Solution

The correct answer is {12n - 6 | n is a natural number}

Key Points Set A generates numbers of the form 4n + 2, and set B generates numbers of the form 3n, both when n is a natural number. The intersection of the two sets will be numbers that can be generated by both formats.

Set A generates number:

n = 1, 2, 3, 4, ...

• 4*1 + 2 = 6
• 4*2 + 2 = 10
• 4*3 + 2 = 14
• 4*4 + 2 = 18

​Set B generates number:

n = 1, 2, 3, 4, ...

• 3*1 = 3
• 3*2 = 6
• 3*3 = 9
• 3*4 = 12
• 3*5 = 15
• 3*5 = 18

 A ⋂ B = (6, 10, 14, 18) ⋂ (3, 6, 9, 12, 15, 18)

 A ⋂ B = (6, 18)

Upon closer inspection, the option 4) {12n - 6 | n is a natural number} can indeed generate the sequence of numbers that belong to both A and B.

This sequence starts with ... when n = 1, 12*1 - 6 = 6 when n = 2, 12*2 - 6 = 18 when n = 3, 12*3 - 6 = 30 and so on.

These are indeed the terms that are common to sets A and B.

So, the correct answer is indeed option 4) {12n - 6 | n is a natural number}.