**Solution:**

Let, the first term of two APs be a_{1} and A_{1} respectively.

The common difference of these two APs is d.

For the first AP, we know,

a_{n} = a + (n − 1) d

Therefore,

a_{100} = a_{1 }+ (100 − 1) d

= a_{1} + 99d

a_{1000} = a_{1 }+ (1000 − 1) d

a_{1000} = a_{1 }+ 999 d

For second A.P., we know,

A_{n} = A + (N − 1)d

Therefore,

A_{100} = A_{2 }+ (100 − 1)d

= A_{2 }+ 99d

A_{1000} = A_{2 }+ (1000 − 1)d

= A_{2 }+ 999d

Given that, difference between 100^{th} term of the two APs = 100

Therefore, ( a_{1 }+ 99d) − (A_{1 }+ 99 d) = 100

a_{1 }− A_{1} = 100……………………………………………………………….. **(i)**

Difference between the 1000^{th} terms of the two APs

(a_{1 }+ 999 d ) − (A_{1 }+ 999d) = a_{1 }− A_{1}

From equation **(i)**,

This difference, a_{1 }− A_{1 }= 100

Hence, the difference between the 1000^{th} terms of the two APs is 100.