Proof for arithmetic progression with different indices

Question

How do I show that $\left(a_n\right)_{n\geq1}$ is an arithmetic progression if and only if $a_i+a_j-a_k = a_{\left(i+j-k\right)}$.

I tried by using the different definitions for arithmetic progressions and equating them, but I have no clue how to solve this. Any ideas would be greatly appreciated!

Answer 1

Suppose $(a_n)_{n \geq 1}$ is an AP. Then $a_n = a_1 + (n - 1)d$ for some $d$, so: \begin{align*} a_i + a_j - a_k &= (a_1 + (i - 1)d) + (a_1 + (j - 1)d) - (a_1 + (k - 1)d) \\ &= a_1 + (i + j - k - 1)d \\ &= a_{i+j-k} \end{align*} Now suppose $a_i + a_j - a_k = a_{i+j-k} \; \forall i,j,k \geq 1$. Then: $$ a_{i} + a_{i+2} - a_{i+1} = a_{i + (i+2) - (i+1)} = a_{i+1} \implies a_{i+2} - a_{i+1} = a_{i+1} - a_i $$ Thus, $(a_n)_{n \geq 1}$ is an AP.

Answer 2

Given $a_i + a_j - a_k = a_{(i+j-k)}$:
So, $a_i - a_k = a_{(i+j-k)} - a_j$
Take $i-k=d$ Then
$a_{k+d} - a_k = a_{j+d} - a_j$. Therefore it is A.P.

Other way round is even easier. You can try.