If the first, second and the last terms of an A.P. are a,b,c respectively, then the sum of the A.P. is

Options

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

Explanation

To find the sum of an arithmetic progression (A.P.) where the first term is \( a \), the second term is \( b \), and the last term is \( c \), we can follow these steps:

### Step 1: Identify the common difference
The common difference \( d \) of an A.P. can be calculated using the first and second terms:
\[ d = b - a \]

### Step 2: Express the last term using the common difference
The last term \( c \) of an A.P. can be expressed in terms of the first term \( a \), the number of terms \( n \), and the common difference \( d \):
\[ c = a + (n - 1)d \]
Substituting the value of \( d \) from Step 1:
\[ c = a + (n - 1)(b - a) \]

### Step 3: Rearrange to find \( n \)
Rearranging the equation gives:
\[ c - a = (n - 1)(b - a) \]
Now, solving for \( n \):
\[ n - 1 = \frac{c - a}{b - a} \]
\[ n = \frac{c - a}{b - a} + 1 \]

### Step 4: Simplify \( n \)
We can combine the terms:
\[ n = \frac{c - a + b - a}{b - a} = \frac{b + c - 2a}{b - a} \]

### Step 5: Use the formula for the sum of the first \( n \) terms of an A.P.
The formula for the sum \( S_n \) of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} (a + c) \]
Substituting the value of \( n \) from Step 4:
\[ S_n = \frac{\frac{b + c - 2a}{b - a}}{2} (a + c) \]

### Step 6: Final expression for the sum
Thus, we can write the sum of the A.P. as:
\[ S_n = \frac{(b + c - 2a)(a + c)}{2(b - a)} \]

### Conclusion
The sum of the A.P. with first term \( a \), second term \( b \), and last term \( c \) is:
\[ S_n = \frac{(b + c - 2a)(a + c)}{2(b - a)} \]