Let $d$ be the distance.
Let $t_1$ be the time it takes to cover this distance at the rate $r_1$.
Let $t_2$ be the time it takes to cover this distance at the rate $r_2$.
Define $t_a$ to be the A.M. of the two times, that is, $t_a := (t_1+t_2)/2$.
Define $r_a$ to be the H.M. of the two rates, that is, $r_a := 2/((1/r_1)+(1/r_2))$.
Theorem. If one wants to cover distance $d$ in $t_a$ time, one should go at
the rate $r_a$. Conversely, if one covers the distance at the rate $r_a$, one
will use up $t_a$ time.
Proof. It's enough to show that $t_a\cdot r_a = d$. We have
\begin{eqnarray*}
t_a\cdot r_a
&=& \dfrac{t_1+t_2}{2}\cdot \dfrac{2}{\dfrac{1}{r_1} + \dfrac{1}{r_2}} \\
&=& \dfrac{t_1+t_2}{2}\cdot \dfrac{2}{\dfrac{t_1}{d} + \dfrac{t_2}{d}} \\
&=& \dfrac{t_1+t_2}{2}\cdot \dfrac{2}{\dfrac{t_1+t_2}{d}} \\
&=& \dfrac{t_1+t_2}{2}\cdot \dfrac{2d}{t_1+t_2} \\
&=& d.
\end{eqnarray*}
The theorem extends to any positive integer $n$.
Let $d$ be the distance.
Let $t_1$ be the time it takes to cover this distance at the rate $r_1$.
Let $t_2$ be the time it takes to cover this distance at the rate $r_2$.
...
Let $t_n$ be the time it takes to cover this distance at the rate $r_n$.
Define $t_a$ to be the A.M. of the $n$ times.
Define $r_a$ to be the H.M. of the $n$ rates.
Theorem. If one wants to cover distance $d$ in $t_a$ time, one should go at
the rate $r_a$. Conversely, if one covers the distance at the rate $r_a$, one
will use up $t_a$ time.
With this theorem, one can pose some seemingly more complicated questions like this one
Question: Dr. Worm leaves his house at exactly 7:20 a.m. every morning.
When he averages 30 miles per hour, he arrives at his workplace ten minutes late.
When he averages 40 miles per hour, he arrives at his workplace five minutes late.
When he averages 60 miles per hour, he arrives fifteen minutes early.
What speed should Dr. Worm average to arrive at his workplace precisely on time?
