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
(C)
I. Multiple Choice Question
1.【Analysis and Solution】From the problem, $a_{n}=a_{1} q^{n-1}$, we have
$a_{n}=a_{1} q^{n-1} \frac{b_{n+1}}{b_{n}}=\frac{a_{3 n+1}+a_{3 n+2}+a_{3 n+3}}{a_{3 n-2}+a_{3 n-1}+a_{3 n}}=\frac{a_{1} q^{3 n}+a_{1} q^{3 n+1}+a_{1} q^{3 n+2}}{a_{1} q^{3 n-3}+a_{1} q^{3 n-2}+a_{1} q^{3 n-1}}=q^{3}, \therefore\left\{b_{n}\right\}$ is a geometric sequence with a common ratio of $q^{3}$.