Write the first six terms of an A.P. in which
`a= 7 (1)/(2) , d= 1 (1)/(2)`

Answer

To find the first six terms of an arithmetic progression (A.P.) where the first term \( a = 7 \frac{1}{2} \) and the common difference \( d = 1 \frac{1}{2} \), we will follow these steps:

### Step 1: Convert Mixed Numbers to Improper Fractions
First, we need to convert the mixed numbers into improper fractions for easier calculations.

- For \( a = 7 \frac{1}{2} \):
\[ a = 7 + \frac{1}{2} = \frac{14}{2} + \frac{1}{2} = \frac{15}{2} \]

- For \( d = 1 \frac{1}{2} \):
\[ d = 1 + \frac{1}{2} = \frac{2}{2} + \frac{1}{2} = \frac{3}{2} \]

### Step 2: Write the Formula for the n-th Term of A.P.
The n-th term of an A.P. can be calculated using the formula:
\[ T_n = a + (n - 1)d \]

### Step 3: Calculate the First Six Terms
Now we will calculate the first six terms using the formula.

1. **First term \( T_1 \)**:
\[ T_1 = a = \frac{15}{2} = 7 \frac{1}{2} \]

2. **Second term \( T_2 \)**:
\[ T_2 = a + d = \frac{15}{2} + \frac{3}{2} = \frac{18}{2} = 9 \]

3. **Third term \( T_3 \)**:
\[ T_3 = a + 2d = \frac{15}{2} + 2 \times \frac{3}{2} = \frac{15}{2} + \frac{6}{2} = \frac{21}{2} = 10 \frac{1}{2} \]

4. **Fourth term \( T_4 \)**:
\[ T_4 = a + 3d = \frac{15}{2} + 3 \times \frac{3}{2} = \frac{15}{2} + \frac{9}{2} = \frac{24}{2} = 12 \]

5. **Fifth term \( T_5 \)**:
\[ T_5 = a + 4d = \frac{15}{2} + 4 \times \frac{3}{2} = \frac{15}{2} + \frac{12}{2} = \frac{27}{2} = 13 \frac{1}{2} \]

6. **Sixth term \( T_6 \)**:
\[ T_6 = a + 5d = \frac{15}{2} + 5 \times \frac{3}{2} = \frac{15}{2} + \frac{15}{2} = \frac{30}{2} = 15 \]

### Step 4: List the First Six Terms
The first six terms of the A.P. are:
1. \( 7 \frac{1}{2} \)
2. \( 9 \)
3. \( 10 \frac{1}{2} \)
4. \( 12 \)
5. \( 13 \frac{1}{2} \)
6. \( 15 \)

### Final Answer
The first six terms of the A.P. are:
- \( 7 \frac{1}{2}, 9, 10 \frac{1}{2}, 12, 13 \frac{1}{2}, 15 \)