jmc

Since $10,a,b$ is an arithmetic progression, we have $a=\frac{1}{2}(10+b)$. Also, we have $a+ab=2b$, and so $a(1+b)=2b$. Substituting the expression for $a$ gives $(10+b)(1+b)=4b$. Solving this quadratic equation gives the solutions $b=-2$ and $b=-5$. The corresponding values for $a$ can be found by $a=\frac{1}{2}(10+b)$, giving solutions $(4,-2)$ $\left(\frac{5}{2},-5\right),$ for a total of $\boxed{2}$ solutions.