2014 AMC 12A Problem 14

Let $d=b-a\gt0,$ so $b=a+d$ and $c=a+2d.$ Since $a,c,b$ is geometric, $$\begin{gathered} \dfrac{c}{a}=\dfrac{b}{c}\\ \implies(a+2d)^2=a(a+d), \end{gathered}$$ which simplifies to $3ad+4d^2=0,$ so $3a+4d=0.$ Then $a=-4k$ and $d=3k$ for a positive integer $k,$ giving $c=a+2d=2k.$ The smallest value is $c=2$ (with $a=-4,$ $b=-1,$ $c=2$). Thus, the correct answer is C.