Three numbers $a,b,c$, none zero, form an arithmetic progression. Increasing $a$ by $1$ or increasing $c$ by $2$ results 
in a geometric progression. Then $b$ equals:
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 8$
Let $d$ be the common difference of the [arithmetic sequence](https://artofproblemsolving.com/wiki/index.php/Arithmetic_sequence), so $a = b-d$ and $c = b+d$.
Since increasing $a$ by $1$ or $c$ by $2$ results in a [geometric sequence](https://artofproblemsolving.com/wiki/index.php/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](https://artofproblemsolving.com/wiki/index.php/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 $\boxed{\textbf{(C)}}$.