1. Let the elements of set $A$ be three real numbers, and by appropriately adjusting the order of these three numbers, they can form both a geometric sequence and an arithmetic sequence. Then the common ratio of the geometric sequence is $\square$ $\qquad$
1. $-\frac{1}{2}$ or -2.
From the problem, let the three numbers be $a, ra, r^{2}a (r>0, r \neq 1)$.
Since $a, b, c$ can form an arithmetic sequence in some order, we have,
$a+r^{2}a=2ra$ or $a+ra=2r^{2}a$ or $r^{2}a+ra=2a$.
Solving these, we get $r=1$ or $-\frac{1}{2}$ or -2.
Thus, the required common ratio is $-\frac{1}{2}$ or -2.