Prove that $(x_n)_{n\geq1}$ is an arithmetic progression Let $(x_n)_{n\geq1}$ be a sequence of integers. Define $y_n=\frac{x_n}{n},n\geq1$. The sequence $(y_n)_{n\geq1}$ is convergent and $n$ divides the sum of any $n$ consecutive terms of the sequence $(x_n)_{n\geq1}, \forall n\geq2$. Prove that  $(x_n)_{n\geq1}$ is an arithmetic progression with ratio even number. Any ideas?
 A: I prove this in 3 steps: First I show that the sequence is arithmetic starting from some point. Then I show that the whole sequence must be arithmetic. Finally I show that the difference between two consecutive terms is even.
By assumption, there exists a sequence $p_n$ of integers, such that $x_1 + x_2 + \dots + x_n = n p_n$. Similarly there exists a sequence $q_n$ of integers, such that $x_2 + x_3 + \dots + x_{n+1} = n q_n$. In particular we have $x_{n+1} - x_1 = n q_n - n p_n$.
We see that $\frac{x_{n+1} - x_1}{n}$ is an integer that converges as $n \to \infty$ to some integer $L$. Therefore for large enough $n$ we have that $x_{n+1} - x_1 = n L$, so that the sequence is arithmetic from some point on.
Next assume that the sequence is arithmetic starting from some index $m+1$. We will prove that it is actually arithmetic starting from the index $m$. By induction the whole sequence will then be arithmetic. Indeed, we have by assumption that
$$\begin{eqnarray*}x_m + x_{m+1} + \dots + x_{m+n-1} & = & x_m + x_{m+1} + (x_{m+1} + L) + \dots + (x_{m+1} + (n-2) L) \\ & = & x_m + (n-1) x_{m+1} + \frac{(n-2)(n-1)}{2}L \\ & = & x_m - x_{m+1} + L + n x_{m+1} + \frac{n(n-3)}{2} L.\end{eqnarray*}$$
Moreover this is divisible by $n$ for all $n$. It follows that $x_m - x_{m+1} + L = 0$, or $x_{m+1} - x_m = L$.
Now we know that the sequence is arithmetic. Thus $x_n = a + n L$ for some $a, L$. We have that $x_1 + x_2 = 2a + 3L$, which should be divisible by $2$. Thus $L$ is even.
