2. For the geometric sequence $\left\{a_{n}\right\}$, the first term $a_{1}=1536$, and the common ratio $q=$ $-\frac{1}{2}$. Let $\prod_{n}$ represent the product of its first $n$ terms. Then the largest $\prod_{n}(n \in N)$ is (. ).
(A) $\Pi$,
(B) $\prod_{11}$
(C) $\prod_{12}$
(D) $\prod_{13}$
2. (C).

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
$$
\prod_{n}=1536^{n} \times\left(-\frac{1}{2}\right)^{\frac{n(n-1)}{2}} \text {. }
$$

It is easy to see that $\Pi_{9}, \Pi_{12}, \Pi_{13}$ are positive, and $\Pi_{10}, \Pi_{11}$ are negative, so we only need to compare $\prod_{2}, \prod_{12}, \prod_{13}$.

Since
$$
\begin{array}{l}
a_{10}=1536 \times\left(-\frac{1}{2}\right)^{9}=-3, \\
a_{11}=a_{10} \cdot\left(-\frac{1}{2}\right)=\frac{3}{2}, \\
a_{12}=a_{11} \cdot\left(-\frac{1}{2}\right)=-\frac{3}{4}, \\
a_{13}=a_{12} \cdot\left(-\frac{1}{2}\right)=\frac{3}{8},
\end{array}
$$

and $a_{10} \cdot a_{11} \cdot a_{12}=(-3)\left(\frac{3}{2}\right)\left(-\frac{3}{4}\right)=\frac{27}{8}>1$,
$$
\therefore \prod_{12}=a_{10} \cdot a_{11} \cdot a_{12} \cdot \prod_{0} \text {. }
$$

Also, $0<a_{13}<1$ and $\prod_{13}=a_{13} \cdot \prod_{12}$,
$$
\therefore \prod_{13}<\prod_{12} \text {. }
$$
(In fact, when $n \geqslant 13$, $\left|a_{n}\right| \leqslant a_{13}=\frac{3}{8} . \prod_{12}$ is the maximum value in $\Pi$. )