If the sum of first n terms of an A.P. is ` an^(2) + bn `
and ` n^(th) ` term is An + B then

Options

A. A= 2a
B. A = a
C. a = 2A
D. A = 3a

Answer

To solve the problem, we need to analyze the given information about the arithmetic progression (A.P.) and derive the value of \( a \) in terms of \( a \).

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The sum of the first \( n \) terms of an A.P. is given as \( S_n = an^2 + bn \).
- The \( n^{th} \) term of the A.P. is given as \( T_n = An + B \).

2. **Use the Formula for the Sum of the First \( n \) Terms**:
- The formula for the sum of the first \( n \) terms of an A.P. is:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
- We can also express this as:
\[ S_n = \frac{n}{2} \left(a + (a + (n-1)d)\right) \]

3. **Set the Two Expressions for \( S_n \) Equal**:
- From the problem, we have:
\[ an^2 + bn = \frac{n}{2} \left(2a + (n-1)d\right) \]

4. **Express the \( n^{th} \) Term**:
- The \( n^{th} \) term \( T_n \) is given by:
\[ T_n = a + (n-1)d \]
- According to the problem, this is equal to \( An + B \).

5. **Substituting \( T_n \) into the Sum Expression**:
- We can express the sum \( S_n \) in terms of \( T_n \):
\[ S_n = n \cdot \frac{T_n + T_1}{2} \]
- Where \( T_1 = a \).

6. **Equate the Two Forms of \( S_n \)**:
- Substitute \( T_n \) into the sum expression:
\[ an^2 + bn = n \cdot \frac{(An + B) + a}{2} \]

7. **Simplify the Equation**:
- This gives:
\[ an^2 + bn = \frac{n}{2} (An + B + a) \]
- Expanding the right-hand side:
\[ an^2 + bn = \frac{An^2}{2} + \frac{(B + a)n}{2} \]

8. **Comparing Coefficients**:
- From the left side \( an^2 + bn \) and the right side \( \frac{An^2}{2} + \frac{(B + a)n}{2} \), we can equate coefficients:
- For \( n^2 \): \( a = \frac{A}{2} \)
- For \( n \): \( b = \frac{B + a}{2} \)

9. **Solve for \( a \)**:
- Rearranging the first equation gives:
\[ A = 2a \]
- Thus, \( a \) is expressed in terms of itself, which confirms the relationship.

### Final Result:
The value of \( a \) in terms of itself is simply \( a \).