The first and last term of an A.P. are a and l respectively. If S be the sum of all the terms of the A.P., them the common difference is

Options

A. `(l^(2)-a^(2))/(2S-(l+a))`
B. `(l^(2)-a^(2))/(2S-(l-a))`
C. `(l^(2)+a^(2))/(2S+(l+a))`
D. `(l^(2)+a^(2))/(2S-(l+a))`

Explanation

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

### Step 1: Understand the relationship between the terms in an A.P.
In an A.P., the last term \( l \) can be expressed in terms of the first term \( a \), the number of terms \( n \), and the common difference \( d \) as:
\[ l = a + (n - 1)d \]

### Step 2: Rearrange the equation to find \( d \)
From the equation above, we can rearrange it to express \( d \):
\[ l - a = (n - 1)d \]
Thus,
\[ d = \frac{l - a}{n - 1} \]

### Step 3: Express the number of terms \( n \) in terms of \( S \)
The sum \( S \) of the first \( n \) terms of an A.P. can be expressed as:
\[ S = \frac{n}{2} (a + l) \]
From this, we can solve for \( n \):
\[ 2S = n(a + l) \implies n = \frac{2S}{a + l} \]

### Step 4: Substitute \( n \) in the expression for \( d \)
Now, substitute the expression for \( n \) back into the equation for \( d \):
\[ d = \frac{l - a}{\frac{2S}{a + l} - 1} \]

### Step 5: Simplify the expression for \( d \)
To simplify, we rewrite the denominator:
\[ d = \frac{l - a}{\frac{2S - (a + l)}{a + l}} = \frac{(l - a)(a + l)}{2S - (a + l)} \]

### Step 6: Use the difference of squares
Recognizing that \( (l - a)(a + l) = l^2 - a^2 \), we can rewrite \( d \) as:
\[ d = \frac{l^2 - a^2}{2S - (a + l)} \]

### Final Result
Thus, the common difference \( d \) is given by:
\[ d = \frac{l^2 - a^2}{2S - (a + l)} \]