If {Nₐ} = { an:n ∈ N} , then {N₃} cap {N₄} =

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1.Set Theory

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If ${N_a} = \{ an:n \in N\} ,$ then ${N_3} \cap {N_4} = $

A

${N_7}$

B

${N_{12}}$

C

${N_3}$

D

${N_4}$

Solution

(b) ${N_3} \cap {N_4} = \{ 3,\,6,\,9,\,12,15……\} \cap \{ 4,\,8,\,12,\,16,\,20,…..\} $

$= \{12, 24, 36……\} =$ ${N_{12}}$.

Trick : ${N_3} \cap {N_4} = {N_{12}}$

[ $\because $ $3, 4$ are relatively prime numbers].

Standard 11

Mathematics

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