Let $\{a_1,a_2,\cdots{}\}$ be a strictly increasing sequence of positive integers in arithmetic progression . Prove that there exists an infinite subsequence of the given sequence whose terms are in geometric progression.

I started by letting $a_1=a,a_2=a+d,\cdots{}$. Im not sure how to continue from here. Any hints or solutions would be helpful.

  • $\begingroup$ The question seems to suggest that the numbers are in arithmetic progression. But it is an important detail that must be clearly mentioned $\endgroup$ – GoodDeeds Oct 10 '16 at 7:27
  • $\begingroup$ @Crostul: Fixed! Its in AP $\endgroup$ – RMO2016 Oct 10 '16 at 7:31

Consider the subsequence $$a,a(d+1),a(d+1)^2,a(d+1)^3,\cdots$$

As the sequence is strictly increasing, $d\gt0$, and $(d+1)\gt1$. Hence, the above sequence is strictly increasing. On expanding, we get, $$a,a+ad,a+d(a(d+2)),a+d(a(d^2+3d+3),\cdots,a+ad\frac{((d+1)^{k-1}-1)}{d},\cdots$$

As the original sequence consists of positive integers, each of the above element is guaranteed to be an integer and to be an element of the original sequence.

Hence, let the subsequence be $a_{n_k}$ where $$a_{n_k}=a+a{((d+1)^{k-1}-1)}$$

where $k\in\mathbb N$.

In other words, the $k^{th}$ element in the subsequence is the $r^{th}$ element of the original sequence where $$r=\frac{a((d+1)^{k-1}-1)}{d}+1$$


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