$\because$ Let $\mid a_{n} \vdots(n \geqslant 1)$ be a geometric sequence composed of real numbers, and let $S_{n}=a_{1}+\cdots+a_{n}(n \geqslant 1)$, then in the sequence $\left|S_{n}\right|$ $(n \geqslant 1)$,
(A) there is no term that is 0
(B) there are but only a finite number of terms that are 0
(C) either there is no term that is 0, or there are only a finite number of terms that are 0
(D) either there is no term that is 0, or there are infinitely many terms that are 0
1. (I)).

Let the common ratio of the sequence $\vdots a_{n}$ be $q$. If there exists $m$ such that $S_{m}=0$, then $q \neq 1$, hence $S_{n n}=\frac{a_{1}\left(1-q^{m}\right)}{1-q}$. Therefore, $1-q^{m}=0$. Since $q$ is a real number, it must be that $q=-1$ (if $m$ is even), thus $S_{2 n}=0(n \geqslant 1)$, meaning that the sequence $\left\{S_{n}\right\}$ contains infinitely many zeros.