Solution: We know that $S_n=\frac{n}{2}(2a_1+(n-1)d)$ where $d=a_{i+1}-a_i$ than $A=\frac{d}{2}((i-1)(j-k)+(j-1)(k-i)+(k-1)(i-j))+\frac{a_1}{2}((j-k)+(k-i)+(i-j))=0$Q.E.D

It is easily shown that $S_i=\frac i 2 (2a_1+(i-1)d)$ where $d$ is the common difference of the arithmetic progression. Then the above espression collapses to $0$, and the thesis follows.

Proposed problem. Let $S_k$ be the sum of the first $k$ terms of the arithmetic sequence $a_1\ ,\ a_2\,\ a_3\ \ldots .$ Show that the value of the expression $\frac{S_m}{m}\cdot (n-p) + \frac{S_n}{n}\cdot (p-m) +\ \frac{S_p}{p}\cdot (m-n)$ does not depend on the numbers $m\ ,\ n\ ,\ p$ nor on the choice of the arithmetic sequence. Proof. $\frac {a_m-a_n}{m-n}=\frac {a_n-a_p}{n-p}=\frac {a_p-a_m}{p-m}=r\implies$ $\boxed {p\left(a_m-a_n\right)+m\left(a_n-a_p\right)+n\left(a_p-a_m\right)=0}\ (*)\implies$ $\boxed {\ (p-n)\cdot a_m+(m-p)\cdot a_n+(n-m)\cdot a_p=0\ }$ - first invariant . Since $\frac {S_m}{m}=\frac{a_1+a_m}{2}$ a.s.o. obtain that $\frac {\frac {S_m}{m}-\frac {S_n}{n}}{a_m-a_n}=\frac {\frac {S_n}{n}-\frac {S_p}{p}}{a_n-a_p}=\frac {\frac {S_p}{p}-\frac {S_m}{m}}{a_p-a_m}=\frac 12\stackrel{(*)}{\implies}$ $p\cdot\left(\frac {S_m}{m}-\frac {S_n}{n}\right)+$ $m\cdot\left(\frac {S_n}{n}-\frac {S_p}{p}\right)+$ $n\cdot \left(\frac {S_p}{p}-\frac {S_m}{m}\right)=0\implies$ $\boxed {\ (p-n)\cdot \frac {S_m}{m}+(m-p)\cdot\frac {S_n}{n}+(n-m)\cdot\frac {S_p}{p}=0\ }$ - second invariant .

mhet49 wrote: Let $S_i$ be the sum of the first $i$ terms of the arithmetic sequence $a_1,a_2,a_3\ldots $. Show that the value of the expression \[\frac{S_i}{i}(j-k) + \frac{S_j}{j}(k-i) +\ \frac{S_k}{k}(i-j)\] does not depend on the numbers $i,j,k$ nor on the choice of the arithmetic sequence $a_1,a_2,a_3,\ldots$. HERE is another type of solution: Let ${i}\le{j}\le{k}$ then note that, ${\frac{S_i}{i}}\le{\frac{S_j}{j}}\le{\frac{S_k}{k}}$ so we can write: ${\frac{S_i}{i}(j-k+k-i+i-j)}\le{\frac{S_i}{i}(j-k) + \frac{S_j}{j}(k-i) +\ \frac{S_k}{k}(i-j)}\le{\frac{S_k}{k}(j-k+k-i+i-j)}$ it follows $\frac{S_i}{i}(j-k) + \frac{S_j}{j}(k-i) +\ \frac{S_k}{k}(i-j)=0$

Observe that $\frac{S_i}{i}$, is just the average of the first $i$ terms. We then compute $\frac{S_i}{i}(j-k) + \frac{S_j}{j}(k-i) +\ \frac{S_k}{k}(i-j)=\frac{a_1+a_i}{2}(j-k) + \frac{a_1+a_j}{2}(k-i) +\ \frac{a_1+a_k}{2}(i-j)$. Upon expansion, it becomes clear that this sum is indeed zero. $\square$

Let $X=\frac{S_i}{i}(j-k) + \frac{S_j}{j}(k-i) +\ \frac{S_k}{k}(i-j)$ Note that $\frac{S_i}{i}=\frac{a_1+a_i}{2}$ this is due to: $S_i=\frac{(a_1+a_i)}{2}i \implies \frac{S_i}{i}=\frac{(a_1+a_i)}{2}$ Then $X=\frac{S_i}{i}(j-k) + \frac{S_j}{j}(k-i) +\ \frac{S_k}{k}(i-j)=\frac{(a_1+a_i)}{2}(j-k)+\frac{(a_1+a_j)}{2}(k-i)+\frac{(a_1+a_k)}{2}(i-j)$ $\implies X=\frac{(a_1j-a_1k+a_jj-a_ik)}{2}+\frac{(a_1k-a_1i+a_jk-a_ji)}{2}+\frac{(a_1i-a_1j+a_ki-a_kj)}{2}$ $\implies X=0$ does not depend on the numbers $i,j,k$ nor on the choice of the arithmetic sequence $a_1,a_2,a_3,\ldots$. $\blacksquare$