II. (50 points)
From an $n$-term non-constant rational arithmetic sequence, what is the maximum number of terms that can be selected to form a geometric sequence $(n \in \mathbf{N})$?
At most, $1+\left[\log _{2} n\right]$ terms can be taken.
( $[x]$ is the largest integer not exceeding $x$) For example, when the arithmetic sequence is $1,2, \cdots, n$, take 1, 2, $\cdots, 2^{2} \cdots, 2^{\left[\ln _{2} n\right]}$, a total of $1+\left[\log _{2} n\right]$ terms. We will now prove that $1+\left[\log _{2} n\right]$ is optimal.

Let $a_{1}, a_{2}, \cdots a_{n}$ be the arithmetic sequence with a common difference $d$. Take $a_{m 1}, a_{m 2}, \cdots a_{m 4}$ to form a geometric sequence with a common ratio $r$, then $r=\frac{a_{m 2}}{a_{m 1}} \in \mathbf{Q}$.
(because $a_{m 1}, a_{m 2} \in \mathbf{Q}$ )

We can set $r=\frac{p}{q}(p, q \in \mathbf{N},(p, q)=1)$, then $p \neq q$, otherwise $a_{m 1}=a_{n 2}$, which is a contradiction.
Then $\frac{\left(\frac{p}{q}\right)^{t-1}-1}{\frac{p}{q}-1}=\frac{a_{m t}-a_{m 1}}{a_{m 2}-a_{m 1}}=\frac{\left(m_{t}-m_{1}\right) d}{\left(m_{2}-m_{1}\right) d}=\frac{m_{t}-m_{1}}{m_{2}-m_{1}}$,

which means $\left(m_{2}-m_{1}\right)\left(\sum_{i=1}^{t-1} p^{i-1} q^{t-i-1}\right)=q^{t-2}\left(m_{t}-m_{1}\right)$.
Take any prime $q_{0} \mid q^{t-2}$,
(if $q_{0}$ does not exist, then $\left(\sum_{i=1}^{t-1} p^{t-1} q^{t-i-1}, q^{t-2}\right)=1$ similarly)
then $q_{0} \mid q \Rightarrow q_{0} \chi p$ and $q_{0} \mid \sum_{i=1}^{t-2} p^{i-1} q^{i-i-1}$.
(because $(p, q)=1$ ) so $q_{0} \times p^{i-2}$
$$
\Rightarrow q_{0} \times \sum_{i=1}^{i-1} p^{i-1} q^{t-i-1}
$$

So $\left(\sum_{i=1}^{t-1} p^{i-1} q^{t-i-1}, q^{t-1}\right)=1$.
So $\sum_{i=1}^{i-1} p^{2-1} q^{t-i-1} \mid m_{t}-m_{1}$
$$
\Rightarrow \sum_{i=1}^{t-1} p^{t-1} q^{t-i-1} \leqslant m_{t}-m_{1} \leqslant n-1 \\
\Rightarrow n \geqslant 1+\sum_{i=1}^{t-1} p^{t-1} q^{t-j-1} \\
\geqslant 1+\sum_{i=1}^{t-1} 2^{i-1} \cdot 1^{t-i-1} \text { (because } p \neq q \text { ) } \\
=2^{t-1} \\
\Rightarrow t \leqslant \log _{2} n+1 \\
$$

Since $t \in \mathbf{N}$, we have $t \leqslant 1+\left[\log _{2} n\right]$.
Therefore, at most $1+\left[\log _{2} n\right]$ terms can be taken to satisfy the condition.