Write the first six terms of an A.P. in which
a=98 , d=-3

Answer

To find the first six terms of an arithmetic progression (A.P.) where the first term \( a = 98 \) and the common difference \( d = -3 \), we can use the formula for the \( n \)-th term of an A.P.:

\[ A_n = a + (n-1) \cdot d \]

### Step-by-Step Solution:

1. **First Term \( A_1 \)**:
\[ A_1 = a = 98 \]

2. **Second Term \( A_2 \)**:
\[ A_2 = a + (2-1) \cdot d = 98 + 1 \cdot (-3) = 98 - 3 = 95 \]

3. **Third Term \( A_3 \)**:
\[ A_3 = a + (3-1) \cdot d = 98 + 2 \cdot (-3) = 98 - 6 = 92 \]

4. **Fourth Term \( A_4 \)**:
\[ A_4 = a + (4-1) \cdot d = 98 + 3 \cdot (-3) = 98 - 9 = 89 \]

5. **Fifth Term \( A_5 \)**:
\[ A_5 = a + (5-1) \cdot d = 98 + 4 \cdot (-3) = 98 - 12 = 86 \]

6. **Sixth Term \( A_6 \)**:
\[ A_6 = a + (6-1) \cdot d = 98 + 5 \cdot (-3) = 98 - 15 = 83 \]

### Summary of the First Six Terms:
- \( A_1 = 98 \)
- \( A_2 = 95 \)
- \( A_3 = 92 \)
- \( A_4 = 89 \)
- \( A_5 = 86 \)
- \( A_6 = 83 \)

Thus, the first six terms of the A.P. are: **98, 95, 92, 89, 86, 83**.