In an A.P. if `t_(7) = 15` then the value of common difference d that would make `t_(2) t_(7) t_(12)` greatest is

Options

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

Answer

To solve the problem step by step, we need to find the value of the common difference \( d \) that maximizes the product \( t_2 t_7 t_{12} \) in an arithmetic progression (A.P.) where \( t_7 = 15 \).

### Step 1: Understand the terms of the A.P.
In an arithmetic progression, the \( n \)-th term is given by:
\[ t_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Write expressions for \( t_2 \), \( t_7 \), and \( t_{12} \)
From the formula for the \( n \)-th term:
- For \( t_7 \):
\[ t_7 = a + 6d = 15 \]
- For \( t_2 \):
\[ t_2 = a + d \]
- For \( t_{12} \):
\[ t_{12} = a + 11d \]

### Step 3: Express \( a \) in terms of \( d \)
From the equation \( t_7 = 15 \):
\[ a + 6d = 15 \implies a = 15 - 6d \]

### Step 4: Substitute \( a \) into \( t_2 \) and \( t_{12} \)
Now substitute \( a \) in the expressions for \( t_2 \) and \( t_{12} \):
- For \( t_2 \):
\[ t_2 = (15 - 6d) + d = 15 - 5d \]
- For \( t_{12} \):
\[ t_{12} = (15 - 6d) + 11d = 15 + 5d \]

### Step 5: Write the product \( P = t_2 t_7 t_{12} \)
Now we can express the product:
\[ P = t_2 \cdot t_7 \cdot t_{12} = (15 - 5d) \cdot 15 \cdot (15 + 5d) \]

### Step 6: Simplify the product
Let's simplify the product:
\[ P = 15 \cdot (15 - 5d)(15 + 5d) \]
Using the difference of squares:
\[ P = 15 \cdot (15^2 - (5d)^2) = 15 \cdot (225 - 25d^2) = 3375 - 375d^2 \]

### Step 7: Maximize the product
To maximize \( P \), we need to take the derivative and set it to zero:
\[ \frac{dP}{dd} = -750d \]
Setting the derivative to zero for maximization:
\[ -750d = 0 \implies d = 0 \]

### Conclusion
The value of the common difference \( d \) that maximizes the product \( t_2 t_7 t_{12} \) is:
\[ \boxed{0} \]