The 4th term of A.P. is equal to 3 times the first term and 7th term exceeds twice the third term by 1. Find its nth term.

Options

A. `n+2`
B. `3n+1`
C. `2n+1`
D. `3n+2`

Answer

To solve the problem step by step, we will use the properties of an Arithmetic Progression (A.P.) and the information given in the question.

### Step 1: Understand the terms of A.P.
In an A.P., the terms can be expressed as:
- First term (a)
- Second term (a + d)
- Third term (a + 2d)
- Fourth term (a + 3d)
- Fifth term (a + 4d)
- Sixth term (a + 5d)
- Seventh term (a + 6d)

### Step 2: Set up the equations based on the given conditions.
1. The 4th term is equal to 3 times the 1st term:
\[ a + 3d = 3a \quad \text{(Equation 1)} \]

2. The 7th term exceeds twice the 3rd term by 1:
\[ a + 6d = 2(a + 2d) + 1 \quad \text{(Equation 2)} \]

### Step 3: Simplify Equation 1.
From Equation 1:
\[ a + 3d = 3a \]
Rearranging gives:
\[ 3d = 3a - a \]
\[ 3d = 2a \quad \Rightarrow \quad d = \frac{2a}{3} \quad \text{(Equation 3)} \]

### Step 4: Simplify Equation 2.
From Equation 2:
\[ a + 6d = 2(a + 2d) + 1 \]
Expanding the right side:
\[ a + 6d = 2a + 4d + 1 \]
Rearranging gives:
\[ a + 6d - 4d = 2a + 1 \]
\[ a + 2d = 2a + 1 \]
Rearranging gives:
\[ 2d = 2a + 1 - a \]
\[ 2d = a + 1 \quad \text{(Equation 4)} \]

### Step 5: Substitute Equation 3 into Equation 4.
Substituting \(d = \frac{2a}{3}\) into Equation 4:
\[ 2\left(\frac{2a}{3}\right) = a + 1 \]
This simplifies to:
\[ \frac{4a}{3} = a + 1 \]
Multiplying through by 3 to eliminate the fraction:
\[ 4a = 3a + 3 \]
Rearranging gives:
\[ 4a - 3a = 3 \]
\[ a = 3 \]

### Step 6: Find the value of d.
Now substituting \(a = 3\) back into Equation 3:
\[ d = \frac{2(3)}{3} = 2 \]

### Step 7: Write the nth term formula.
The nth term of an A.P. is given by:
\[ T_n = a + (n - 1)d \]
Substituting the values of \(a\) and \(d\):
\[ T_n = 3 + (n - 1) \cdot 2 \]
Simplifying this:
\[ T_n = 3 + 2n - 2 = 2n + 1 \]

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