# 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?

• The reason you can't solve for $d$, by the way, is because there are infinitely many arithmetic progressions containing these terms. The answer given by the textbook is just an example of such progression; in particular, it's the one with the greatest common difference. – Deathkamp Drone Apr 16 '14 at 16:38

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.

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 $\big(\frac{n}{M}\big)_{n=-\infty}^{\infty}$.

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 $d = (–1/30)$.