*Given that 2 ap's have same common difference*

*given that their 100th terms difference is 100*

*let the first no. of first series be a1 and second series be a2*

*then, a(1)100 - a(2)100=100 ---- 1*

*for 1st series ---- a100=a1+99d*

* 2nd series ---- a100 = a2+99d*

*keep these values in (1)*

*then, *

*a1+99d - (a2+99d) = 100*

*a1+99d-a2-99d=100*

*therefore, a1-a2 =100 ------------------------------------------- 2*

*then the difference between their 1000th terms is*

*for 1st series --- a1000 = a1+999d*

*for 2nd series --- a1000 = a2+999d*

*their 100th terms difference is*

*a(1)1000-a(2)1000*

*a1+999d-(a2+999d)*

*a1+999d-a2-999d*

*therefore we get the value a1-a2*

*from (2) a1-a2 = 100*

*therefore the difference between their 1000th terms is 100*