More three-term arithmetic progression questions This question was inspired by a recent question about whether
$\frac1{2}$, $\frac1{3}$ and $\frac1{15}$
can be (possibly non-consecutive) terms
in an arithmetic progression.
My question(s): Which of the following sets of
three values can be (possibly non-consecutive) terms
in an arithmetic progression,
which can not, and which are you sure of the answer
but have no hope of a proof?


*

*$1, \sqrt{2}, 2$

*$1, \sqrt{2}, \sqrt{3}$

*$\sqrt{2}, \sqrt{3}, \sqrt{5}$

*$1, e, e^2$

*$1, \sqrt{2}, e$

*$1, e, \pi$

*$\gamma, e, \pi$

*$\pi, \pi^2, \pi^3$

*$\pi, \pi^2, \pi^4$

*$\pi, \pi^2, \pi^5$
 A: Generically, suppose that $(a, b, c)$ are three terms of an arithmetic progression with common difference $d$.  Then $b-a$, $c-b$, and $c-a$ are all integer multiples of $d$ (and in fact, if any two of these quantities are, then by the appropriate addition or subtraction the third is as well); call them $k_1d$, $k_2d$ and $k_3d$, with $k_1, k_2, k_3\in\mathbb{Z}$ (and $k_3=k_2+k_1$).  Then we have $\frac{k_2}{k_1} = \frac{c-b}{b-a}\in\mathbb{Q}$; contrariwise, if $\frac{c-b}{b-a}=\frac{m}{n}\in\mathbb{Q}$, then by choosing $d=\frac{b-a}n=\frac{c-b}m$, $b=a+nd$ and $c=b+md=a+(m+n)d$ are clearly members of an arithmetic progression with common difference $d$.
Applying this to several of your cases:
1) $1,\ \sqrt{2},\ 2$: $\displaystyle\frac{c-b}{b-a} = \frac{2-\sqrt{2}}{\sqrt{2}-1} = \frac{2-\sqrt{2}}{\sqrt{2}-1}\cdot\frac{\sqrt{2}+1}{\sqrt{2}+1} = 2\cdot\sqrt{2}+2-\sqrt{2}\cdot\sqrt{2}-\sqrt{2}) = \sqrt{2}$.  Since this ratio is not rational, then they can't possibly be part of an a.p.
4) $1,\ e,\ e^2$: $\frac{c-b}{b-a} = \frac{e^2-e}{e-1}=e$.  Again, since the ratio isn't rational, they can't possibly be part of an a.p.
10) $\pi,\ \pi^2,\ \pi^5$: Here the ratio is $\frac{\pi^5-\pi^2}{\pi^2-\pi} = \pi^3+\pi^2+\pi$.  If this were a rational number $r=\frac{p}{q}$, then $\pi$ would be a solution of the equation $q\pi^3+q\pi^2+q\pi-p=0$; in other words, $\pi$ would be algebraic.  But since this is known not to be so, then once again the terms cannot possibly be part of an a.p.
Can you see how to handle the rest in similar fashion?
A: Looks like it boils down to the following general question:
Can numbers $a_1$, $a_2$, and $a_3$ be terms of a arithmetic progression. Assume they are. Then
$a_1=a_0+n_1d$
$a_2=a_0+n_2d$
$a_3=a_0+n_3d$
It follows that $$\frac{a_i-a_j}{a_k-a_l}=\frac{n_i-n_j}{n_k-n_l}=q,$$
where $q$ is a rational number and the indexes $(i,j,k,l)$ can take values $1,2,3$ such that no index can be repeated more than twice and avoid dividing by zero. So if all such possible combinations of $\frac{a_i-a_j}{a_k-a_l}$ are rational numbers then the answer is positive. 
You can examine all cases provided. Let's take a look at $1$, $e$, and $e^2$ for example:
$$\frac{1-e^2}{1-e}=1+e\not\in\mathbb{Q}$$
(i.e. not rational) therefore $1$, $e$, and $e^2$ cannot be members of an arithmetic progression. Similar argument applies to the rest, although it is not trivial to establish that $\frac{1-\pi}{1-e}$ is not rational per se .. 
A: $a,b,c$ are of same arithmetic progression, if either


*

*$a=b=c$;

*$\dfrac{c-b}{b-a} \in \mathbb{Q}$ $\quad$($a\ne b$ and $b\ne c$).



1, 4, 8: no, because fractions are: $\sqrt{2}, e, \pi$;
2, 3: no, because fractions are quadratic irrational;
I don't see any arithmetic progression above.
