parmenides51 wrote: Let $n$ be a natural number, $n \ge 5$, and $a_1, a_2, . . . , a_n$ real numbers such that all possible sums $a_i + a_j$, where $1 \le i < j \le n$, form $\frac{n(n-1)}{2}$ consecutive members of an arithmetic progression when taken in some order. Prove that $a_1 = a_2 = . . . = a_n$. Let $\rm a_1\leq a_2 \leq ...\leq a_{n-1}\leq a_n$ Clearly the first term is the $\rm a_1+a_2$ and the second term is the $\rm a_1+a_3$. So ,if $\rm k$ is the order of the arithmetic progression then $\rm k=(a_3+a_1)-(a_2+a_1)=a_3-a_2$. If $\rm a_3=a_2$ then $\rm k=0$ so obviously $\rm a_1=a_2=..=a_n$. If $\rm a_2\neq a_3$ then the last term is $\rm a_n+a_{n-1}$ and the previous term is the $\rm a_n+a_{n-2}$ so $\rm (a_n+a_{n-1})-(a_n+a_{n-2})=k\Leftrightarrow a_{n-1}-a_{n-2}=a_3-a_2\Leftrightarrow a_{n-1}+a_2=a_3+a_{n-2}$. This is contradiction because since $\rm k\neq 0$ there are no exist two equal terms in arithmetic progression .