Japan Mathematical Olympiad

360 The assumption implies that there holds $i{a}_i=a_1+(i-1)(2a_2-a_1)$ for any integer $2\le i\le 7$. Subtracting $a_1$ from both sides and dividing by $i$, one obtains $$a_i-a_1=\frac{i{a}_i-i{a}_1}{i}=\frac{(i-1)(2a_2-a_1)-(i-1)a_1}{i}=\frac{2(i-1)(a_2-a_1)}{i}.$$ Since $i$ and $i-1$ are coprime, $2(a_2-a_1)$ is divided by $i$ for any $2\le i\le 7$, thus divided by $420$. By the assumption $a_2-a_1\ne 0$, thus $|2(a_2-a_1)|\ge 420$. Hence $$|a_7-a_1|=\left|\frac{2\cdot 6\cdot (a_2-a_1)}{7}\right|\ge \frac{6}{7}\cdot 420=360.$$ On the other hand, the sequence $(a_1,a_2,a_3,a_4,a_5,a_6,a_7)=(420,210,140,105,84,70,60)$ satisfies the assumption and $|a_7-a_1|=360$, thus the answer is $360$.