If we have two numbers $a$ and $b$ with $a<b$, and let the arithmetic mean of $a$ and $b$ be $A$, we can create an arithmetic sequence {$a, A, b$}. I know that we can't do exactly the same for three numbers, but how can we create an arithmetic sequence using three numbers, or in general, $n$ numbers, and their arithmetic mean?
At first, I thought this question would be answered here, but the answer provided did not answer my question.
My answer would be that this is impossible in general.
I interpret "this" to mean finding a finite arithmetic sequence containing the original terms. For example, given three numbers $\frac13$, $\frac12$, and $\frac34$, we could insert a few terms to create the arithmetic sequence $\frac13,\frac5{12},\frac12,\frac7{12},\frac23,\frac34$.
However, in this situation with three numbers $a<b<c$, it's not too hard to show that this is possible if and only if $\frac{c-a}{b-a}$ is rational.
For example, given the numbers $0$, $1$, and $\sqrt2$, there is no number of terms we could insert to produce a finite arithmetic sequence. And indeed, if we could do so with insertions $r_1<r_2<\cdots<r_k$, then we would quickly find that $\sqrt 2=(k+2)r_1$, contradicting the irrationality of $\sqrt2$.
For many original numbers $a_1,\dots,a_j$, the criterion for producing a finite arithmetic sequence containing them is that $\frac{a_i-a_1}{a_2-a_1}$ must be rational for all $3\le i\le j$.