Let the inradius of $\triangle ABC$ be $r$. According to the Law of Sines, we have $b \sin A = a \sin B$. Also, $a \cos A = b \cos B$, so $\sin 2A = \sin 2B$. Since $A \neq B$, then $A + B = 90^{\circ}$. Therefore, $\triangle ABC$ is a right triangle, $\angle C = 90^{\circ}$, and $a^2 + b^2 = c^2$.

Given that $c = 10$, and $a, b, c$ form an arithmetic sequence, we have $a = 6, b = 8, (8 - r) + (6 - r) = 10$, solving for $r$ gives $r = 2$.
* In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. If $a, b, c$ form an arithmetic sequence, and $c=10, a \cos A=b \cos B, A \neq B$, then the inradius of $\triangle A B C$ is equal to