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Intersection of Sets

A = 4n − 3n −...

Question

$A = {(4^{n} - 3 n - 1) | n \in N},$ $B = {9 (n - 1) | n \in N},$ then $A \cap B$ is

A

A

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B

B

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C

N

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D

ϕ

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Solution

The correct option is A $A$
$A = {0, 9, 54, . . .} B = {0, 9, 18, 27, . . .} \Rightarrow A \cap B = A$

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Similar questions

A = {4^{n} - 3 n - 1 : n \in N}

B = {9 (n - 1) : n \in N}

A = {(4^{n} - 3 n - 1) | n \in N},

B = {9 (n - 1) | n \in N},

A \cap B

U = {x \in N ∣ x < 20}

be the universal set. Let

A = {x \in N ∣ x is prime less than 20}

B = {x \in N ∣ x is 3 n, n \in N, n \leq 6}

C = {x \in N ∣ x = 2 n - 1, n \in N, n \leq 10}

n [(A \cup B)^{'} \cup (A \cap C)] =

X = {4^{n} - 3 n - 1; n \in R}

Y = {9 (n - 1); n \in N}

X \cap Y =

A = {n \in N ∣ ∣ n^{2} \leq n + 10, 000},

B = {3 k + 1 | k \in N}

C = {2 k | k \in N}

. Then the sum of all the elements of the set

A \cap (B - C)

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Intersection of Sets

Standard XI Mathematics

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