The fourth term of an A.P. is equal to 3 times the first term and seventh term exceeds twice the third term by 1. find the first term and the common difference.

Answer

To solve the problem step by step, we will use the properties of an arithmetic progression (A.P.).

### Step 1: Define the first term and common difference
Let the first term of the A.P. be \( A \) and the common difference be \( D \).

### Step 2: Write the expressions for the terms
The \( n \)-th term of an A.P. can be expressed as:
\[ T_n = A + (n-1)D \]
Thus, we can write:
- The fourth term \( T_4 \):
\[ T_4 = A + (4-1)D = A + 3D \]
- The seventh term \( T_7 \):
\[ T_7 = A + (7-1)D = A + 6D \]
- The third term \( T_3 \):
\[ T_3 = A + (3-1)D = A + 2D \]

### Step 3: Set up the equations based on the given conditions
From the problem, we have two conditions:
1. The fourth term is equal to 3 times the first term:
\[ A + 3D = 3A \quad \text{(Equation 1)} \]
2. The seventh term exceeds twice the third term by 1:
\[ A + 6D = 2(A + 2D) + 1 \quad \text{(Equation 2)} \]

### Step 4: Simplify Equation 1
Rearranging Equation 1:
\[ A + 3D = 3A \implies 3D = 3A - A \implies 3D = 2A \implies A = \frac{3D}{2} \]

### Step 5: Substitute \( A \) in Equation 2
Now, substitute \( A = \frac{3D}{2} \) into Equation 2:
\[ A + 6D = 2(A + 2D) + 1 \]
Substituting for \( A \):
\[ \frac{3D}{2} + 6D = 2\left(\frac{3D}{2} + 2D\right) + 1 \]
Simplifying the left-hand side:
\[ \frac{3D}{2} + \frac{12D}{2} = \frac{15D}{2} \]
Now simplifying the right-hand side:
\[ 2\left(\frac{3D}{2} + \frac{4D}{2}\right) + 1 = 2\left(\frac{7D}{2}\right) + 1 = 7D + 1 \]
So we have:
\[ \frac{15D}{2} = 7D + 1 \]

### Step 6: Clear the fraction and solve for \( D \)
Multiply through by 2 to eliminate the fraction:
\[ 15D = 14D + 2 \]
Rearranging gives:
\[ 15D - 14D = 2 \implies D = 2 \]

### Step 7: Find \( A \) using \( D \)
Now substitute \( D = 2 \) back into the equation for \( A \):
\[ A = \frac{3D}{2} = \frac{3 \times 2}{2} = 3 \]

### Conclusion
The first term \( A \) is 3, and the common difference \( D \) is 2.

### Final Answer
- First term \( A = 3 \)
- Common difference \( D = 2 \)