If there are any multitude whatsoever of numbers in continued proportion (which are) the least of those (numbers) having the same ratio as them then the outermost of them are prime to one another. * Let $A$, $B$, $C$, $D$ be any multitude whatsoever of numbers in continued proportion (which are) the least of those (numbers) having the same ratio as them. * I say that the outermost of them, $A$ and $D$, are prime to one another.
If $\frac AB=\frac BC=\frac CD$ are a geometric progression and the least of these numbers, then $A$ and $D$ are co-prime.
Proofs: 1