(related to Proposition: 8.02: Construction of Geometric Progression in Lowest Terms)
So it is clear, from this, that if three numbers in continued proportion are the least of those (numbers) having the same ratio as them then the outermost of them are square, and, if four (numbers), cube.
Obviously, in the geometric progression $A^nq^i$, $i=0,\ldots,n$ with $q=\frac AB$ ($A$ and $B$ being co-prime), the first and the last numbers are the powers $A^n$ and $B^n.$
Proofs: 1