These numbers can be written as $\frac{k}{3}-d, \frac{k}{3}, \frac{k}{3}+d$, so we must have that $\frac{k}{3}-d, \frac{k}{3}, \frac{k}{2}+d$ is a geometric sequence. Hence, $(\frac{k}{3})^2 = (\frac{k}{3}-d)(\frac{k}{2}+d)$ $\frac{k^2}{9} = \frac{k^2}{6} - \frac{k}{6}d - d^2$ $d^2 + \frac{k}{6}d - \frac{k^2}{18}=0$ $d = \frac{-\frac{k}{6} \pm \sqrt{(\frac{k}{6})^2 + 4(\frac{k^2}{18})}}{2}$ $d \in \{ \frac{k}{6}, -\frac{k}{3}\}$ Therefore, the numbers can be $\boxed{(\frac{k}{6},\frac{k}{3},\frac{k}{2})}$ or $\boxed{(\frac{2k}{3},\frac{k}{3},0)}$.