# Prove that there exists an infinite subsequence in geometric progression given a strictly increasing sequence of integers in an arithmetic progression

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.

• The question seems to suggest that the numbers are in arithmetic progression. But it is an important detail that must be clearly mentioned – GoodDeeds Oct 10 '16 at 7:27
• @Crostul: Fixed! Its in AP – 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, $$a,a+ad,a+d(a(d+2)),a+d(a(d^2+3d+3),\cdots,a+ad\frac{((d+1)^{k-1}-1)}{d},\cdots$$
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 $$r=\frac{a((d+1)^{k-1}-1)}{d}+1$$