If Na - an - n - N - then N3 - N4 -

The correct option is B N_12 nbsp; nbsp; N_a = { an : n ∈ N } N_3 ∩ N_4 = {3,6,9,12,15 ……}∩{4,8,12,16,20, ……} nbsp;= {12,24,36 ……} = N_12 nbsp; [∵ 3,4 ...

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Byju's Answer

Standard VIII

Mathematics

Intersection of Sets

If Na = an : ...

Question

If $N_{a} = {a n : n \in N}$ , then $N_{3} \cap N_{4}$ =

$N_{7}$

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$N_{12}$

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$N_{3}$

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$N_{4}$

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Solution

The correct option is B
$N_{12}$

$N_{a} = {a n : n \in N}$
$N_{3} \cap N_{4} = {3, 6, 9, 12, 15 \dots \dots} \cap {4, 8, 12, 16, 20, \dots \dots}$

$= {12, 24, 36 \dots \dots} = N_{12}$

[ $∵$ 3,4 are relatively prime numbers ]
$∴$ $N_{3} \cap N_{4}$ = $N_{12}$

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If Na={an:n∈N} , then N3∩N4 =

The correct option is B N12 Na={an:n∈N} N3∩N4={3,6,9,12,15……}∩{4,8,12,16,20,……} ={12,24,36……}=N12 [∵ 3,4 are relatively prime numbers ] ∴ N3∩N4 = N12

If $N_{a} = {a n : n \in N}$ , then $N_{3} \cap N_{4}$ =

The correct option is B
$N_{12}$

$N_{a} = {a n : n \in N}$
$N_{3} \cap N_{4} = {3, 6, 9, 12, 15 \dots \dots} \cap {4, 8, 12, 16, 20, \dots \dots}$

$= {12, 24, 36 \dots \dots} = N_{12}$

[ $∵$ 3,4 are relatively prime numbers ]
$∴$ $N_{3} \cap N_{4}$ = $N_{12}$