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

<p>$$\begin{array}{ll} 1^{\text {st }} \text { A.P. }: 1,6,11 \ldots & \Rightarrow T_n=S_n-4 \\ 2^{\text {nd }} \text { A.P.: } 9,16,23 \ldots & \Rightarrow T_m=2+7 m \end{array}$$</p> <p>Let's find when they are equal for the first time:</p> <p>$$\begin{aligned} & 5 n-4=2+7 m \\ & \Rightarrow 5 n-7 m=6 \\ & \Rightarrow n=4, m=2 \end{aligned}$$</p> <p>$\Rightarrow 16$ is the first term, common difference will be</p> <p>$\operatorname{LCM}\left(d_1, d_2\right)=\operatorname{LCM}(5,7)=35$</p> <p>$\Rightarrow$ Common terms will be 16, 51, $86 \ldots$</p> <p>The last term of $1^{\text {st }}$ A.P.</p> <p>$=T_{2025}=5 \times 2025-4=10121$</p> <p>$$\begin{aligned} &\Rightarrow \text { Common term must be less than that }\\ &\begin{aligned} & \Rightarrow 35 n-19 \\ & \Rightarrow 35 n-19 \leq 10121 \Rightarrow 35 n \leq 10140 \\ & \quad \Rightarrow n \leq 289.7 \\ & \quad \Rightarrow n=289 \\ & \Rightarrow \operatorname{in} n(A \cup B)=n(A)+n(B)-n(A \cap B) \\ & \quad=2025+2025-289 \\ & \quad=3761 \end{aligned} \end{aligned}$$</p>