Mathematics · Sequence and Series

JEE Main 2026 — 2 April, Morning Shift — Question 24

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 :

Option A:
4
Option B:
5
Correct
Option C:
6
Option D:
7

Answer: B

Step-by-step solution

$\mathrm{AP}_{1}$ Set $\mathrm{A}=\{1,6,11,16 \ldots \ldots \ldots . .101$ terms $\}$ $\mathrm{AP}_{2}$ Set $\mathrm{B}=\{9,16$ 71 terms $\}$ $\mathrm{D}=\mathrm{L} . \mathrm{C} . \mathrm{M}\left\{\mathrm{d}_{1}, \mathrm{~d}_{2}\right\}=35$ $1^{\text {st }}$ Common term is 16 $16+(\mathrm{n}-1) 35 \leq 499$ $\mathrm{n} \leq 14.8$ $\Rightarrow \mathrm{n}=14$ $\mathrm{A} \cap \mathrm{B}=\{16,51,86,121,156,191,226,261,296, 331, 366,401,436,471 \}$ terms divisible by ' 3 ' $=\{51,156,261,366,471\}$ Ans. 5 terms

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Exam: JEE Main 2026
Paper: 2 April, Morning Shift
Subject: Mathematics
Chapter: Sequence and Series
Topic: Arithmetic Progression