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(B) Let the first terms of the two $APs$ be $a_1$ and $a_2$ respectively,and let the common difference be $d$.For the first $AP$,the $n^{th}$ term is

Vedclass · Accepted Answer

$100$

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(B) Let the first terms of the two $APs$ be $a_1$ and $a_2$ respectively,and let the common difference be $d$.For the first $AP$,the $n^{th}$ term is $a_n = a_1 + (n-1)d$.For the second $AP$,the $n^{th}$ term is $b_n = a_2 + (n-1)d$.The $100^{th}$ terms are $a_{100} = a_1 + 99d$ and $b_{100} = a_2 + 99d$.Given that the difference between the $100^{th}$ terms is $100$:$(a_1 + 99d) - (a_2 + 99d) = 100$$a_1 - a_2 = 100$ (Equation $1$)Now,the $1000^{th}$ terms are $a_{1000} = a_1 + 999d$ and $b_{1000} = a_2 + 999d$.The difference between the $1000^{th}$ terms is:$(a_1 + 999d) - (a_2 + 999d) = a_1 - a_2$Substituting the value from Equation $1$,the difference is $100$.

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

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