The value of the common difference of an A.P. which makes `T_(1) T_(2) T_(7)` least, given that `T_(7) = 9`, is

Options

A. `d = (33)/(2)`
B. `d = (5)/(4)`
C. `d = (33)/(20)`
D. none of these

Answer

To solve the problem, we need to find the value of the common difference \( D \) of an arithmetic progression (A.P.) that minimizes the sum \( T_1 + T_2 + T_7 \), given that \( T_7 = 9 \).

### Step-by-Step Solution:

1. **Define the Terms of the A.P.**:
- Let \( A \) be the first term of the A.P.
- The terms can be expressed as:
- \( T_1 = A \)
- \( T_2 = A + D \)
- \( T_7 = A + 6D \)

2. **Use the Given Information**:
- We know that \( T_7 = 9 \).
- Therefore, we can write:
\[ A + 6D = 9 \]
- From this, we can express \( A \) in terms of \( D \):
\[ A = 9 - 6D \]

3. **Express \( T_1 \) and \( T_2 \)**:
- Substitute \( A \) into the expressions for \( T_1 \) and \( T_2 \):
- \( T_1 = 9 - 6D \)
- \( T_2 = (9 - 6D) + D = 9 - 5D \)

4. **Formulate the Function to Minimize**:
- We need to minimize the sum:
\[ S = T_1 + T_2 + T_7 = (9 - 6D) + (9 - 5D) + 9 \]
- Simplifying this gives:
\[ S = 27 - 11D \]

5. **Differentiate the Function**:
- To find the minimum, we differentiate \( S \) with respect to \( D \):
\[ \frac{dS}{dD} = -11 \]
- Since the derivative is constant, it indicates that \( S \) is a linear function decreasing in \( D \).

6. **Determine the Minimum Value**:
- As \( D \) increases, \( S \) decreases. Therefore, to minimize \( S \), we should consider the smallest value of \( D \) that keeps \( T_1 \) non-negative (since \( T_1 \) represents a term in the A.P.):
- Setting \( T_1 \geq 0 \):
\[ 9 - 6D \geq 0 \implies D \leq \frac{9}{6} = \frac{3}{2} \]

7. **Conclusion**:
- The common difference \( D \) that minimizes \( T_1 + T_2 + T_7 \) while satisfying the condition \( T_7 = 9 \) is:
\[ D = \frac{3}{2} \]

### Final Answer:
The value of the common difference \( D \) is \( \frac{3}{2} \).