JEE Main 2025 — 4 April, Morning Shift — Question 22

$1^{\text {st }}$ A.P. : $1,6,11 \ldots \quad \Rightarrow T_{n}=S_{n}-4$

$2^{\text {nd }}$ A.P.: 9, 16, 23.. $\quad \Rightarrow T_{m}=2+7 m$

Let's find when they are equal for the first time: $5 n-4=2+7 m$

$\Rightarrow 5 n-7 m=6$

$\Rightarrow n=4, m=2$

$\Rightarrow 16$ is the first term, common difference will be $\operatorname{LCM}\left(d_{1}, d_{2}\right)=\operatorname{LCM}(5,7)=35$

$\Rightarrow$ Common terms will be $16,51,86 \ldots$ The last term of $1^{\text {st }}$ A.P. $=T_{2025}=5 \times 2025-4=10121$

$\Rightarrow$ Common term must be less than that $\Rightarrow 35 n-19$

$\Rightarrow 35 n-19 \leq 10121 \Rightarrow 35 n \leq 10140$

$$\begin{aligned} & \Rightarrow n \leq 289.7 \\& \Rightarrow n=289 \end{aligned}$$ $$\begin{aligned} & \Rightarrow \text { in } n(A \cup B)=n(A)+n(B)-n(A \cap B) \\&=2025+2025-289 \\&=3761 \end{aligned}$$