1. Given a geometric sequence $\left\{a_{n}\right\}$ with a common ratio of $q(q \neq 1)$, let $b_{1}=a_{1}+a_{2}+a_{3}, b_{2}=a_{4}+a_{5}+$ $a_{6}, \cdots, b_{n}=a_{3 n-2}+a_{3 n-1}+a_{3 n}, \cdots$, then the sequence $\left\{b_{n}\right\}$
A. is an arithmetic sequence
B. is a geometric sequence with a common ratio of $q$
C. is a geometric sequence with a common ratio of $q^{3}$
D. is neither an arithmetic sequence nor a geometric sequence
Solve:
\[
\frac{b_{n+1}}{b_{n}}=\frac{a_{3 n+1}+a_{3 n+2}+a_{3 n+3}}{a_{3 n-2}+a_{3 n-1}+a_{3 n}}=\frac{q^{3}\left(a_{3 n-2}+a_{3 n-1}+a_{3 n}\right)}{a_{3 n-2}+a_{3 n-1}+a_{3 n}}=q^{3}.
\]
Therefore, the answer is $C$.