Solution:

Let a1 and a2 be the initial terms of two APs, respectively.

And let d be the common difference between these APs.

For the first A.P.,we know,

a_n = a+(n−1)d

Therefore,

a₁₀₀ = a₁+(100−1)d

= a₁ + 99d

a₁₀₀₀ = a₁+(1000−1)d

a₁₀₀₀ = a₁+999d

For second A.P., we know that,

a_n = a+(n−1)d

Therefore,

a₁₀₀ = a₂+(100−1)d

= a₂+99d

a₁₀₀₀ = a₂+(1000−1)d

= a₂+999d

Given that, the difference between the two APs’ 100th terms equals 100.

Therefore, (a₁+99d) − (a₂+99d) = 100

a₁−a₂ = 100……………………………………………………………….. (i)

Difference between the two APs’ 1000th terms

(a₁+999d) − (a₂+999d) = a₁−a₂

From equation (i),

This difference, a₁−a₂ = 100

As a result, the difference between the two A.P. 1000th terms will be 100.