Showing that a progression is arithmetic this one is from Gelfand's book "Algebra".
Problem 204. Is it possible that numbers $\frac{1}{2}$, $\frac{1}{3}$, and $\frac{1}{5}$ are (not necessarily adjacent) terms of the same arithmetic progression?
Is there a method to show if they're from same progression? or should I just try different differences?
All that came to my mind was to write system of equations:
$$\left\{\begin{array}\frac{1}{2}-nd=\frac{1}{3}\\\frac{1}{3}-kd=\frac{1}{5}\end{array}\right.$$ But it can't be solved for $d$ (difference).
By the way, answer is $d=-\frac{1}{30}$, which is $-2*5*3$, so maybe the difference depends on denominators of progression?
 A: Multiplying by the common denominator, our three terms are $15,10,6$.  These are (non-adjacent) terms in the sequence $15,14,13,12,11,10,9,8,7,6$.  Then divide them all by that same common denominator.
A: If the progression has the $n$-th term of
$a+bn$,
then the difference of any two terms
is a multiple of $b$.
So we want
$\frac1{2}-\frac1{3} = ib$
and $\frac1{3}-\frac1{15} = jb$
for some integers $i$ and $j$.
Then
$\frac1{6} = ib$ and $jb = \frac{4}{15}$.
Dividing these,
$\frac{i}{j} = \frac{15}{6} = \frac{5}{2}$.
Since $i$ and $j$ are integers,
$i = 5m$ and $j = 2m$ for some integer $m$.
Putting this into
 $\frac1{6} = ib$,
$\frac1{6} = 5mb$,
or $b = \frac1{30m}$.
If $\frac1{2} = a+ub$,
then
$a = \frac1{2} - ub = \frac1{2}-\frac{u}{30m}
= \frac{15m-u}{30m}$.
We can thus arbitrarily choose integers
$m$ and $u$ to get $a$ and $b$
which will define the sequence.
From this,
$\frac1{2} = a+ub$,
$\frac1{3}= \frac1{2}-ib=a+(u-i)b$
and $\frac1{15} = \frac1{3}-jb=a+(u-i-j)b$.
This obviously generalizes to a set of rational numbers:
Let the numbers be
$S = \big(\frac{a_i}{b_i}\big)_{i=1}^n$.
Then, if $M = lcm((b_i)_{i=1}^n)$
(so that $\frac{M}{b_i}$ is an integer for all $i$)
and $k$ is any positive integer,
a linear sequence with difference
$\frac1{kM}$ can be found that contains all 
elements of $S$.
The simplest one would be
$\big(\frac{n}{M}\big)_{n=-\infty}^{\infty}$.
A: Suppose that $\{T(1), T(2), …, T(n), …, T(m)\}$ is an AP with common difference $d$.
Then, 
$$\tag 1 T(n) = T(1) + (n – 1)d \, .$$
And, 
$$\tag 2 T(m) = T(1) + (m – 1)d \, .$$
Therefore, 
$$
\tag{*} T(m) – T(n) = d(m – n) \, ..
$$
We first arrange the given three terms in descending order of magnitude.
Without loss of generality we can further assume that:
$1/2 = T(1)$, $1/3 = T(n)$, for some n > 1; $1/5 = T(m)$, for some $m > n$.
$$T(n) – T(1) =-1/6 = \dots = (–1/30)[5] = (–1/30)[6 – 1] \, ;$$ (some imagination required).
Comparing with (*), we have the right to guess that $d = (–1/30)$ and $n = 6$.
Similarly, 
$$
T(m) – T(1) = \dots = (–1/30)[10 – 1].
$$
Comparing with (*), we have the right to guess that $d = (–1/30)$ and $m = 10$.
And also, 
$$
T(m) – T(n) = (1/5) – (1/3) = \dots = (–1/30)[10 – 6].
$$
Comparing with (*), we have the right to guess that d = (–1/30) n = 6 and m = 10.
The last guess confirms that: (i) the difference is common; and (ii) the choices of n and m are consistent.
Therefore, the three terms are members of some arithmetic sequences with $d = (–1/30)$.
