Two APs have the same common difference.

Solution:

Let the two APs with same common difference d be

$a, a+d, a+2 d, \ldots$

$b, b+d, b+2 d, \ldots .(a>b)$

we are given that

$\{100$ th term of the first $A P\}$

$-\{100$ th term of the second $A P\}=100$

$\Rightarrow\{a+99 d\}-\{b+99 d\}=100$

$\Rightarrow a-b=100$ ...(1)

Now, $\{1000$ th term of the first $A P\}$

$-\{1000$ th term of the second $A P\}$

$=\{a+999 d\}-\{b+999 d\}=a-b=100$

$\{$ By (1) $\}$