If for each `n in N, S_(n)=nA+(1)/(2)n(n+1) B` is sum of the first n terms of an A.P., the common difference of the A.P. is

Options

A. A+B
B. A+2B
C. B
D. B/2

Answer

To find the common difference of the arithmetic progression (A.P.) given the sum of the first n terms \( S_n = nA + \frac{1}{2}n(n+1)B \), we can follow these steps:

### Step 1: Understand the formula for the sum of the first n terms of an A.P.
The formula for the sum of the first n terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Rewrite the given expression for \( S_n \)
The given expression for \( S_n \) is:
\[ S_n = nA + \frac{1}{2}n(n+1)B \]
We can expand this to:
\[ S_n = nA + \frac{1}{2}n^2B + \frac{1}{2}nB \]

### Step 3: Compare coefficients from both expressions
From the standard formula, we have:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) = \frac{n}{2} \left(2a + nd - d\right) \]
This can be rewritten as:
\[ S_n = nA + \frac{n^2}{2}d - \frac{nd}{2} \]

### Step 4: Set the coefficients equal
Now we compare the coefficients of \( n^2 \) and \( n \) from both expressions:
1. For \( n^2 \):
\[ \frac{d}{2} = \frac{B}{2} \implies d = B \]

2. For \( n \):
\[ A - \frac{d}{2} = A + \frac{B}{2} \]
This confirms that \( d = B \).

### Step 5: Conclusion
Thus, the common difference \( d \) of the A.P. is equal to \( B \).

### Final Answer
The common difference of the A.P. is:
\[ d = B \]