Let $a_k=a_1+(k-1)d, b_k=a_1+(k-1)p$, where $d,p>0$ $\frac{a_k}{b_k}-\frac{d}{p}=\frac{(a_1+(k-1)d)*p-(b_1+(k-1)p)*d}{b_k p}=\frac{a_1p-b_1d}{b_kp}$ If $a_1p=b_1d$ then $\frac{a_k}{b_k}=\frac{d}{p}$ is constant If $a_1p \neq b_1d$ then $\frac{a_k}{b_k}$ is decreasing sequence of positive integers, so starting from some index it is constant. So $\frac{a_k}{b_k}$ is constant

As above, let $a_{k+1}=a_1+k d_a, b_{k+1}=a_1+k d_b$, for $k$, $d_a$, $d_b>0$. For $a_1 d_b - b_1 d_a>0$, \begin{align*} a_1 d_b > b_1 d_a &\implies a_1 b_1+k a_1 d_b > a_1 b_1+k b_1 d_a \\ & \frac{a_1}{b_1} > \frac{a_1+k d_a}{b_1+kd_a} \\ &\therefore \frac{a_{k+1}}{b_{k+1}} \text{ is bounded from above and decreasing so it is eventually constant} \\ \end{align*} For $a_1 d_b - b_1 d_a<0$, \begin{align*} a_1 d_b < b_1 d_a &\implies a_1 d_b +k d_a d_b < b_1 d_a +k d_a d_b \\ & \frac{a_1+k d_a}{b_1+kd_a} < \frac{d_a}{d_b} \\ &\therefore \frac{a_{k+1}}{b_{k+1}} \text{ is bounded from above and decreasing so it is eventually constant} \\ \end{align*}For $a_1 d_b - b_1 d_a=0$, $\frac{a_k}{b_k}=\frac{a_1}{b_1}$ is constant.