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Answer :
Let first term of \(1^{st}\) AP = a
Let first term of \(2^{nd}\) AP = a'
It is given that their common difference is same.
Let their common difference be d.
It is given that difference between their \(100^{th}\) terms is 100.
Using formula \(a_n = a + (n - 1)d\), to find nth term of arithmetic progression,
\(\Rightarrow \) a + (100 - 1) d – [a' + (100 - 1) d]= 100
\(\Rightarrow \) a + 99d - a' - 99d = 100
\(\Rightarrow \) a - a' = 100… (1)
We want to find difference between their \(1000^{th}\) terms which means we want to calculate:
a + (1000 - 1) d – [a' + (1000 - 1) d]
= a + 999d - a' - 999d = a – a'
Putting equation (1) in the above equation,
a + (1000 - 1) d – [a' + (1000 - 1) d]
= a + 999d - a' + 999d = a - a' = 100
Therefore, difference between their \(1000^{th}\) terms would be equal to 100.