) $\frac{5}{3}$
2. A.

Let the common difference be $d$.
From $a_{3}^{2}=a_{2} a_{9}$, we get $\left(a_{2}+d\right)^{2}=a_{2}\left(a_{2}+7 d\right)$.
Combining $d \neq 0$, we know $d=5 a_{2} \neq 0$.
Thus, $a_{4}+a_{5}+a_{6}=3 a_{5}=3\left(a_{2}+3 d\right)=48 a_{2}$,
$$
a_{2}+a_{3}+a_{4}=3 a_{3}=3\left(a_{2}+d\right)=18 a_{2} \text {, }
$$

The ratio of the two is $\frac{48}{18}=\frac{8}{3}$.