If the `6^(th)` term of an A.P. is equal to four times its first term and the sum of first six terms is 75, find the first term and the common difference.

Answer

To solve the problem, we need to find the first term (A) and the common difference (D) of the arithmetic progression (A.P.) given the conditions provided.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The 6th term of the A.P. is equal to four times the first term.
- The sum of the first six terms is 75.

2. **Define the Terms:**
- Let the first term be \( A \).
- Let the common difference be \( D \).

3. **Expression for the 6th Term:**
- The formula for the nth term of an A.P. is given by:
\[ T_n = A + (n-1)D \]
- For the 6th term (\( T_6 \)):
\[ T_6 = A + 5D \]
- According to the problem, \( T_6 = 4A \):
\[ A + 5D = 4A \]

4. **Rearranging the Equation:**
- From the equation \( A + 5D = 4A \), we can rearrange it to find \( D \):
\[ 5D = 4A - A \]
\[ 5D = 3A \]
\[ D = \frac{3A}{5} \quad \text{(Equation 1)} \]

5. **Expression for the Sum of the First Six Terms:**
- The sum of the first n terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2A + (n-1)D) \]
- For the first six terms (\( S_6 \)):
\[ S_6 = \frac{6}{2} \times (2A + 5D) = 3(2A + 5D) \]
- According to the problem, \( S_6 = 75 \):
\[ 3(2A + 5D) = 75 \]

6. **Simplifying the Sum Equation:**
- Dividing both sides by 3:
\[ 2A + 5D = 25 \quad \text{(Equation 2)} \]

7. **Substituting Equation 1 into Equation 2:**
- Substitute \( D = \frac{3A}{5} \) into Equation 2:
\[ 2A + 5\left(\frac{3A}{5}\right) = 25 \]
\[ 2A + 3A = 25 \]
\[ 5A = 25 \]

8. **Finding the First Term:**
- Dividing both sides by 5:
\[ A = 5 \]

9. **Finding the Common Difference:**
- Substitute \( A = 5 \) back into Equation 1 to find \( D \):
\[ D = \frac{3 \times 5}{5} = 3 \]

### Final Answer:
- The first term \( A \) is 5.
- The common difference \( D \) is 3.