If the sum of `n` terms of an A.P. be `3n^2-n` and its common difference is 6, then its first term is

Options

A. `2`
B. `3`
C. `1`
D. `4`

Explanation

To solve the problem step by step, we will use the formula for the sum of the first \( n \) terms of an Arithmetic Progression (A.P.) and the information given in the question.

### Step-by-Step Solution:

1. **Identify the formula for the sum of the first \( n \) terms of an A.P.**:
The formula for the sum of the first \( n \) terms \( S_n \) of an A.P. is given by:
\[ S_n = \frac{n}{2} \left(2A + (n-1)D\right) \]
where \( A \) is the first term and \( D \) is the common difference.

2. **Substitute the given sum into the formula**:
We are given that the sum of the first \( n \) terms \( S_n = 3n^2 - n \). Therefore, we can set up the equation:
\[ 3n^2 - n = \frac{n}{2} \left(2A + (n-1)D\right) \]

3. **Substitute the known common difference**:
We know that the common difference \( D = 6 \). Substitute \( D \) into the equation:
\[ 3n^2 - n = \frac{n}{2} \left(2A + (n-1) \cdot 6\right) \]
Simplifying the right side gives:
\[ 3n^2 - n = \frac{n}{2} \left(2A + 6n - 6\right) \]

4. **Multiply both sides by 2 to eliminate the fraction**:
\[ 2(3n^2 - n) = n(2A + 6n - 6) \]
This simplifies to:
\[ 6n^2 - 2n = n(2A + 6n - 6) \]

5. **Cancel \( n \) from both sides** (assuming \( n \neq 0 \)):
\[ 6n - 2 = 2A + 6n - 6 \]

6. **Rearrange to isolate \( 2A \)**:
Subtract \( 6n \) from both sides:
\[ -2 = 2A - 6 \]
Now, add 6 to both sides:
\[ 4 = 2A \]

7. **Solve for \( A \)**:
Divide both sides by 2:
\[ A = 2 \]

### Conclusion:
The first term \( A \) of the A.P. is \( 2 \).