Sum of first n terms of an A.P. whose last term is
l and common difference is d, is

Options

A. `(n)/(2) [ l + (n-1)d]`
B. `(n)/(2) [ l-(n-1)d]`
C. `(n)/(2) [2l + (n-1) d]`
D. `(n)/(2) [2l-(n-1)d]`

Answer

To find the sum of the first n terms of an arithmetic progression (A.P.) where the last term is \( l \) and the common difference is \( d \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formula for the nth term of an A.P.**:
The nth term \( T_n \) of an A.P. can be expressed as:
\[ T_n = A + (n - 1)d \]
where \( A \) is the first term, \( n \) is the number of terms, and \( d \) is the common difference.

2. **Set the nth term equal to the last term \( l \)**:
Since the last term is given as \( l \), we can set:
\[ l = A + (n - 1)d \]

3. **Rearrange to find the first term \( A \)**:
Rearranging the equation gives:
\[ A = l - (n - 1)d \]

4. **Use the formula for the sum of the first n terms of an A.P.**:
The sum \( S_n \) of the first n terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2A + (n - 1)d) \]

5. **Substitute the expression for \( A \)**:
Substitute \( A \) from step 3 into the sum formula:
\[ S_n = \frac{n}{2} \times \left(2(l - (n - 1)d) + (n - 1)d\right) \]

6. **Simplify the expression**:
Expanding the expression inside the parentheses:
\[ S_n = \frac{n}{2} \times \left(2l - 2(n - 1)d + (n - 1)d\right) \]
This simplifies to:
\[ S_n = \frac{n}{2} \times \left(2l - (n - 1)d\right) \]

7. **Final formula for the sum of the first n terms**:
Thus, the sum of the first n terms of the A.P. is:
\[ S_n = \frac{n}{2} \times \left(2l - (n - 1)d\right) \]