1. The problem states that two arithmetic progressions (APs) have the same common difference and the difference between their 100th terms is 100. 2. Let the first AP be $a_n = a + (n-1)d$ and the second AP be $b_n = b + (n-1)d$, where $a$ and $b$ are the first terms and $d$ is the common difference (same for both). 3. The difference between their 100th terms is given by: $$a_{100} - b_{100} = [a + 99d] - [b + 99d] = a - b = 100$$ 4. Since the common difference $d$ is the same, the difference between the $n$th terms is always: $$a_n - b_n = (a + (n-1)d) - (b + (n-1)d) = a - b = 100$$ 5. Therefore, the difference between their 1000th terms is also: $$a_{1000} - b_{1000} = 100$$ **Final answer:** The difference between their 1000th terms is 100.