Deduce a formula in terms of $V_1$ and $V_2$ The average speed for a journey is the distance covered divided by the time taken.
If a journey is completed by travelling at speed $V_1$ for half the 
distance and at speed $V_2$ from the second half. Find the average speed $V_b$ for the journey in terms of $V_1$ and $V_2$.
I.e. Show that $V_b=\dfrac{2V_1V_2}{V_1+V_2}$
 A: Distance = Velocity * Time.
Let $D$ be the total distance travelled, and this takes time $T=t_1+t_2$, so $D=V_b T$, so $T=\dfrac D{V_b}$.
But $\dfrac D2=V_1 t_1=V_2 t_2$, so that $t_1=\dfrac D{2V_1}$ and $t_2=\dfrac D{2V_2}$.
Therefore $\dfrac D{V_b}=\dfrac D{2V_1}+\dfrac D{2V_2}$.
So $\dfrac 1{V_b}=\dfrac 1{2V_1}+\dfrac 1{2V_2}$,
$\dfrac 1{V_b}=\dfrac {V_1+V_2}{2V_1V_2}$,
$V_b=\dfrac {2V_1V_2}{V_1+V_2}$,
A: let d be the total distance covered, and $t_1$ and  $t_2$ the time needed to cover the first half of the distance and the second half respectively, then:
$t_1 = \frac{d}{2v_1}$ and $t_2 = \frac{d}{2v_2}$
so $T = t_1 + t_2$  and $v_b = \frac{d}{T}$
you should get the result mentioned.
A: $$ \frac d v = \frac{d}{2v^\prime} + \frac{d}{2v^{\prime\prime}} $$
$$ \frac d v = d\left( \frac{1}{2v^\prime} + \frac{1}{2v^{\prime\prime}} \right) $$
$$ \frac 1 v = \left( \frac{1}{2v^\prime} + \frac{1}{2v^{\prime\prime}} \right) $$
$$ \frac 1 v = \frac{v^\prime + v^{\prime\prime}}{2v^\prime v^{\prime\prime}} $$
$$ v = \frac{2v^\prime v^{\prime\prime}}{v^\prime + v^{\prime\prime}} $$
A: @JonMarkPerry 's accepted answer is right.
It's worth pointing out that this line
$$
\frac{1}{V_b} =
\frac{1}{2V_1} + \frac{1}{2V_2} =
\frac{1}{2}\left( \frac{1}{V_1} + \frac{1}{V_2}\right)
$$
shows that the answer tells you that the reciprocal of the average speed is the ordinary average of the reciprocals of the speeds for the journey halves. That makes sense if you do the calculations using the units hours/mile rather than miles per hour.
The fact that you can't just average two speeds makes for several nice puzzles. If you go from here to there at 60 miles per hour and back at 30 miles per hour your average speed isn't 45 miles per hour. In fact, if you teleported back instanteously at infinitely many miles per hour your average speed wouldn't be the average of $60$ and infinity, it would be $120$ miles per hour. That fits the formula: $1/120$ is (in an appropriate sense) the average of $1/60$ and $1/\infty = 0$.
This computation has a name: it's the harmonic mean. At that wikipedia page you can read

Typically, it is appropriate for situations when the average of rates
  is desired.

