However, one may avoid the use of the harmonic mean for the case of "weighting by distance". Pose the problem as finding "slowness" of the trip where "slowness" (in hours per kilometre) is the inverse of speed. When trip slowness is found, invert it so as to find the "true" average trip speed. For each trip segment i, the slowness si = 1/speedi. Then take the weighted arithmetic mean of the si's weighted by their respective distances (optionally with the weights normalized so they sum to 1 by dividing them by trip length). This gives the true average slowness (in time per kilometre). It turns out that this procedure, which can be done with no knowledge of the harmonic mean, amounts to the same mathematical operations as one would use in solving this problem by using the harmonic mean. Thus it illustrates why the harmonic mean works in this case.
However, one may avoid the use of the harmonic mean for the case of "weighting by distance". Pose the problem as finding "slowness" of the trip where "slowness" (in hours per kilometre) is the inverse of speed. When trip slowness is found, invert it so as to find the "true" average trip speed. For each trip segment i, the slowness si = 1/speedi. Then take the weighted arithmetic mean of the si's weighted by their respective d.istances (optionally with the weights normalized so they sum to 1 by dividing them by trip length) This gives the true average slowness (in time per kilometre). It turns out that this procedure, which can be done with no knowledge of the harmonic mean, amounts to the same mathematical operations as one would use in solving this problem by using the harmonic mean. Thus it illustrates why the harmonic mean works in this case.