JEE Main20254 Apr 2025Morning ShiftMathematicsSequences and SeriesActual
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
Options
- A3814
- B4027
- C3761
- D4003
Correct answer
C. 3761
Step-by-step solution
aligned & 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) T_n=16+(n-1)(35) 10121 & (n-1) 10121-16 35 =288.71 & n 289.71 n=289 & n(A B)=n(A)+n(B)-n(A B) & =2025+2025-289=3761 aligned