A) 100
B) 1000
C) 1200
D) 10000
Correct Answer: A
Solution :
(a): Let the two APs be \[{{a}_{1}},{{a}_{2}},{{a}_{3}},\].......... and\[{{b}_{1}},{{b}_{2}},{{b}_{3}},.....{{b}_{n}}\]. Also, let d be the same common difference of two APs, then The nth term of first AP \[{{a}_{n}}={{a}_{1}}+(n-1)d\] And the nth term of second AP \[{{b}_{n}}={{b}_{1}}+(n-1)d\] Now, \[{{a}_{n}}-{{b}_{n}}=\left[ {{a}_{1}}+(n-1)d \right]-\left[ {{b}_{1}}+(n-1)d \right]\] \[\Rightarrow {{a}_{n}}-{{b}_{n}}={{a}_{1}}-{{b}_{1}}\forall n\in N\] \[\Rightarrow {{a}_{100}}-{{b}_{100}}={{a}_{1}}-{{b}_{1}}=100\] (Given) \[\therefore \,\,\,{{a}_{1000}}-{{b}_{1000}}={{a}_{1}}-{{b}_{1}}\]
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