Which are the finite groups $G$ such that the element orders of $G$ form an arithmetic progression? Several remarks:


1. $S\_3$, $A\_4$ and any $p$-group of exponent $p$ satisfy this property.
2. If $G$ satisfies this property and $p\_1<p\_2<...<p\_k$ are the prime divisors of $n=\mid G\mid$, then $p\_2=2p\_1-1$, and consequently $(p\_1,p\_2)\in \{(2,3),(3,5),(7,13),(19,37),...\}$.
3. If $G$ satisfies this property and there is $a\in G$ such that $o(a)=\exp(G)$ (in particular, if $G$ is nilpotent), then $G$ is a $p$-group of exponent $p$.


My impression is that the non-nilpotent groups whose element orders form a progression are of order $2^\alpha3^\beta$ (and consequently are solvable), but I failed in proving this. 



Finite groups for which the element orders form an arithmetic progression
No, your impression is not correct. -- A counterexample is the group ${\rm S}\_5$, whose elements have orders $1, 2, 3, 4, 5$ and $6$.


As to groups with *spectrum* (i.e. set of element orders) equal to $\{1, \dots, n\}$: the maximum here is 8, and for $n = 8$ the only such group is an extension of ${\rm PSL}(3,4)$ by a unitary automorphism -- see [here](https://mathoverflow.net/questions/119095/the-set-of-orders-of-elements-in-a-group/119099#119099).


