$13 \cdot 67$ is a geometric sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=1536$, and the common ratio $q=-\frac{1}{2}$. Let $\Pi_{n}$ represent the product of its first $n$ terms, then the largest $\Pi_{n}(n \in N)$ is.
(A) $\mathrm{II}_{9}$.
(B) $\mathrm{II}_{11}$.
(C) $\mathrm{Il}_{12}$.
(D) $\mathrm{II}_{13}$.
(China High School Mathematics League, 1996)
[Solution] The general term formula of the geometric sequence $\left\{a_{n}\right\}$ is
$$
a_{n}=1536 \times\left(-\frac{1}{2}\right)^{n-1} \text {. }
$$

The product of the first $n$ terms is
$$
I I_{n}=1536^{n} \times\left(\frac{1}{2}\right)^{\frac{n(n-1)}{2}} .
$$

It is easy to see that $\Pi_{9}, \Pi_{12}, \mathrm{II}_{13}$ are positive, and $\Pi_{10}, \mathrm{I}_{11}$ are negative, so we only need to compare $\mathrm{II}_{9}, \mathrm{I}_{12}$, $\mathrm{II}_{13}$. Note that
$$
\begin{array}{l}
a_{10}=1536 \times\left(-\frac{1}{2}\right)^{9}=-3, a_{11}=a_{10} \cdot\left(-\frac{1}{2}\right)=\frac{3}{2}, \\
a_{12}=a_{11} \cdot\left(-\frac{1}{2}\right)=-\frac{3}{4}, a_{13}=a_{12} \cdot\left(-\frac{1}{2}\right)=\frac{3}{8},
\end{array}
$$

and $a_{10} \cdot a_{11} \cdot a_{12}=(-3)\left(\frac{3}{2}\right)\left(-\frac{3}{4}\right)=\frac{27}{8}>1$.
$$
\therefore \quad \mathrm{I}_{12}=a_{10} \cdot a_{11} \cdot a_{12} \cdot \mathrm{II}_{9} \text {. }
$$

Also, $0<a_{13}<1$ and $\Pi_{13}=a_{13} \cdot \mathrm{II}_{12}$,
$\therefore \quad \Pi_{13}<\Pi_{12}$.
(In fact, for $n \geqslant 13$, $\left|a_{n}\right| \leqslant a_{13}=\frac{3}{8} . \Pi_{12}$ is the maximum value among $\Pi_{n}$.) Therefore, the answer is $(C)$.