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 :
- 4
- 5
- 6
- 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)