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Sum of First 100 Terms of Two Added APs | JEE

JEE Maths question with a full step-by-step solution.

Question
Let a1,a2,…a_1, a_2, \ldots and b1,b2,…b_1, b_2, \ldots be arithmetic progressions with a1=25a_1=25, b1=75b_1=75 and a100+b100=100a_{100}+b_{100}=100. Then the sum of the first hundred terms of a1+b1, a2+b2, …a_1+b_1,\ a_2+b_2,\ \ldots is
A10001000
B100000100000
C1000010000correct
D2400024000
Solution
Step 1: ak+bka_k+b_k is again an A.P., with first term a1+b1=25+75=100.a_1+b_1=25+75=100. Step 2: For common difference dd, the 100100th term minus the first is 99d99d:
(a100+b100)−(a1+b1)=99d ⇒ 100−100=99d ⇒ d=0.(a_{100}+b_{100})-(a_1+b_1)=99d\ \Rightarrow\ 100-100=99d\ \Rightarrow\ d=0.
 Step 3: Every term equals 100100, so
Sum=100×100=10000.\text{Sum}=100\times 100=10000.
 Correct answer: (3)
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