4th term of an A.P. is equal to 3 times its first term and 7th term exceeds twice the 3rd 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 terms of the A.P.
Let the first term of the A.P. be \( A \) and the common difference be \( D \).

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

### Step 3: Set up the equations based on the problem statements.
1. According to the first statement, the 4th term is equal to 3 times the first term:
\[ T_4 = 3A \implies A + 3D = 3A \]
Rearranging gives:
\[ 3D = 2A \quad \text{(Equation 1)} \]

2. According to the second statement, the 7th term exceeds twice the 3rd term by 1:
\[ T_7 = 2T_3 + 1 \implies A + 6D = 2(A + 2D) + 1 \]
Expanding and rearranging gives:
\[ A + 6D = 2A + 4D + 1 \]
\[ 6D - 4D = 2A - A + 1 \implies 2D = A + 1 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations simultaneously.
From Equation 1, we have:
\[ A = \frac{3D}{2} \]
Substituting this value of \( A \) into Equation 2:
\[ 2D = \frac{3D}{2} + 1 \]
To eliminate the fraction, multiply the entire equation by 2:
\[ 4D = 3D + 2 \]
Subtract \( 3D \) from both sides:
\[ D = 2 \]

### Step 5: Find the first term \( A \).
Now substitute \( D = 2 \) back into Equation 1:
\[ 3D = 2A \implies 3(2) = 2A \implies 6 = 2A \implies A = 3 \]

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

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