jmc

If we can find such a number whose hundreds digit is 9, then this number will be greater than any with a hundreds digit lower than 9. We want the common ratio to be as small as possible, so that the other digits are as large as possible. If $r$ is the common ratio, then the ones digit is $\frac{9}{r^2}$. One might therefore expect that $r$ has a $3$ in its numerator. We could set $r=3$ for $931$. But $r=\frac{3}{2}$ also works and is actually smaller, giving $\boxed{964}$.

(Note that $r=1$ would not give distinct digits, and $r<1$ would make the ones digit too large.)