10.51 Given a geometric sequence $\left\{a_{n}\right\}$ with a common ratio of $q(q \neq 1)$, set $b_{1}=a_{1}+a_{2}+$ $a_{3}, b_{2}=a_{4}+a_{5}+a_{6}, \cdots, b_{n}=a_{3 n-2}+a_{3 n-1}+a_{3 n}, \cdots$, 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.
(China High School Mathematics League, 1999)
[Solution] Given $a_{n}=a_{1} q^{n-1}$, then $\frac{b_{n}+1}{b_{n}}=\frac{a_{3 n+1}+a_{3 n+2}+a_{3 n+3}}{a_{3 n-2}+a_{3 n-1}+a_{3 n}}=\frac{a_{1} q^{3 n}+a_{1} q^{3 n+1}+a_{1} q^{3 n+2}}{a_{1} q^{3 n-3}+a_{1} q^{3 n-2}+a_{1} q^{3 n-1}}$ $=\frac{a_{1} q^{3 n}\left(1+q+q^{2}\right)}{a_{1} q^{3 n-3}\left(1+q+q^{2}\right)}=q^{3}$,

Therefore, $\left\{b_{n}\right\}$ is a geometric sequence with a common ratio of $q^{3}$. Hence, the answer is $(C)$.