Find if terms are terms of the same arithmetic progression 
Is it possible that numbers $\frac{1}{2}, \frac{1}{3}, \frac{1}{5}$ are (not necessarily adjacent) terms of the same arithmetic progression?
  Hint: Yes. Try $\frac{1}{30}$ as a difference.

I  was going back and forth how they found out that difference.
My idea was since we have an arithmetic sequence defined as $a, a+d, a+2d,...$ I thought I could solve for the difference $d=\frac{1}{30}$. Since:
$$\frac{1}{3} = \frac{1}{2}+nd$$
And
$$\frac{1}{5} = \frac{1}{3}+md$$
Then $nd = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6}$ and $md = \frac{1}{5} - \frac{1}{3} = -\frac{2}{15}$
Since it is also part of the same sequence we can find: 
$$nd + md = -\frac{1}{6} -\frac{2}{15} = -\frac{3}{10}$$
Now I'm stuck since I can't see how this brings me any closer to find $m, n, d$. 

The key is to find the greatest common divisor of the two differences $\frac{1}{6}$ and $\frac{2}{15}$. How? Exactly like we would for integers.
If we divide $\frac{1}{6}$ by $\frac{2}{15}$, we get $1$ with a remainder of $\frac{1}{6}-\cdot\frac{2}{15} = \frac{5}{30}-\frac{4}{30}=\frac{1}{30}$.
Next, we have the pair $\frac{2}{15}$ and $\frac{1}{30}$. Dividing the former by the latter, we get $4$ with a remainder of zero.
That remainder of zero marks an end to the algorithm. The greatest common divisor is the last number we had before we reached zero - in this case, $\frac1{30}$.
So that's $d$. To find $m$ and $n$, we divide: $m=\frac 16/ \frac{1}{30} = 5$ and $n=\frac{2}{15}/ \frac{1}{30} = 4$.
The Euclidean algorithm works to find the greatest common divisor of any two rational numbers, not just any two integers as usually presented.
And yes, that means that if we write down three different rational numbers, there's always some arithmetic progression that has all three of them as terms. The common difference can be the greatest common divisor, or any further divisor of that number.
