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 $m, n, d$. 
 A: 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.
A: Suggestion: Following the given hint, try writing each fraction with a common denominator of $30$:
$\frac12=\frac{15}{30}$
$\frac13=\frac{10}{30}$
$\frac15 = \frac{6}{30}$
Does that help?
A: 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
$.
A: Let the terms be $a_1=\frac12, a_m=\frac13,a_n=\frac15$. Then:
$$d=\frac{a_m-a_1}{m-1}=\frac{a_n-a_1}{n-1} \Rightarrow \\
d=\frac{\frac13-\frac12}{m-1}=\frac{\frac15-\frac12}{n-1} \Rightarrow \\
-\frac16(n-1)=-\frac3{10}(m-1) \Rightarrow \\
n=\frac{9}{5}(m-1)$$
As $m,n\in \mathbb N_+$ and $n>m$, we get $m=5k+1,n=9k,d=-\frac1{30k},k\in\mathbb N_+$:
$$\begin{array}{c|c|c|c|c}
m&6&11&16&\cdots\\
\hline
n&9&18&27&\cdots\\
\hline
d&-\frac1{30}&-\frac1{60}&-\frac1{90}&\cdots
\end{array}$$
Note: It was considered a decreasing AP. It can also be considered an increasing AP.
A: 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.
