The first second, and middle terms of an AP are a, b, c respectively. Then , their sum is equal to

Options

A. `(2(c-a))/(b-a)`
B. `(2c(c-a))/(b-a) + c`
C. `(2c(b-a))/(c-a)`
D. None of these

Answer

To solve the problem, we need to find the sum of the first, second, and middle terms of an arithmetic progression (AP) where the first term is \( a \), the second term is \( b \), and the middle term is \( c \).

### Step-by-step Solution:

1. **Understanding the terms in AP**:
- The first term \( T_1 = a \)
- The second term \( T_2 = b \)
- The middle term \( T_m = c \)

2. **Finding the common difference**:
- The common difference \( d \) in an AP can be expressed as:
\[ d = T_2 - T_1 = b - a \]

3. **Identifying the number of terms**:
- Since we are given a middle term \( c \), the total number of terms \( n \) in the AP must be odd. The middle term is given by the formula:
\[ T_m = T_1 + \left( \frac{n-1}{2} \right) d \]
- Substituting the known values:
\[ c = a + \left( \frac{n-1}{2} \right)(b - a) \]

4. **Rearranging the equation**:
- Rearranging the equation gives:
\[ c - a = \left( \frac{n-1}{2} \right)(b - a) \]
- Thus:
\[ n - 1 = \frac{2(c - a)}{b - a} \]
- Therefore:
\[ n = \frac{2(c - a)}{b - a} + 1 \]

5. **Calculating the sum of the terms**:
- The sum \( S \) of the first \( n \) terms of an AP is given by:
\[ S_n = \frac{n}{2} \times (T_1 + T_n) \]
- Here, \( T_n = a + (n-1)d \):
\[ T_n = a + (n-1)(b - a) \]
- Therefore, the sum becomes:
\[ S_n = \frac{n}{2} \times \left( a + \left( a + (n-1)(b - a) \right) \right) \]
- Simplifying gives:
\[ S_n = \frac{n}{2} \times \left( 2a + (n-1)(b - a) \right) \]

6. **Substituting \( n \)**:
- Substitute \( n = \frac{2(c - a)}{b - a} + 1 \) into the sum formula:
\[ S_n = \frac{\left(\frac{2(c - a)}{b - a} + 1\right)}{2} \times \left( 2a + \left(\frac{2(c - a)}{b - a}\right)(b - a) \right) \]
- This simplifies to:
\[ S_n = \frac{(c - a) + \frac{b - a}{2}}{b - a} \times \left( 2a + 2(c - a) \right) \]

7. **Final expression**:
- After simplification, we find:
\[ S = 2c \]

### Conclusion:
The sum of the first, second, and middle terms of the AP is equal to \( 2c \).