Your average speed must be twice that of the first lap Edit: the problem would've gone something like (I saw it years ago): "find the speed you must run on a second lap around the same track in order to double your average speed for the 2 laps combined."
I have 2 approaches to this problem that give different answers, in both cases we let
$v_1=$ velocity of first lap, $v_2=$ be the velocity of 2nd,
$t_1=$ time of 1st lap, $t_2=$ time of 2nd, and
$d=$ distance of 1 lap, $v_a=$ average speed.
First approach: $$v_a=\frac{v_1+v_2}{2}=\frac{\frac{d}{t_1}+\frac{d}{t_2}}{2}$$
we want this to equal $$2v_1=\frac{2d}{t_1}$$
solving: $$\frac{\frac{d}{t_1}+\frac{d}{t_2}}{2}=\frac{2d}{t_1}$$
$$\frac{d}{t_1}+\frac{d}{t_2}=\frac{4d}{t_1}$$
$$\frac{d}{t_2}=\frac{3d}{t_1}$$
$$3dt_2=dt_1$$
$$t_2=\frac{1}{3}t_1$$
Second approach: $$v_a=\frac{2d}{t_1+t_2}$$
we want this to equal $$2v_1=\frac{2d}{t_1}$$
solving: $$\frac{2d}{t_1+t_2}=\frac{2d}{t_1}$$
$$t_1+t_2=t_1$$
$$t_2=0$$
Obviously, it's not possible for $t_2$ to be $0$, that would mean infinite speed. How come my first approach had a different conclusion?
 A: By definition, the average speed over a path $\gamma_n$ is $$v_n = \frac{d_n}{t_n}$$ where $d_n$ is the length of the path and $t_n$ is the amount of time taken to travel the length of the path. (Subscript $n$ included here so that we can talk about different paths in the same equation.)
Another way to put it, the average speed is $v_n$ such that
$  d_n = v_n t_n. $
A path of two laps has length $d_0 = d_1 + d_2$ where $d_1$ is the length of the first lap and $d_2$ is the length of the second lap.
Of course $d_1 = d_2$ it both laps are around the same track.
So if it takes time $t_1$ to travel the first lap,
and it takes time $t_2$ to travel the second lap, then if the two laps together are traveled at an average speed of $v_0$ in time $t_0$, we have
\begin{align}
d_0 &= v_0 t_0, \\
d_1 + d_2 &= v_0 (t_1 + t_2), \\
v_1 t_1 + v_2 t_2 &= v_0 (t_1 + t_2).
\end{align}
Solving for the average speed $v_0,$
$$ v_0 =  \frac{v_1 t_1 + v_2 t_2}{t_1 + t_2}. $$
This is a weighted average. The weights are $t_1$ and $t_2.$
The "heavier" $t_1$ is in comparison to $t_2,$ the more the average $v_0$
will look like $v_1,$ and vice versa.

Aside:
Imagine you have two jobs, the first of which pays $\$10$ per hour and the second of which pays $\$50$ per hour. Your two employers agree to give you $8$ hours total work in one day; the first employer gives you $x$ hours ($0 \leq x \leq 8$) and the other employer gives you $8 - x$ hours.
If $x = 1$ (you work $1$ hour in the first job and $7$ hours in the second), you make $\$360$ that day, which averages to $\$45$/hour for the $8$ hours you worked. But if $x = 7$ you make $\$120$, which averages to $\$15$/hour.
The average rate of pay here is a weighted average, just like the average speed, and clearly the relative sizes of the weights can make a big difference.

If you start with two different speeds $v_1$ and $v_2,$ the only way you can get a weighted average like $\dfrac{v_1 t_1 + v_2 t_2}{t_1 + t_2}$ to come out to $\frac12(v_1 + v_2)$ is to make the two weights equal.
That is, you set $t_2 = t_1$ so that
$$
v_0 = \frac{v_1 t_1 + v_2 t_2}{t_1 + t_2}
= \frac{v_1 t_1 + v_2 t_1}{t_1 + t_1}
= \frac{(v_1 + v_2) t_1}{2t_1}
= \frac{v_1 + v_2}{2}.
$$
Your first approach, by starting with $v_0 = \dfrac{v_1 + v_2}{2},$
is implicitly assuming from the start that $t_1 = t_2.$
And that's a very bad assumption, because while you were guaranteed that the two laps would have the same length, you were not guaranteed that they would take the same time to traverse.
In fact, if they did take the same time to traverse, you'd have the same speed on each of them, and the average would be $v_1,$ not $2v_1$. That doesn't work at all!
A: The average speed of the two-lap journey, i.e., the runner's average speed over the two laps, $$\frac{2d}{t_1+t_2}\\=\frac{t_1}{t_1+t_2}\left(\frac{d}{t_1}\right)+\frac{t_2}{t_1+t_2}\left(\frac{d}{t_2}\right)\\=\frac{t_1}{t_1+t_2}\,v_1+\frac{t_2}{t_1+t_2}\,v_2,$$ is the weighted average of the two speeds (their weighted arithmetic mean), proportionally accounting for the lap durations.

First approach: $$v_a=\frac{v_1+v_2}{2}$$

You have instead taken the average of the two lap speeds (their ordinary arithmetic mean).
This equals the average speed over the two laps if and only if both laps have the same duration, i.e., $$t_1=t_2.$$
P.S. Your misinterpretation is akin to falsely claiming that the overall performance on an exam comprising several papers and in which every mark is worth the same value (LHS) equals the simple average of the individual scaled scores (RHS). But of course, $$\frac{55+15}{60+100}\ne\frac12\left(\frac{55}{60}+\frac{15}{100}\right).$$
A: EDIT:
I changed the previous answer completely as it was incorrect...
The second approach still stands correct than the first approach.
However, $t_2 = 0$ is incorrect. This is due to the fact that the first average (first lap) cannot be compared in anyway to the final average (both laps). Plot the first average and final average... they require 2 different set of cartesian graphs because the axes of the first average plot are not applicable for the final average.
Simply saying $2v_1 = v_a$ is incorrect as $d_1 ≠ d_1 + d_1$, $d_1 ≠ 2d_1$, which brings the fact that $k * \frac{d_1}{t_1} ≠ \frac{2d_1}{t_1 + t_2}$, and finally ... $k * v_1 ≠ v_a$, where $k = 2$. You are comparing averages over $2$ different events/domains (might be the difference in distance covered) and you cannot compare them unless you find a way to relate those domains first.
