If `S_n=n P+(n(n-1))/2Q ,w h e r eS_n` denotes the sum of the first `n` terms of an A.P., then find the common difference.

Explanation

To find the common difference of the arithmetic progression (A.P.) given the sum of the first \( n \) terms, we start with the equation provided:

\[ S_n = nP + \frac{n(n-1)}{2}Q \]

### Step 1: Rewrite the equation

We can express the sum \( S_n \) in a more standard form. Let's factor out \( n \):

\[ S_n = n \left( P + \frac{(n-1)}{2}Q \right) \]

### Step 2: Compare with the standard formula for the sum of the first \( n \) terms of an A.P.

The standard 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 3: Set the two expressions equal

From our rewritten equation, we have:

\[ S_n = n \left( P + \frac{(n-1)}{2}Q \right) \]

We can also express this as:

\[ S_n = \frac{n}{2} \left( 2 \left( P + \frac{(n-1)}{2}Q \right) \right) \]

Now, we compare this with the standard formula:

\[ \frac{n}{2} \left( 2a + (n-1)d \right) \]

### Step 4: Identify coefficients

From the two expressions, we can identify:

1. \( 2a = 2P \) which implies \( a = P \)
2. \( (n-1)d = (n-1)Q \)

From the second equation, we can equate the coefficients:

\[ d = Q \]

### Conclusion

Thus, the common difference \( d \) of the A.P. is:

\[ d = Q \]