2. For the arithmetic sequence $\left\{a_{n}\right\}$, the sum of the first $n$ terms is $S_{n}$. If $S_{1}, S_{3}, S_{2}$ form a geometric sequence with common ratio $q$, then
$$
q=
$$
$\qquad$ (Li Hong)
2. $\frac{1 \pm \sqrt{37}}{6}$.

According to the problem, we know $S_{3}^{2}=S_{1} \cdot S_{2}$. Also, $S_{1}=a_{1}, S_{2}=a_{1}+a_{2}, S_{3}=a_{1}+a_{2}+a_{3}=3 a_{2}$, so we have
$$
\left(3 a_{2}\right)^{2}=a_{1}\left(a_{1}+a_{2}\right),
$$

Solving this, we get $a_{2}=\frac{1 \pm \sqrt{37}}{18} a_{1}$, therefore
$$
q=\frac{S_{3}}{S_{1}}=\frac{3 a_{2}}{a_{1}}=\frac{1 \pm \sqrt{37}}{6} .
$$