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: Calculate \( S_1 \)
To find \( S_1 \), substitute \( n = 1 \) into the formula for \( S_n \):

\[ S_1 = 1^2 a + \frac{1}{4}(1-1)d = a + 0 = a \]

### Step 2: Calculate \( S_2 \)
Next, substitute \( n = 2 \) into the formula for \( S_n \):

\[ S_2 = 2^2 a + \frac{2}{4}(2-1)d = 4a + \frac{2}{4}d = 4a + \frac{1}{2}d \]

### Step 3: Find \( a_1 \) and \( a_2 \)
From the properties of an arithmetic progression:
- \( a_1 = S_1 = a \)
- \( a_2 = S_2 - S_1 \)

Calculating \( a_2 \):

\[ a_2 = S_2 - S_1 = \left(4a + \frac{1}{2}d\right) - a = 3a + \frac{1}{2}d \]

### Step 4: Calculate the common difference \( d \)
The common difference \( d \) is given by:

\[ d = a_2 - a_1 = (3a + \frac{1}{2}d) - a \]

Simplifying this:

\[ d = 3a + \frac{1}{2}d - a = 2a + \frac{1}{2}d \]

### Step 5: Rearranging to solve for \( d \)
Now, isolate \( d \):

\[ d - \frac{1}{2}d = 2a \]

This simplifies to:

\[ \frac{1}{2}d = 2a \]

Multiplying both sides by 2:

\[ d = 4a \]

### Conclusion
Thus, the common difference \( d \) is:

\[ \boxed{4a} \]

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