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Sequences and Series · Question 13

JEE Main 2026 (Online) 2nd April Morning ShiftMCQMedium📈 Topic in 55% papers+4−1
1

Let A be the set of first 101 terms of an A.P., whose first term is 1 and the common difference is 5 and let B be the set of first 71 terms of an A.P., whose first term is 9 and the common difference is 7. Then the number of elements in A B, which are divisible by 3, is :

  1. 4
  2. 5
  3. 6
  4. 7

Answer

(2) 5

Solution

A=\1,6,11, ,501\: terms 5m+1, 0 m 100

B=\9,16,23, ,499\: terms 7p+9, 0 p 70

Smallest common term: 16 (∵ 16=5 3+1=7 1+9)

Common terms form an A.P. with common difference LCM(5,7)=35:

A B=\16,51,86, \, general term 16+35t 499 t=0,1, ,13

Divisible by 3: 16+35t 1+2t 0 3 t 1 3

t=1,4,7,10,13, i.e. the terms 51,156,261,366,471

∴ Number of such elements =5

Answer: (2)

JEE Main 2026 (Online) 2nd April Morning Shift

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