Please check my solution. It appears to be wrong because i didn't have to use that, $a_1 < a_2 < \cdots < a_n$ and $b_1 < b_2 < \cdots < b_n$, or that they are positive integers Denote by $c_i$ the difference $a_i-b_i$. We are given that $c_i+c_j+c_l=0$ for all integers $1 \le j < k < l \le n$. Now if$n<3,$ the problem is meaningless, and in view of the condition $gcd(n,6)=1$, n cant be 3,4. So $n>4$. If $i$ and $j$ are such that at least none of them belongs to at least one of {1,2},{n-1,n} (WLOG suppose the latter), then we have $c_1+c_2+c_i=0=c_1+c_2+c_j$ giving $c_i=c_j$. This implies that all $c_i$'s are equal and thus we easily get $c_i=0 \forall 1 \le i \le n$, as required.

Ankoganit wrote: Pls check my solution. It appears to be wrong because i didnt have to use that $gcd(n, 6) = 1$, $a_1 < a_2 < \cdots < a_n$ and $b_1 < b_2 < \cdots < b_n$, or that they are positive integers Denote by $c_i$ the difference $a_i-b_i$. We are given that $c_i+c_j+c_l=0$ for all integers $1 \le j < k < l \le n$. If $i$ and $j$ are such that at least none of them belongd to at least one of {1,2},{n-1,n} (WLOG suppose the latter), then we have $c_1+c_2+c_i=0=c_1+c_2+c_j$ giving $c_i=c_j$. This implies that all $c_i$'s are equal and thus we easily get $c_i=0 \forall 1 \le i \le n$, as required. In my solution i didn't use the gcd thing either.

The gcd(n,6)=1 is not strict but is important. Your solution does not work for n=2,3. Thus gcd(n,6)=1 is important.

@utkarshgupta: could u post ur solution? @satvikgupta: thnx for your remark.