Example 2 (1999 National High School League Question) Given a geometric sequence $\left\{a_{n}\right\}$ with common ratio $q(q \neq 1)$, 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 $\left\{b_{n}\right\}$ is ( ).
A. An arithmetic sequence
B. A geometric sequence with common ratio $q$
C. A geometric sequence with common ratio $q^{3}$
D. Neither an arithmetic sequence nor a geometric sequence
Solution: Choose C. Reason: $\frac{b_{n+1}}{b_{n}}=\frac{a_{3 n+1}+a_{3 n+2}+a_{3 n+3}}{a_{3 n-2}+a_{3 n-1}+a_{3 n}}=\frac{q^{3}\left(a_{3 n-2}+a_{3 n-1}+a_{3 n}\right)}{a_{3 n-2}+a_{3 n-1}+a_{3 n}}=q^{3}$.