2. If the positive real numbers $a_{1}: a_{2}: a_{3}: a_{4}=1: q: q^{2}: q^{3}$ $(q \neq 1)$, and the average of $a_{1}$ and $a_{2}$ is $\frac{1}{2} a_{3}$, then the value of $\frac{a_{1}+a_{2}+a_{3}}{a_{2}+a_{3}+a_{4}}$ is $(\quad$.
(A) $\frac{1+\sqrt{5}}{2}$
(B) $\frac{-2+\sqrt{5}}{2}$
(C) $\frac{-1+\sqrt{5}}{2}$
(D) $\frac{3-\sqrt{5}}{2}$
2.C.

It is evident that $q>0$. From the given condition, we have
$$
\frac{a_{1}+a_{2}+a_{3}}{a_{2}+a_{3}+a_{4}}=\frac{1}{q} \text {. }
$$

Also, $a_{1}+a_{1} q=a_{1} q^{2}$, which simplifies to $1+q-q^{2}=0$.
Therefore, $\frac{1}{q^{2}}+\frac{1}{q}-1=0$.
Solving this, we get $\frac{1}{q}=\frac{-1 \pm \sqrt{5}}{2}$.
Since $q>0$, we have $\frac{1}{q}=\frac{-1+\sqrt{5}}{2}$.