The `4^(th)` term of an A.P. is equal to 3 times the first term and `7^(th)` term excess with the `3^(rd)` term by 1. Find its `n^(th)` term.

Answer

To solve the problem, we need to find the nth term of an arithmetic progression (A.P.) given certain conditions about its terms. Let's break down the solution step by step.

### Step 1: Understand the nth term of an A.P.
The nth term of an A.P. is given by the formula:
\[ T_n = A + (n - 1)D \]
where:
- \( A \) is the first term,
- \( D \) is the common difference,
- \( n \) is the term number.

### Step 2: Set up the equations based on the given conditions.
1. The 4th term is equal to 3 times the first term:
\[ T_4 = 3A \]
Using the formula for the 4th term:
\[ T_4 = A + 3D \]
Therefore, we have:
\[ A + 3D = 3A \]
Rearranging gives us:
\[ 3D = 3A - A \implies 3D = 2A \implies A = \frac{3D}{2} \quad \text{(Equation 1)} \]

2. The 7th term exceeds the 3rd term by 1:
\[ T_7 = T_3 + 1 \]
Using the formulas for the 7th and 3rd terms:
\[ T_7 = A + 6D \quad \text{and} \quad T_3 = A + 2D \]
Therefore, we have:
\[ A + 6D = (A + 2D) + 1 \]
Simplifying gives:
\[ A + 6D = A + 2D + 1 \implies 6D - 2D = 1 \implies 4D = 1 \implies D = \frac{1}{4} \quad \text{(Equation 2)} \]

### Step 3: Substitute \( D \) back into Equation 1 to find \( A \).
From Equation 2, we found \( D = \frac{1}{4} \). Now substituting this value into Equation 1:
\[ A = \frac{3D}{2} = \frac{3 \times \frac{1}{4}}{2} = \frac{3}{8} \]

### Step 4: Write the nth term using the values of \( A \) and \( D \).
Now that we have both \( A \) and \( D \):
- \( A = \frac{3}{8} \)
- \( D = \frac{1}{4} \)

We can substitute these values into the formula for the nth term:
\[ T_n = A + (n - 1)D = \frac{3}{8} + (n - 1) \cdot \frac{1}{4} \]

### Step 5: Simplify the expression for \( T_n \).
\[ T_n = \frac{3}{8} + \frac{n - 1}{4} \]
To combine the terms, we convert \( \frac{1}{4} \) to have a common denominator of 8:
\[ \frac{1}{4} = \frac{2}{8} \]
Thus,
\[ T_n = \frac{3}{8} + \frac{2(n - 1)}{8} = \frac{3 + 2(n - 1)}{8} = \frac{3 + 2n - 2}{8} = \frac{2n + 1}{8} \]

### Final Answer:
The nth term of the A.P. is:
\[ T_n = \frac{2n + 1}{8} \]