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)≤3510121−16=288.71n≤289.71⇒n=289∴n(A∪B)=n(A)+n(B)−n(A∩B)=2025+2025−289=3761
Let A=1,6,11,16, … and B=9,16,23,30, … be the sets consisting of the first 2025 terms of two arithmetic progressions. Then n(A ∪ B) is
JEE Main 2025 — Mathematics Algebra
Let A={1,6,11,16,…} and B={9,16,23,30,…} be the sets consisting of the first 2025 terms of two arithmetic progressions. Then n(A∪B) is
Held on 4 Apr 2025 · Verified 6 Jul 2026.
Options
- A
3814
- B
4027
- C
3761
- D
4003
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