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An observation regarding two APs with same common difference

For two APs (Arithmetic Progressions), with initial terms being a and b respectively, and common difference being same, say d, the difference between their nth terms remains same, and is equal to a – b, as, {a + (n – 1)d} – {b + (n – 1)d} = a-b, the constant.

I do a question.

Two APs have the same common difference. The difference between their 100th terms is 100. What is the difference between their 1000th terms?

Given, T2(100) – T1(100) = 100; i.e. {a + (100-1)d} – {b + (100-1)d} = 100, where a and b are initial terms of two APs and d is the same common difference. Thus, a – b = 100

Hence, T2(1000) – T1(1000) = {a + (1000-1)d} – {b + (1000-1)d} = a-b = 100.

The difference between their 1000th terms is 100.


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