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 $m, n, d$. 

My usual naive playing.
Suppose the values are
$u, v, w$
and we want to find
$a, d, m, n$
such that
$u = a,
v = a+md,
w = a+nd$.
This is 3 equations in
4 unknowns,
so we expect this to be
underdetermined
and probably have
a one-parameter set of solutions.
Then
$md = v-u,
nd = w-u$
so
$\dfrac{n}{m}
=\dfrac{w-u}{v-u}
=r
$.
Then
$n = mr$
and
$w-v
=(m-n)d
=(m-mr)d
=md(1-r)
$.
For any $d$,
$m
=\dfrac{w-v}{d(1-r)}
$,
$n = mr$,
$a = u$.
For your case,
$u, v, w 
=\frac12, \frac13, \frac15
$,
$r
=\dfrac{\frac15-\frac12}{\frac13-\frac12}
=\dfrac{6-15}{10-15}
=\dfrac95
$,
$m 
= \dfrac{\frac15-\frac13}{-d\frac45}
= \dfrac{5\cdot-\frac{2}{15}}{-4d}
= -\dfrac{1}{6d}
$,
$n
=mr
=-\dfrac{1}{6d}\dfrac95
=-\dfrac{9}{30d}
=-\dfrac{3}{10d}
$,
$a
=\frac12
$.
Check:
$a+md=
=\dfrac12-\dfrac{1}{6d}d
=\dfrac12-\dfrac16
=\dfrac13
$
and
$a+nd
=\dfrac12-\dfrac{3}{10d}d
=\dfrac12-\dfrac{3}{10}
=\dfrac15
$.
