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If $A=\{4 n+2 \mid n$ is a natural number $\}$ and $\mathrm{B}=\{3 n\rfloor n$ is a natural number\}, then what is $(A \cap B)$ equal to?
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The correct answer is:
$\{12 n-6 \mid n$ is a natural number}
Let $A=\{4 n+2: n \in N\}$
and $B=\{3 n: n \in N\}$
$\Rightarrow \quad A=\{6,10,14,18,22,26,30,34,38,42, \ldots \ldots$
and $\mathrm{B}=\{3,6,9,12,15,18,21,24,27,30, \ldots .\}$
$\mathrm{A} \cap \mathrm{B}=\{6,18,30,42, \ldots . .\}$
$=6+12 n-12=12 n-6 .$
Hence, $\mathrm{A} \cap \mathrm{B}=\{12 n-6: n$ is a natural number $\}$.
and $B=\{3 n: n \in N\}$
$\Rightarrow \quad A=\{6,10,14,18,22,26,30,34,38,42, \ldots \ldots$
and $\mathrm{B}=\{3,6,9,12,15,18,21,24,27,30, \ldots .\}$
$\mathrm{A} \cap \mathrm{B}=\{6,18,30,42, \ldots . .\}$
$=6+12 n-12=12 n-6 .$
Hence, $\mathrm{A} \cap \mathrm{B}=\{12 n-6: n$ is a natural number $\}$.
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