If the sums of the first 8 and 19 terms of an A.P. are 64 and 361 respectively, find (i) the common difference and (ii) the sum of n terms of the series.

Answer

To solve the problem, we need to find the common difference \( D \) and the sum of \( n \) terms of the arithmetic progression (A.P.) given the sums of the first 8 and 19 terms.

### Step-by-Step Solution:

1. **Understanding the Sum of Terms in an A.P.**:
The formula for the sum of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2A + (n - 1)D) \]
where \( A \) is the first term and \( D \) is the common difference.

2. **Using Given Information**:
We know:
- \( S_8 = 64 \) (sum of the first 8 terms)
- \( S_{19} = 361 \) (sum of the first 19 terms)

3. **Setting Up Equations**:
For \( S_8 \):
\[ S_8 = \frac{8}{2} \times (2A + (8 - 1)D) = 64 \]
Simplifying this:
\[ 4 \times (2A + 7D) = 64 \]
\[ 2A + 7D = 16 \quad \text{(Equation 1)} \]

For \( S_{19} \):
\[ S_{19} = \frac{19}{2} \times (2A + (19 - 1)D) = 361 \]
Simplifying this:
\[ 19 \times (2A + 18D) = 722 \]
\[ 2A + 18D = 38 \quad \text{(Equation 2)} \]

4. **Solving the Equations**:
Now we have two equations:
- Equation 1: \( 2A + 7D = 16 \)
- Equation 2: \( 2A + 18D = 38 \)

We can subtract Equation 1 from Equation 2:
\[ (2A + 18D) - (2A + 7D) = 38 - 16 \]
This simplifies to:
\[ 11D = 22 \]
Dividing both sides by 11:
\[ D = 2 \]

5. **Finding the First Term \( A \)**:
Now, substitute \( D = 2 \) back into Equation 1:
\[ 2A + 7(2) = 16 \]
\[ 2A + 14 = 16 \]
\[ 2A = 2 \]
\[ A = 1 \]

6. **Finding the Sum of \( n \) Terms**:
Now that we have \( A = 1 \) and \( D = 2 \), we can find the sum of \( n \) terms:
\[ S_n = \frac{n}{2} \times (2A + (n - 1)D) \]
Substituting \( A \) and \( D \):
\[ S_n = \frac{n}{2} \times (2 \times 1 + (n - 1) \times 2) \]
\[ S_n = \frac{n}{2} \times (2 + 2n - 2) \]
\[ S_n = \frac{n}{2} \times 2n \]
\[ S_n = n^2 \]

### Final Answers:
- (i) The common difference \( D = 2 \)
- (ii) The sum of \( n \) terms \( S_n = n^2 \)