If any multitude whatsoever of numbers is in continued proportion, (starting) from a unit, and the (number) after the unit is square, then all the remaining (numbers) will also be square. And if the (number) after the unit is cube, then all the remaining (numbers) will also be cube. * Let any multitude whatsoever of numbers, $A$, $B$, $C$, $D$, $E$, $F$, be in continued proportion, (starting) from a unit. * And let the (number) after the unit, $A$, be square. * I say that all the remaining (numbers) will also be square. * And so let $A$ be cube. * I say that all the remaining (numbers) are also cube.
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Proofs: 1