<s>[INST] 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 $m, n, d$. 
 [/INST] Thinking about my previous answer,
it becomes obvious that
any monotone sequence 
of rational numbers
can be embedded
in a linear sequence.
Let the sequence be
$(a_k)|_{k=1}^n$
with
$a_k < a_{k+1}$
and
$a_k = \dfrac{u_k}{v_k}$
with $u_k, v_k$ integers
and $v_k > 0$.
To make the
$a_k$ in the sequence
$a+nd$
with $n$ a positive integer
and $d > 0$,
let $V$ be the
least common multiple
of the $v_k$,
so that
$\dfrac{V}{v_k}
=w_k
$
is an integer.
Set $a = a_1$.
We want there to be
a $d$ and,
for each $k$,
an integer $n_k$
such that
$a+dn_k 
=a_k
=\dfrac{u_k}{v_k}
$.
Multiplying by $V$,
$u_kw_k
=V(a+dn_k)
=Va_1+Vdn_k
$.
If we set
$d = \dfrac1{V}$,
then
$n_k
= u_kw_k-Va_1
$.
Done.
</s>