JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Sets Relations and Functions

Let

A = {1, 6, 11, 16, \dots}

and

B = {9, 16, 23, 30, \dots}

be the sets consisting of the first 2025 terms of two arithmetic progressions. Then

n (A \cup B)

Answer: (C)

Solution

Set

A

has common difference

5

with last term

1 + 2024 (5) = 10121

, and set

B

has common difference

7

with last term

9 + 2024 (7) = 14177

. Terms common to both are

\equiv 1 (mod 5)

and

\equiv 2 (mod 7)

, starting at

16

with common difference

lcm (5, 7) = 35

. Those not exceeding

10121

number

⌊ \frac{10121 - 16}{35} ⌋ + 1 = 289

. By inclusion and exclusion,

n (A \cup B) = 2025 + 2025 - 289 = 3761

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Single correct option