74th Romanian Mathematical Olympiad

We will prove the requirement by induction on $k$.

We notice that $$\frac{1}{1+ma}+\frac{1}{(1+ma)(1+2ma)}=\frac{2}{1+2ma},$$ therefore there is an arithmetic progression consisting of three terms of the sequence: $x_m,x_{2m},x_{2m^{2}a+3m}$.

We assume that there are $k$ non-zero natural numbers $n_1<n_2<\cdots <n_k$ such that the numbers $x_{n_1},x_{n_2},\dots ,x_{n_k}$ are the consecutive terms of an arithmetic progression. Considering $y=2x_{n_1}-x_{n_2}$, the numbers $y,x_{n_1},x_{n_2},\dots ,x_{n_k}$ are in arithmetic progression (in the number of $k+1$).

We have: $$y=\frac{2}{1+n_{1}a}-\frac{1}{1+n_{2}a}=\frac{1+pa}{(1+n_{1}a)(1+n_{2}a)}=\frac{1+pa}{1+(n_1+n_2+n_1{n}_{2}a)a},$$ where $p=2n_2-n_1$.

Therefore $$\frac{y}{1+pa},\frac{x_{n_1}}{1+pa},\frac{x_{n_2}}{1+pa},\dots ,\frac{x_{n_k}}{1+pa}$$ are $k+1$ terms $x_{m_1},x_{m_2},\dots ,x_{m_{k+1}}$ of the sequence $(x_n{)}_{n\ge 1}$.

They are in arithmetic progression, because when we divide the terms of an arithmetic progression by a non-zero real number, we still get an arithmetic progression.

In addition, we have $m_1<m_2<\cdots <m_{k+1}$, because the sequence $(x_n{)}_{n\ge 1}$ is strictly monotone. With this, the proof is complete.

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Alternative solution.

We notice that $(1+na)(1+ma)=1+(mn+m+n)a$, for any nonzero natural numbers $m$ and $n$.

For $k\ge 3$ we consider the arithmetic progression $1+a,1+2a,\dots ,1+ka$. We divide each term by their product, thus obtaining $k$ terms of the sequence $(x_n{)}_{n\ge 1}$ in arithmetic progression, because when we divide the terms of an arithmetic progression by a non-zero real number, we still get an arithmetic progression.