5. If the infinite decreasing geometric sequence $\left\{a_{n}\right\}$ satisfies
$$
\lim _{n \rightarrow \infty} \frac{a_{1}+a_{4}+a_{7}+\cdots+a_{3 n-2}}{a_{1}+a_{2}+\cdots+a_{n}}=\frac{3}{4},
$$

then, the common ratio $q$ of the sequence is ().
(A) $\frac{\sqrt{21}-3}{6}$
(B) $\frac{-3-\sqrt{21}}{6}$
(C) $\frac{\sqrt{21} \pm 3}{6}$
(D) $\frac{-3 \pm \sqrt{21}}{6}$
5. A.
$$
\begin{array}{l}
\lim _{n \rightarrow \infty} \frac{a_{1}+a_{4}+\cdots+a_{3 n-2}}{a_{1}+a_{2}+\cdots+a_{n}}=\frac{a_{1}}{1-q^{3}} \cdot \frac{1-q}{a_{1}} \\
=\frac{1}{q^{2}+q+1} . \\
\text { Also } \frac{1}{1+q+q^{2}}=\frac{3}{4},
\end{array}
$$

We have $3 q^{2}+3 q-1=0$.
Then $q=\frac{-3 \pm \sqrt{21}}{6}$, discard $q=\frac{-3-\sqrt{21}}{6}$. Hence $q=\frac{\sqrt{21}-3}{6}$.