If `a_n` be the term of an A.P. and if `a_7=15`, then the value of the common difference that could makes `a_2a_7a_12` greatest is:

Options

A. 9
B. `9//4`
C. 0
D. 18

Explanation

To solve the problem, we need to find the value of the common difference \( d \) that maximizes the product \( a_2 a_7 a_{12} \) given that \( a_7 = 15 \).

### Step-by-Step Solution:

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

2. **Finding \( a \)**:
Given \( a_7 = 15 \):
\[ a_7 = a + 6d = 15 \]
Rearranging gives:
\[ a = 15 - 6d \]

3. **Expressing \( a_2 \) and \( a_{12} \)**:
Now, we can express \( a_2 \) and \( a_{12} \):
- For \( a_2 \):
\[ a_2 = a + d = (15 - 6d) + d = 15 - 5d \]
- For \( a_{12} \):
\[ a_{12} = a + 11d = (15 - 6d) + 11d = 15 + 5d \]

4. **Finding the Product \( a_2 a_7 a_{12} \)**:
Now, we can write the product:
\[ P = a_2 \cdot a_7 \cdot a_{12} = (15 - 5d) \cdot 15 \cdot (15 + 5d) \]

5. **Simplifying the Product**:
We can simplify this expression:
\[ P = 15 \cdot (15 - 5d)(15 + 5d) \]
Using the difference of squares:
\[ (15 - 5d)(15 + 5d) = 15^2 - (5d)^2 = 225 - 25d^2 \]
Therefore:
\[ P = 15 \cdot (225 - 25d^2) = 3375 - 375d^2 \]

6. **Maximizing the Product**:
The expression \( P = 3375 - 375d^2 \) is a downward-opening parabola (as the coefficient of \( d^2 \) is negative). This means it achieves its maximum value at the vertex.

The vertex of a parabola given by \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \). Here, \( a = -375 \) and \( b = 0 \):
\[ d = -\frac{0}{2 \cdot -375} = 0 \]

7. **Conclusion**:
The value of the common difference \( d \) that maximizes \( a_2 a_7 a_{12} \) is:
\[ \boxed{0} \]