If in an A.P. ` a_(7)=9` and if `a_(1)a_(2)a_(7)` is least, then common difference is _____

Answer

To solve the problem step by step, we need to find the common difference \( d \) in an arithmetic progression (A.P.) where \( a_7 = 9 \) and the product \( a_1 \cdot a_2 \cdot a_7 \) is minimized.

### Step 1: Write the formula for the nth term of an A.P.
The nth term of an A.P. is given by:
\[ a_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Find \( a_7 \)
For \( n = 7 \):
\[ a_7 = a + 6d \]
Given \( a_7 = 9 \):
\[ a + 6d = 9 \quad \text{(1)} \]

### Step 3: Express \( a_1 \) and \( a_2 \)
Using the formula for the first and second terms:
- \( a_1 = a \)
- \( a_2 = a + d \)

### Step 4: Write the product \( P = a_1 \cdot a_2 \cdot a_7 \)
Substituting the values of \( a_1 \), \( a_2 \), and \( a_7 \):
\[ P = a \cdot (a + d) \cdot 9 \]
Substituting \( a \) from equation (1):
\[ a = 9 - 6d \]
Thus,
\[ P = (9 - 6d) \cdot (9 - 6d + d) \cdot 9 \]
This simplifies to:
\[ P = (9 - 6d) \cdot (9 - 5d) \cdot 9 \]

### Step 5: Expand the product
Expanding \( P \):
\[ P = 9 \cdot (9 - 6d)(9 - 5d) \]
Calculating \( (9 - 6d)(9 - 5d) \):
\[ = 81 - 45d - 54d + 30d^2 = 81 - 99d + 30d^2 \]
Thus,
\[ P = 9(81 - 99d + 30d^2) = 729 - 891d + 270d^2 \]

### Step 6: Differentiate \( P \) to find critical points
To minimize \( P \), we differentiate with respect to \( d \):
\[ \frac{dP}{dd} = -891 + 540d \]
Setting the derivative equal to zero for critical points:
\[ -891 + 540d = 0 \]
Solving for \( d \):
\[ 540d = 891 \quad \Rightarrow \quad d = \frac{891}{540} = \frac{33}{20} = 1.65 \]

### Step 7: Verify if it's a minimum
To confirm that this critical point is a minimum, we take the second derivative:
\[ \frac{d^2P}{dd^2} = 540 \]
Since \( 540 > 0 \), this indicates that \( P \) has a minimum at \( d = \frac{33}{20} \).

### Conclusion
Thus, the common difference \( d \) is:
\[ \boxed{\frac{33}{20}} \quad \text{or} \quad 1.65 \]