jmc

It follows that the common ratio of the geometric sequence is equal to $\frac{5}{3}$. Thus, $D=\frac{5}{3}\cdot C=\frac{5}{3}\cdot \frac{5}{3}\cdot B=\frac{25B}{9}$. Since $D$ is an integer, it follows that $B$ must be divisible by $9$. The lowest possible value of $B$ is $B=9$, which yields a value of $C=15$ and $D=25$. The common difference between the first three terms is thus $15-9=6$, so it follows that $A=B-6=3$. The sum $A+B+C+D=3+9+15+25=\boxed{52}$.

If $B=9k$ for $k>1$, then $C=\frac{5}{3}\cdot B=15k$ and $D=\frac{5}{3}\cdot C=25k$. Then, $A+B+C+D>B+C+D\ge 49k\ge 98$, so it follows that $52$ is indeed the smallest possible value of $A+B+C+D$.