For any multitude whatsoever of given ratios, (expressed) in the least numbers, to find the least numbers in continued proportion in these given ratio. * Let the given ratio, (expressed) in the least numbers, be the (ratios) of $A$ to $B$, and of $C$ to $D$, and, further, of $E$ to $F$. * So it is required to find the least numbers in continued proportion in the ratio of $A$ to $B$, and of $C$ to $B$, and, further, of $E$ to $F$.
Let the proportions $\frac AB,$ $\frac CD,$ and $\frac EF$ be given. It is to find least numbers $N,O,M,P$ with $\frac AB=\frac NO,$ $\frac CD=\frac OM,$ and $\frac EF=\frac MP.$
To continue the continued proportion $$\cfrac{a}{\cfrac{\vdots}{b}}$$ with the ratio $\frac cd.$ * Multiply each of the numbers in the proportion by $\frac{\operatorname{lcm}(b,c)}{b}$ * Continue the proportion with the number $\frac{d\cdot \operatorname{lcm}(b,c)}{c}.$
To take $\cfrac{4}{5}$ and continue it by $\cfrac 23.$ * Factor to multiply all numbers with: $\frac{\operatorname{lcm}(5,2)}{5}=\frac{10}{5}=2.$ * It follows $\cfrac{4\cdot 2}{5\cdot 2}=\cfrac{8}{10}.$ * New number to continue with: $\frac{3\cdot \operatorname{lcm}(5,2)}{2}=\frac{3\cdot 10}{2}=15.$ * New continued proportion: $$\cfrac{8}{\cfrac{10}{15}}.$$
To take $\cfrac{8}{\cfrac{10}{15}}$ and continue it by $\cfrac 47.$ * Factor to multiply all numbers with: $\frac{\operatorname{lcm}(15,4)}{15}=\frac{60}{15}=4.$ * It follows $$\cfrac{8\cdot 4}{\cfrac{10\cdot 4}{15\cdot 4}}=\cfrac{32}{\cfrac{40}{60}}.$$ * New number to continue with: $\frac{7\cdot \operatorname{lcm}(15,4)}{4}=\frac{7\cdot 60}{4}=105.$ * New continued proportion: $$\cfrac{32}{\cfrac{40}{\cfrac{60}{105}}}.$$
Proofs: 1