$j=1,i=k+1,k>1: km_k-(k-1)m_{k+1}-m_1=c$ $(a_1+...+a_k)-(a_1+...+a_{k+1})+2m_{k+1}-a_1=c$ $2m_{k+1}-a_{k+1}-a_1-c=0$ $2a_{k+1}=2(k+1)m_{k+1}-2km_k=(k+1)a_{k+1}-ka_k+a_1+c$ $(k-1)a_{k+1}-ka_k=-(a_1+c)$ $\frac{a_{k+1}}{k} -\frac{a_{k}}{k-1}= -(a_1+c) (\frac{1}{k-1} -\frac{1}{k})$ $\frac{a_{k+1}}{k}-a_2= -(a_1+c) (1-\frac{1}{k}$ $a_{k+1}=ka_2-(k-1)(a_1+c)=k(a_2-a_1-c)+(a_1+c)$ $a_3=2a_2-a_1-c$ $c=-m_3-m_1+2m_2= \frac{c-3a_2}{3}-a_1+(a_1+a_2) \to c=0$ so for $d=a_2-a_1$ we have $a_{k+1}=a_1+kd$

The main idea is to fix $i,j$ and increment $k$. Solution. $$(i-j)m_k+(j-k)m_i+(k-i)m_j=(i-j)m_{k+1}+(j-(k+1))m_i+((k+1)-i)m_j=c$$$$(i-j)(m_{k+1}-m_k)+m_j-m_i=0$$Switching $i$ and $j$, and increment $k+1$ to $k+2$, we have $$(j-i)(m_{k+2}-m_{k+1})+m_i-m_j=0$$Adding gives $$(i-j)(2m_{k+1}-m_k-m_{k+2})=0$$Plugging $(i,j,k)=(k,k+1,k+2)$ in the question gives $c=2m_{k+1}-m_k-m_{k+2}$, so $c=0$. There are many ways to continue from here. It can also be finished by induction of $n$. Remark. It can also be finished by noting that $(i-j)(m_{k+1}-m_k)+m_j-m_i=0$ implies that $m_{i+1}-m_i$ is constant for all $i$, which implies that the sequence $\{a_n\}$ is an arithmetic sequence after induction or other methods.