The common value $r$ of the ratios can be only a divisor of $24$, different from $1$; these divisors are $2$, $3$, $4$, $6$, $8$, $12$ and $24$.
If $r = 2$, then $2 = \frac{24}{12} = \frac{8}{4} = \frac{6}{3} = \frac{2}{1}$;
if $r = 3$, then $3 = \frac{24}{8} = \frac{12}{4} = \frac{6}{2} = \frac{3}{1}$;
if $r = 4$, then $4 = \frac{24}{6} = \frac{12}{3} = \frac{8}{2} = \frac{4}{1}$.
There are no other sequences for $r = 2$, $r = 3$ or $r = 4$, because if we order decreasingly the divisors and use them one by one, we have to put at the numerator the largest divisor $d$ still unused, and at the denominator $\frac{d}{r}$, getting the sequences from above.
There is no sequence for $r \ge 8$, respectively $r = 6$, because none of the equalities $\frac{d}{6} = r$ and $\frac{6}{d} = r$, respectively $\frac{d}{3} = 6$ and $\frac{3}{d} = 6$, can be fulfilled.