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JEE Main 20254 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2025 · 4 April · Morning Shift · Mathematics · Sequences and Series

Let A=\ 1,6,11,16, \ and B=\ 9,16,23,30, \ be the sets consisting...

Let A={1,6,11,16,…} and B={9,16,23,30,…}A=\{1,6,11,16, \ldots\} \text{ and }B=\{9,16,23,30, \ldots\} be the sets consisting of the first 2025 terms of two arithmetic progressions. Then n(A∪B)n(A \cup B) is

Options

  1. A3814
  2. B4027
  3. C3761
  4. D4003

Correct answer

C. 3761

Step-by-step solution

A={1,6,11,16,21,26,31,36,41,46,51,56,61,66,71,76,81,86,91,……}B={9,16,23,30,37,44,51,58,65,72,79,86,93,100,……}A∩B={16,51,86,……} For set ’A’ ⇒T2025=1+(2025−1)(5)=10121 For set ’ B′⇒T2025=9+(2025−1)(7)=14177 So, for (A∩B)⇒Tn=16+(n−1)(35)≤10121(n−1)≤10121−1635=288.71n≤289.71⇒n=289∴n(A∪B)=n(A)+n(B)−n(A∩B)=2025+2025−289=3761\begin{aligned} & A=\{1,6,11,16,21,26,31,36,41,46,51,56,61, \\ & 66,71,76,81,86,91, \ldots \ldots\} \\ & B=\{9,16,23,30,37,44,51,58,65,72,79,86, \\ & 93,100, \ldots \ldots\} \\ & A \cap B=\{16,51,86, \ldots \ldots\} \\ & \text{ For set 'A' } \Rightarrow T_{2025}=1+(2025-1)(5)=10121 \\ & \text{ For set ' } B^{\prime} \Rightarrow T_{2025}=9+(2025-1)(7)=14177 \\ & \text{ So, for }(A \cap B) \Rightarrow T_n=16+(n-1)(35) \leq 10121 \\ & (n-1) \leq \frac{10121-16}{35}=288.71 \\ & n \leq 289.71 \Rightarrow n=289 \\ & \therefore n(A \cup B)=n(A)+n(B)-n(A \cap B) \\ & =2025+2025-289=3761\end{aligned}
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