JEE Main 2025 (Online) 4th April Morning ShiftQuestion 1
Let $A=\{1,6,11,16,\ldots\}$ and $B=\{9,16,23,30,\ldots\}$ be the sets consisting of the first $2025$ terms of two arithmetic progressions. Then $n(A\cup B)$ is
Sets, Relations and Function
Solution
$A=\{1+5i\}$, $B=\{9+7j\}$, $i,j=0..2024$. Common terms: $\equiv1 \pmod 5$ and $\equiv2\pmod7$, i.e. $\equiv16\pmod{35}$, starting at $16$. Terms of $A$ up to $1+5\cdot2024=10121$, of $B$ up to $9+7\cdot2024=14177$; common must be $\le10121$, $\ge16$: $16+35t\le10121\Rightarrow t\le288.7$, so $289$ terms (all are in $B$ as $\ge9$). $n(A\cup B)=2025+2025-289=3761$. Option B.