The `6th` term of an `AP` is equal to `2`, the value of the common difference of the `AP` which makes the product `a_1a_4a_5` least is given by

Options

A. `(8)/(5)`
B. `(5)/(4)`
C. `(2)/(3)`
D. None of these

Explanation

To solve the problem step by step, we will follow the logic presented in the video transcript and derive the solution systematically.

### Step 1: Understand the given information
We know that the 6th term of an arithmetic progression (AP) is equal to 2. The formula for the nth term of an AP is given by:
\[ a_n = a_1 + (n-1)d \]
For the 6th term:
\[ a_6 = a_1 + 5d = 2 \]

### Step 2: Express \( a_1 \) in terms of \( d \)
From the equation \( a_1 + 5d = 2 \), we can express \( a_1 \) as:
\[ a_1 = 2 - 5d \]

### Step 3: Write the product \( P = a_1 \cdot a_4 \cdot a_5 \)
The terms \( a_4 \) and \( a_5 \) can be expressed as follows:
\[ a_4 = a_1 + 3d = (2 - 5d) + 3d = 2 - 2d \]
\[ a_5 = a_1 + 4d = (2 - 5d) + 4d = 2 - d \]
Now, we can write the product \( P \):
\[ P = a_1 \cdot a_4 \cdot a_5 = (2 - 5d)(2 - 2d)(2 - d) \]

### Step 4: Expand the product
Now, we will expand \( P \):
\[ P = (2 - 5d)(2 - 2d)(2 - d) \]
First, calculate \( (2 - 5d)(2 - 2d) \):
\[ (2 - 5d)(2 - 2d) = 4 - 4d - 10d + 10d^2 = 4 - 14d + 10d^2 \]
Now multiply this result by \( (2 - d) \):
\[ P = (4 - 14d + 10d^2)(2 - d) = 8 - 4d - 28d + 14d^2 + 20d^2 - 10d^3 \]
Combining like terms, we get:
\[ P = 8 - 32d + 34d^2 - 10d^3 \]

### Step 5: Find the value of \( d \) that minimizes \( P \)
To minimize \( P \), we will differentiate it with respect to \( d \):
\[ \frac{dP}{dd} = -32 + 68d - 30d^2 \]
Setting the derivative to zero for critical points:
\[ -30d^2 + 68d - 32 = 0 \]
This is a quadratic equation in \( d \). We can use the quadratic formula:
\[ d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-68 \pm \sqrt{68^2 - 4 \cdot (-30) \cdot (-32)}}{2 \cdot (-30)} \]
Calculating the discriminant:
\[ 68^2 = 4624 \quad \text{and} \quad 4 \cdot 30 \cdot 32 = 3840 \]
Thus:
\[ d = \frac{68 \pm \sqrt{4624 - 3840}}{-60} = \frac{68 \pm \sqrt{784}}{-60} = \frac{68 \pm 28}{-60} \]
This gives us two solutions:
\[ d_1 = \frac{96}{-60} = -\frac{8}{5}, \quad d_2 = \frac{40}{-60} = -\frac{2}{3} \]

### Step 6: Determine which value minimizes \( P \)
We need to check the second derivative to determine which value minimizes \( P \):
\[ \frac{d^2P}{d^2} = -60d + 68 \]
Evaluating at \( d = \frac{2}{3} \):
\[ \frac{d^2P}{d^2} = -60 \cdot \frac{2}{3} + 68 = -40 + 68 = 28 > 0 \quad \text{(indicates a minimum)} \]
Evaluating at \( d = \frac{8}{5} \):
\[ \frac{d^2P}{d^2} = -60 \cdot \frac{8}{5} + 68 = -96 + 68 = -28 < 0 \quad \text{(indicates a maximum)} \]

### Conclusion
Thus, the value of the common difference \( d \) that minimizes the product \( P = a_1 \cdot a_4 \cdot a_5 \) is:
\[ \boxed{\frac{2}{3}} \]