smc

Let $d$ be the common difference of the arithmetic sequence, so $a=b-d$ and $c=b+d$. Since increasing $a$ by $1$ or $c$ by $2$ results in a geometric sequence, $$\frac{b}{b-d+1}=\frac{b+d}{b}$$ $$\frac{b}{b-d}=\frac{b+d+2}{b}$$ Cross-multiply in both equations to get a system of equations. $$b^2=b^2-d^2+b+d$$ $$b^2=b^2-d^2+2b-2d$$ Rearranging terms results in $$d^2=b+d$$ $$d^2=2b-2d$$ Substitute and solve for $d$. $$b+d=2b-2d$$ $$d=\frac{b}{3}$$ Finally, substitute $d$ and solve for $b$. Since $b\ne 0$, dividing by $b$ is allowed. $$(\frac{b}{3}{)}^2=b+\frac{b}{3}$$ $$\frac{b^2}{9}=\frac{4b}{3}$$ $$\frac{b}{9}=\frac{4}{3}$$ $$b=12$$ The answer is $\boxed{12}$.