For an A.P. if `t_(4)=20` and `t_(7)=32`, find a,d and `t_(n)`.

Answer

To solve the problem, we need to find the first term (A), the common difference (D), and the general term (Tn) of the arithmetic progression (A.P.) given that \( T_4 = 20 \) and \( T_7 = 32 \).

### Step-by-Step Solution:

1. **Write the formulas for T4 and T7**:
- The nth term of an A.P. can be expressed as:
\[ T_n = A + (n-1)D \]
- For \( T_4 \):
\[ T_4 = A + 3D = 20 \quad \text{(Equation 1)} \]
- For \( T_7 \):
\[ T_7 = A + 6D = 32 \quad \text{(Equation 2)} \]

2. **Set up the equations**:
- From Equation 1:
\[ A + 3D = 20 \]
- From Equation 2:
\[ A + 6D = 32 \]

3. **Subtract Equation 1 from Equation 2**:
- This will eliminate A:
\[ (A + 6D) - (A + 3D) = 32 - 20 \]
- Simplifying gives:
\[ 3D = 12 \]

4. **Solve for D**:
- Divide both sides by 3:
\[ D = 4 \]

5. **Substitute D back into Equation 1 to find A**:
- Substitute \( D = 4 \) into Equation 1:
\[ A + 3(4) = 20 \]
- This simplifies to:
\[ A + 12 = 20 \]
- Therefore:
\[ A = 20 - 12 = 8 \]

6. **Write the formula for Tn**:
- The general term \( T_n \) is given by:
\[ T_n = A + (n-1)D \]
- Substitute A and D:
\[ T_n = 8 + (n-1) \cdot 4 \]
- Simplifying gives:
\[ T_n = 8 + 4n - 4 = 4n + 4 \]

### Final Results:
- The first term \( A = 8 \)
- The common difference \( D = 4 \)
- The nth term \( T_n = 4n + 4 \)