10. For the geometric sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=1536$, common ratio $q=-\frac{1}{2}$, $\Pi_{n}$ represents the product of its first $n$ terms $\left(n \in \mathbf{N}^{*}\right)$, then the value of $n$ that maximizes $\mathrm{I}_{n}$ is $\qquad$.
10. 12 The general term formula of the geometric sequence $\left\{a_{n}\right\}$ is $a_{n}=1536 \cdot\left(-\frac{1}{2}\right)^{n-1}$, to make the product of the first $n$ terms $\Pi_{n}$ maximum, $n$ must be an odd number or a multiple of 4. Let $\left|a_{n}\right| \geqslant 1$ get $1536 \cdot\left(\frac{1}{2}\right)^{n-1} \geqslant 1$, so $n$ is an odd number and $2^{n-1} \leqslant 1536$. By $2^{11}=2048$, hence $n=1,3,4,5,7,8,9,11$, so when $n=11$ $\Pi_{n}$ is maximum.