An arithmetic progression or arithmetic sequence is a number’s sequence such that the difference between the consecutive terms is constant.

Solution:

Let the two A.Ps be $A.{{P}_{1}}$ and $A.{{P}_{2}}$

For $A.{{P}_{1}}$ the first term $=a$ and the common difference $=d$

And for $A.P{}_{2}$ the first term $=b$ and the common difference $=d$

So, from the question we have

${{a}_{100}}-{{b}_{100}}=100$

$\left( a+99d \right)-\left( b+99d \right)=100$

$a-b=100$

Now, the difference between their ${{1000}^{th}}$ terms is,

$\left( a+999d \right)-\left( b+999d \right)=a-b=100$

Therefore, the difference between their ${{1000}^{th}}$ terms is also $100.$