If `S_(n)=n^(2)a+(n)/(4)(n-1)d` is the sum of the first n terms of an arithmetic progression, then the common difference is

Options

A. `a+2d`
B. `2a+d`
C. `(a+d)/(2)`
D. `2a+(d)/(2)`

Answer

To find the common difference \( d \) of the arithmetic progression given the sum of the first \( n \) terms \( S_n = n^2 a + \frac{n}{4}(n-1)d \), we can follow these steps:

### Step 1: Understand the formula for the sum of the first \( n \) terms of an AP
The sum of the first \( n \) terms \( S_n \) of an arithmetic progression can be expressed as:
\[ S_n = \frac{n}{2} [2a + (n-1)d] \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Set up the equations
We are given:
\[ S_n = n^2 a + \frac{n}{4}(n-1)d \]
We will find \( S_1 \) and \( S_2 \) to help us derive \( d \).

### Step 3: Calculate \( S_1 \)
For \( n = 1 \):
\[ S_1 = 1^2 a + \frac{1}{4}(1-1)d = a \]

### Step 4: Calculate \( S_2 \)
For \( n = 2 \):
\[ S_2 = 2^2 a + \frac{2}{4}(2-1)d = 4a + \frac{1}{2}d \]

### Step 5: Use the relationship to find \( d \)
The common difference \( d \) can be expressed as:
\[ d = S_2 - 2S_1 \]
Substituting the values we found:
\[ d = \left(4a + \frac{1}{2}d\right) - 2(a) \]
This simplifies to:
\[ d = 4a + \frac{1}{2}d - 2a \]
\[ d = 2a + \frac{1}{2}d \]

### Step 6: Solve for \( d \)
Rearranging gives:
\[ d - \frac{1}{2}d = 2a \]
\[ \frac{1}{2}d = 2a \]
Multiplying both sides by 2:
\[ d = 4a \]

### Conclusion
Thus, the common difference \( d \) is:
\[ \boxed{4a} \]