Geometric Sequence Proof Problem:

(a) Determine all nonnegative integers $r$ such that it is possible for an infinite arithmetic sequence to contain exactly $r$ terms that are integers. Prove your answer.
(b) Determine all nonnegative integers $r$ such that it is possible for an infinite geometric sequence to contain exactly $r$ terms that are integers. Prove your answer.

My Solution for $a)$:

$a)$ First, we can analyze the case when $r \geq 2$. In this case, there will be at least $2$ terms in the arithmetic sequence which are integers; $a+md$ and $a+nd$. WLOG, $m>n$. These terms are both integers, so $(m-n)d$ is also an integer. Adding $(m - n)d$ to $a + md$ gives $a + (2m - n)d$, which is also a member of the arithmetic sequence and an integer, so this process can continue, giving the arithmetic sequence an infinite number of terms. Therefore, $r$ can only be $r=1$.

I am currently stuck on $b)$, but I know that we could do a similar approach, but how would I do it?
 A: For part (b):
Take a positive integer $m$. 
Construct a GP with first term $m^{r-1}$ and common ratio $\frac 1m$. The GP is
$$\overbrace{ m^{r-1}, m^{r-2},m^{r-3},\cdots\cdots,m^2, m^1, 1}^{r \text{  integer terms}}, \overbrace{\frac 1m, \frac 1{m^2},\cdots\cdots\cdots}^{\text{non-integer terms}}$$
A: Consider $\{a+nt\}$ (where the sequence starts at $n=0$). $r=0$ is achieved when $a,t$ are irrational. $r=1$ is achieved when $a$ is an integer and $t$ is irrational. $r\geq 2$ is impossible because given two integers $n, m$ in the sequence (WLOG $n<m$) , $m+(m-n)$ is in the sequence.
Consider $\{at^n\}$ (where the sequence starts at $n=0$). $r=0$ is achieved when $a$ is irrational and $t$ is an integer. $r=1$ is achieved when $a$ is an integer and $t$ is irrational. $r\ge 2$ is impossible because given integers $n$, $nt^k$ in the sequence $t^k$ is an integer and $nt^{2k}$ is an integer.
What about the arithmetic and geometric sequences gives them this property? Is there a more general class of sequences with this property? What types of sequences have infinitely many $r$ values?
A: Actually, for (b) any nonnegative integer $r$ works. If you want exactly $r$ terms to be integers, take $2^{r-1}, 2^{r-2}, 2^{r-3},..., 2^2, 2, 1, \frac{1}{2}, ...$
example: $r=3$
$\text{ratio}=\frac{1}{3}$
we get $2^{3-1}, 2^{3-2}, 2^{3-3}$, and those are all the integer terms.
I'm only an 8th grader so there may be a flaw in my thinking, if so please correct me!
