The prime factorization of $720$ is $2^4 \cdot 3^2 \cdot 5$. Let $r = \frac{m}{n}$ be the common ratio of the geometric sequence, where $m$ and $n$ are relatively prime positive integers. If $n$ had any prime factor greater than $5$, then $b = 720r$ would not be an integer. Analogously, if $m$ had any prime factor greater than $5$, then $a = \frac{720}{r}$ would not be an integer. It follows that $r = 2^i \cdot 3^j \cdot 5^k$, where $i, j$, and $k$ are (not necessarily positive) integers. Furthermore, $|i| \le 4$, $|j| \le 2$, and $|k| \le 1$.

To minimize the value of $b$, it suffices to minimize the value of $r > 1$. Taking $r = \frac{16}{15} = 2^4 \cdot 3^{-1} \cdot 5^{-1}$ yields the sequence $675$, $720$, $768$. To check that no lesser values of $r$ exist, first observe that $\frac{17}{16}$ is not a possible value for $r$, so both $n$ and $m$ are greater than $17$. This means that $m$ and $n$ must borrow at least two prime factors each from $720$, but $720$ has only three distinct prime factors, so this is impossible. It follows that the least possible value of $b$ is $768$, and the requested sum of digits is $7 + 6 + 8 = 21$.