Why does equal-distance average speed use the harmonic mean?

Average speed is not generally the average of the numbers on the speedometer. By definition, $\bar v=\frac{\text{total distance}}{\text{total time}}.$ Suppose each leg has length $d>0$, with constant speeds $v_1>0$ and $v_2>0$. The total distance is $2d$. The two travel times are $d/v_1$ and $d/v_2$, so $\bar v=\frac{2d}{d/v_1+d/v_2}=\frac{2}{1/v_1+1/v_2}=\frac{2v_1v_2}{v_1+v_2}.$ The distance $d$ cancels: what matters is that the two distances are equal, not their particular length.

Choose two legs of $120$ km. At $60$ km/h, the first takes $120/60=2\text{ hours},$ while at $40$ km/h, the second takes $120/40=3\text{ hours}.$ The trip covers $240$ km in $5$ hours, so $\bar v=240/5=48\text{ km/h}.$ The harmonic-mean formula gives the same result: $\frac{2\cdot60\cdot40}{60+40}=\frac{4800}{100}=48\text{ km/h}.$ The two calculations provide a direct check.

The arithmetic mean of the speeds is $(60+40)/2=50$ km/h, but it gives the two speeds equal time weight. Over equal distances, the $40$ km/h leg lasts $3$ hours, while the $60$ km/h leg lasts only $2$ hours, so the slower speed acts for longer.

The value $50$ km/h would be correct if you travelled for the same amount of time at each speed. For example, one hour at $60$ km/h and one hour at $40$ km/h cover $100$ km in $2$ hours, giving an average of $50$ km/h. Before choosing a mean, ask what is equal: time or distance.

For $n$ equal-distance legs travelled at positive speeds $v_1,\ldots,v_n$, the same reasoning gives the harmonic mean $\bar v=\frac{n}{1/v_1+\cdots+1/v_n}.$ If the distances differ, use the definition directly: $\bar v=\frac{d_1+\cdots+d_n}{d_1/v_1+\cdots+d_n/v_n}.$ Add any stops to total time when the question asks for the average over the entire trip. In every case, the reliable check is total distance divided by total time.