1. **Problem statement:** Two arithmetic progressions (APs) have the same common difference. The difference between their 100th terms is 100. We need to find the difference between their 1000th terms. 2. **Recall the formula for the nth term of an AP:** $$a_n = a_1 + (n-1)d$$ where $a_n$ is the nth term, $a_1$ is the first term, and $d$ is the common difference. 3. **Let the first terms of the two APs be $a_1$ and $b_1$, and the common difference be $d$.** 4. The 100th term of the first AP is: $$a_{100} = a_1 + 99d$$ 5. The 100th term of the second AP is: $$b_{100} = b_1 + 99d$$ 6. The difference between their 100th terms is given as: $$a_{100} - b_{100} = (a_1 + 99d) - (b_1 + 99d) = a_1 - b_1 = 100$$ 7. Since the common difference $d$ is the same, it cancels out when subtracting the terms. 8. Now, find the difference between their 1000th terms: $$a_{1000} - b_{1000} = (a_1 + 999d) - (b_1 + 999d) = a_1 - b_1 = 100$$ 9. **Final answer:** The difference between their 1000th terms is also 100.