Let $A=\{1,6,11,16, \ldots\}$ and $B=\{9,16,23,30, \ldots\}$ be the sets consisting of the first 2025 terms…
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
3814
4027
3761
4003
Solution
$\begin{aligned} & A=\{1,6,11,16,21,26,31,36,41,46,51,56,61, \\ & 66,71,76,81,86,91, \ldots \ldots\} \\ & B=\{9,16,23,30,37,44,51,58,65,72,79,86, \\ & 93,100, \ldots \ldots\} \\ & A \cap B=\{16,51,86, \ldots \ldots\} \\ & \text { For set 'A' } \Rightarrow T_{2025}=1+(2025-1)(5)=10121 \\ & \text { For set ' } B^{\prime} \Rightarrow T_{2025}=9+(2025-1)(7)=14177 \\ & \text { So, for }(A \cap B) \Rightarrow T_n=16+(n-1)(35) \leq 10121 \\ & (n-1) \leq \frac{10121-16}{35}=288.71 \\ & n \leq 289.71 \Rightarrow n=289 \\ & \therefore n(A \cup B)=n(A)+n(B)-n(A \cap B) \\ & =2025+2025-289=3761\end{aligned}$
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