If between two numbers there fall (some) numbers in continued proportion then, as many numbers as fall in between them in continued proportion, so many (numbers) will also fall in between (any two numbers) having the same ratio [as them] in continued proportion. * For let the numbers, $C$ and $D$, fall between two numbers, $A$ and $B$, in continued proportion, and let it have been contrived (that) as $A$ (is) to $B$, so $E$ (is) to $F$. * I say that, as many numbers as have fallen in between $A$ and $B$ in continued proportion, so many (numbers) will also fall in between $E$ and $F$ in continued proportion.
Given some equal continued proportions e.g. $\frac AB=\frac EF=\frac GL$ (which we can also write $B=Aq^n$ , $F=Eq^n$, $L=Gq^n$ for some $q > 0$ and some positive integer $n\ge 3$) there are $n-1$ numbers falling between each of them, e.g, for $n=3$:
Proofs: 1
