2022/066) Let a,b be real numbers (b≠0) and consider the infinite arithmetic sequence a,a+b,a+2b,⋯. How do I show that this sequence contains an infinite geometric subsequence if$\frac{a}{b}$ is rational?

Let $\frac{a}{b}$ be rational and so let it be $\frac{n}{m}$ where n and m are integers

so $\frac{a}{b}= \frac{n}{m}$ or $b= \frac{ma}{n}$

now let $t_n$ the the nth term

so $t_(n+1) = a + ma = a(1+m)$

to get $a(1+m)^2$ we need to add $a(m^2 + 2m)$ or $am(m+2)$ as $nb = ma$ so $nb(m+2) = am(m+2)$ and so we get this term as well

to get $a(1+m)^k$ we need to add $a((1+m)^k -1)$ or $amf(m)$ as $nb = ma$ so $nbf(m) = amf(m)$ and so we get this term as well

so we get the terms in GP