Question

*Set-builder form of the set A = {3, 6, 9, 12, 15 } is:*

1️⃣ A= { x : x = 3 y and 1≤ y ≤ 4 }
2️⃣ A = { x : x = y + 3 and 0≤ y ≤ 5 }
3️⃣ A = { x : x = 3y and 1≤ y ≤ 5}
4️⃣ A = { x : x = y- 3 and 1≤ y ≤ 5 }​

Answer 1

Required Answer

Given Set :

$\sf \: A = \{3,6,9,12,15 \}$

To Find :

$\sf : \rightarrow Roster \: \: Form \: \: of \: \: set \: \: A$

Solution :

we can write those elements of set A→ a + 3 { 0≤ a ≤ 5 }

Hence,

$\odot \: \: \sf \: A = \{x : x = y + 3 \: \& \: \: 0 \leqslant y \leqslant 5 \}$