The first and last terms of an A.P. are a and l respectively. If S be the sum of all the terms of the A.P., then 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))`

Answer

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

### Step 1: Write the formula for the sum of an A.P.
The sum \(S\) of the first \(n\) terms of an A.P. can be expressed as:
\[ S = \frac{n}{2} (a + l) \]
where \(n\) is the number of terms, \(a\) is the first term, and \(l\) is the last term.

### Step 2: Solve for \(n\)
From the equation above, we can rearrange it to find \(n\):
\[ n = \frac{2S}{a + l} \]

### Step 3: Use the formula for the last term of an A.P.
The last term \(l\) can also be expressed in terms of the first term \(a\), the number of terms \(n\), and the common difference \(d\):
\[ l = a + (n - 1)d \]

### Step 4: Substitute \(n\) in the last term equation
Substituting the expression for \(n\) from Step 2 into the equation for \(l\):
\[ l = a + \left(\frac{2S}{a + l} - 1\right)d \]

### Step 5: Rearrange to solve for \(d\)
Rearranging the equation gives:
\[ l - a = \left(\frac{2S}{a + l} - 1\right)d \]
\[ l - a = \frac{2S - (a + l)d}{a + l} \]

### Step 6: Solve for \(d\)
Now, we can express \(d\) in terms of \(a\), \(l\), and \(S\):
\[ d = \frac{(l - a)(a + l)}{2S - (l - a)} \]

### Final Expression for Common Difference
Thus, the common difference \(d\) can be expressed as:
\[ d = \frac{(l - a)(a + l)}{2S - (l - a)} \]