The common difference d of the A.P. in which `T_(7) = 9 and T_(1)T_(2)T_(7)` is least is

Options

A. `33//2`
B. `5//4`
C. `33//20`
D. `7//3`

Answer

To solve the problem, we need to find the common difference \( d \) of the arithmetic progression (A.P.) given that \( T_7 = 9 \) and the product \( T_1 \times T_2 \times T_7 \) is minimized.

### Step-by-Step Solution:

1. **Understanding the Terms of A.P.**:
The \( n \)-th term of an A.P. can be expressed as:
\[ T_n = A + (n - 1) \cdot d \]
where \( A \) is the first term and \( d \) is the common difference.

2. **Finding \( T_7 \)**:
Given \( T_7 = 9 \):
\[ T_7 = A + (7 - 1) \cdot d = A + 6d \]
Setting this equal to 9:
\[ A + 6d = 9 \quad \text{(1)} \]

3. **Finding \( T_1 \) and \( T_2 \)**:
The first term \( T_1 \) is:
\[ T_1 = A + (1 - 1) \cdot d = A \]
The second term \( T_2 \) is:
\[ T_2 = A + (2 - 1) \cdot d = A + d \]

4. **Calculating the Product**:
We need to calculate the product \( T_1 \times T_2 \times T_7 \):
\[ T_1 \times T_2 \times T_7 = A \times (A + d) \times 9 \]
Substituting \( A \) from equation (1):
\[ A = 9 - 6d \]
Therefore:
\[ T_1 \times T_2 \times T_7 = (9 - 6d) \times (9 - 6d + d) \times 9 \]
Simplifying \( T_2 \):
\[ T_2 = (9 - 6d) + d = 9 - 5d \]
Thus:
\[ T_1 \times T_2 \times T_7 = (9 - 6d) \times (9 - 5d) \times 9 \]

5. **Finding the Minimum**:
To minimize \( (9 - 6d)(9 - 5d) \), we can expand this expression:
\[ (9 - 6d)(9 - 5d) = 81 - 45d - 54d + 30d^2 = 81 - 99d + 30d^2 \]
We need to differentiate this expression with respect to \( d \) and set it to zero to find the minimum:
\[ \frac{d}{dd}(81 - 99d + 30d^2) = -99 + 60d = 0 \]
Solving for \( d \):
\[ 60d = 99 \implies d = \frac{99}{60} = \frac{33}{20} \]

6. **Conclusion**:
The common difference \( d \) of the A.P. is:
\[ \boxed{\frac{33}{20}} \]