If for an A.P.
(i) a=6, d=3, find `S_(10)`
(ii) `a=6, d=3`, find `S_(6)`

Answer

To solve the problem, we will use the formula for the sum of the first \( n \) terms of an arithmetic progression (A.P.), which is given by:

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

where:
- \( S_n \) is the sum of the first \( n \) terms,
- \( a \) is the first term,
- \( d \) is the common difference,
- \( n \) is the number of terms.

### Part (i): Finding \( S_{10} \)

1. **Identify the values**:
- Given \( a = 6 \), \( d = 3 \), and \( n = 10 \).

2. **Substitute the values into the formula**:
\[ S_{10} = \frac{10}{2} \left(2 \times 6 + (10 - 1) \times 3\right) \]

3. **Calculate \( \frac{10}{2} \)**:
\[ S_{10} = 5 \left(2 \times 6 + 9 \times 3\right) \]

4. **Calculate \( 2 \times 6 \)**:
\[ 2 \times 6 = 12 \]

5. **Calculate \( 9 \times 3 \)**:
\[ 9 \times 3 = 27 \]

6. **Add the results**:
\[ S_{10} = 5 \left(12 + 27\right) = 5 \times 39 \]

7. **Calculate \( 5 \times 39 \)**:
\[ S_{10} = 195 \]

### Part (ii): Finding \( S_{6} \)

1. **Identify the values**:
- Given \( a = 6 \), \( d = 3 \), and \( n = 6 \).

2. **Substitute the values into the formula**:
\[ S_{6} = \frac{6}{2} \left(2 \times 6 + (6 - 1) \times 3\right) \]

3. **Calculate \( \frac{6}{2} \)**:
\[ S_{6} = 3 \left(2 \times 6 + 5 \times 3\right) \]

4. **Calculate \( 2 \times 6 \)**:
\[ 2 \times 6 = 12 \]

5. **Calculate \( 5 \times 3 \)**:
\[ 5 \times 3 = 15 \]

6. **Add the results**:
\[ S_{6} = 3 \left(12 + 15\right) = 3 \times 27 \]

7. **Calculate \( 3 \times 27 \)**:
\[ S_{6} = 81 \]

### Final Results
- \( S_{10} = 195 \)
- \( S_{6} = 81 \)