jmc

Let $d$ be the common difference, so $a=b-d$ and $c=b+d$. We can assume that $d$ is positive. (In particular, $d$ can't be 0, because the triangle is not equilateral.) Then the perimeter of the triangle is $a+b+c=(b-d)+b+(b+d)=3b=60$, so $b=20$. Hence, the sides of the triangle are $20-d$, 20, and $20+d$.

These sides must satisfy the triangle inequality, which gives us $$(20-d)+20>20+d.$$ Solving for $d$, we find $2d<20$, or $d<10$. Therefore, the possible values of $d$ are 1, 2, $\dots$, 9, which gives us $\boxed{9}$ possible triangles.