2. (1996 National High School Competition Question) In the geometric sequence $\left\{a_{n}\right\}$, $a_{1}=1536$, common ratio $q=-\frac{1}{2}$, use $\pi_{n}$ to denote the product of its first $n$ terms, then the largest $\pi_{n}\left(n \in \mathbf{N}_{+}\right)$ is ( ).
A. $\pi_{9}$
B. $\pi_{10}$
C. $\pi_{12}$
D. $\pi_{13}$
2.C. Given $a_{n}=1536 \times\left(-\frac{1}{2}\right)^{n-1}=-3 \times\left(-\frac{1}{2}\right)^{n-10}$, so $\pi_{n}=(-3)^{n} \cdot$ $\left(-\frac{1}{2}\right)^{(-9)+(-8)+\cdots+(n-10)}=(-3)^{n}\left(-\frac{1}{2}\right)^{\frac{1}{2} n(n-19)}$. It is easy to see that $\pi_{9}, \pi_{12}, \pi_{13}$ are positive, $\pi_{11}$ is negative, and $\pi_{12}=a_{12} \cdot a_{11} \cdot a_{10} \cdot \pi_{9}=\left(-\frac{3}{4}\right)\left(\frac{3}{2}\right)(-3) \pi_{9}>\pi_{9}, \pi_{13}=a_{13} \cdot \pi_{12}=\frac{3}{8} \pi_{12}<\pi_{12}$, so $\pi_{12}$ is the largest, hence the answer is C.