Two A.P's have the same common difference. The difference between their 100^th  terms is 100, then the difference between their 1000^th  terms is

$Q=b_1,b_2,b_3,\cdots ,b_n$  is in A.P

Given that $a_2-a_1=b_2-b_1$

Also given that $\left(a_1+99d\right)-\left(b_1+99d\right)=100\Rightarrow a_1-b_1=100$

Consider $\left(a_1+999d\right)-\left(b_1+999d\right)=a_1-b_1=100$ .