Let the first terms of the two arithmetic progressions (A.P.s) be \(a_1\) and \(a_2\), respectively, and let the common difference for both be \(d\).
For the first A.P., the 100th term is \(a_{100} = a_1 + (100 − 1)d = a_1 + 99d\), and the 1000th term is \(a_{1000} = a_1 + (1000 − 1) d = a_1 + 999d\).
For the second A.P., the 100th term is \(a_{100} = a_2 + (100 − 1) d = a_2 + 99d\), and the 1000th term is \(a_{1000} = a_2 + (1000 − 1) d = a_2 + 999d\).
It is given that the difference between the 100th terms of these A.P.s is 100.
Therefore, \((a_1 + 99d) − (a_2 + 99d) = 100\), which simplifies to \(a_1 − a_2 = 100\) (Equation 1).
The difference between the 1000th terms of these A.P.s is \((a_1 + 999d) − (a_2 + 999d)\), which equals \(a_1 − a_2\).
From Equation (1), this difference, \(a_1 − a_2\), is 100.
Hence, the difference between the 1000th terms of these A.P.s will be 100.