JEE Mains · Maths · STD 11 - 8. sequence and series
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 \(A \cap B\), which are divisible by 3, is :
- A \(4\)
- B \(5\)
- C \(6\)
- D \(7\)
Answer & Solution
Correct Answer
(B) \(5\)
Step-by-step Solution
Detailed explanation
The set \(A\) consists of the first 101 terms of an A.P. with first term \(a_1 = 1\) and common difference \(d_1 = 5\). The terms of \(A\) are \(1, 6, 11, 16, \dots, 1 + (101-1) \times 5 = 501\). The set \(B\) consists of the first 71 terms of an A.P. with first term \(b_1 = 9\)…
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