If A = {4n+ 2| n is a natural number} and B = {3n|n is a natural number}, then what is `(A nn B)` equal to?

Options

A. `{12n^2+6n|`n is a natural number
B. `{24n-12|` n is a natural number
C. `{60n+30|` n is a natural number
D. `{12n-6|` n is a natural number

Answer

To solve the problem, we need to find the intersection of the sets A and B defined as follows:

- Set A = {4n + 2 | n is a natural number}
- Set B = {3n | n is a natural number}

### Step 1: Determine the elements of Set A
Set A consists of elements generated by the formula \(4n + 2\) where \(n\) is a natural number (1, 2, 3, ...).

- For \(n = 1\): \(4(1) + 2 = 6\)
- For \(n = 2\): \(4(2) + 2 = 10\)
- For \(n = 3\): \(4(3) + 2 = 14\)
- For \(n = 4\): \(4(4) + 2 = 18\)
- For \(n = 5\): \(4(5) + 2 = 22\)
- For \(n = 6\): \(4(6) + 2 = 26\)
- For \(n = 7\): \(4(7) + 2 = 30\)

So, the elements of Set A are:
\[ A = \{6, 10, 14, 18, 22, 26, 30, \ldots\} \]

### Step 2: Determine the elements of Set B
Set B consists of elements generated by the formula \(3n\) where \(n\) is a natural number.

- For \(n = 1\): \(3(1) = 3\)
- For \(n = 2\): \(3(2) = 6\)
- For \(n = 3\): \(3(3) = 9\)
- For \(n = 4\): \(3(4) = 12\)
- For \(n = 5\): \(3(5) = 15\)
- For \(n = 6\): \(3(6) = 18\)
- For \(n = 7\): \(3(7) = 21\)
- For \(n = 8\): \(3(8) = 24\)
- For \(n = 9\): \(3(9) = 27\)
- For \(n = 10\): \(3(10) = 30\)

So, the elements of Set B are:
\[ B = \{3, 6, 9, 12, 15, 18, 21, 24, 27, 30, \ldots\} \]

### Step 3: Find the intersection of Sets A and B
The intersection \(A \cap B\) consists of the elements that are common to both sets A and B.

From our lists:
- Common elements: \(6\), \(18\), and \(30\)

Thus, the intersection is:
\[ A \cap B = \{6, 18, 30, \ldots\} \]

### Step 4: Generalize the intersection
We can observe that the common elements are increasing by 12:
- From \(6\) to \(18\) is an increase of \(12\)
- From \(18\) to \(30\) is also an increase of \(12\)

This indicates that the intersection can be expressed in a general form:
\[ A \cap B = \{6 + (n - 1) \times 12 | n \in \mathbb{N}\} \]

### Final Result
Thus, the intersection \(A \cap B\) can be expressed as:
\[ A \cap B = \{6 + 12(n - 1) | n \in \mathbb{N}\} \]