The sixth term of an `A.P., a_1,a_2,a_3,.............,a_n` is `2`. If the quantity `a_1a_4a_5`, is minimum then then the common difference of the `A.P.`

Options

A. `x=8//5`
B. `x=5//4`
C. `x=2//3`
D. `x=4//5`

Answer

To solve the problem step-by-step, we need to find the common difference of the arithmetic progression (A.P.) given that the sixth term is 2 and the product \( a_1 a_4 a_5 \) is minimized.

### Step 1: Express the terms of the A.P.
The \( n \)-th term of an A.P. can be expressed as:
\[ a_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Write the sixth term in terms of \( a \) and \( d \)
Given that the sixth term \( a_6 = 2 \):
\[ a_6 = a + 5d = 2 \]
From this, we can express \( a \) in terms of \( d \):
\[ a = 2 - 5d \]

### Step 3: Write the terms \( a_1, a_4, a_5 \)
Now, we can express \( a_1, a_4, \) and \( a_5 \) in terms of \( a \) and \( d \):
- \( a_1 = a = 2 - 5d \)
- \( a_4 = a + 3d = (2 - 5d) + 3d = 2 - 2d \)
- \( a_5 = a + 4d = (2 - 5d) + 4d = 2 - d \)

### Step 4: Write the product \( a_1 a_4 a_5 \)
Now we need to find the product:
\[ P = a_1 a_4 a_5 = (2 - 5d)(2 - 2d)(2 - d) \]

### Step 5: Expand the product
We will expand this product step by step. First, expand \( (2 - 5d)(2 - 2d) \):
\[ (2 - 5d)(2 - 2d) = 4 - 4d - 10d + 10d^2 = 4 - 14d + 10d^2 \]
Now multiply this result by \( (2 - d) \):
\[ P = (4 - 14d + 10d^2)(2 - d) \]
Expanding this:
\[ P = 8 - 4d - 28d + 14d^2 + 20d^2 - 10d^3 = 8 - 32d + 34d^2 - 10d^3 \]

### Step 6: Find the critical points
To minimize \( P \), we take the derivative \( P' \) and set it to zero:
\[ P' = -32 + 68d - 30d^2 \]
Setting \( P' = 0 \):
\[ -30d^2 + 68d - 32 = 0 \]
Dividing through by -2:
\[ 15d^2 - 34d + 16 = 0 \]

### Step 7: Use the quadratic formula
Using the quadratic formula \( d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ d = \frac{34 \pm \sqrt{(-34)^2 - 4 \cdot 15 \cdot 16}}{2 \cdot 15} \]
Calculating the discriminant:
\[ 34^2 = 1156, \quad 4 \cdot 15 \cdot 16 = 960 \]
Thus:
\[ d = \frac{34 \pm \sqrt{1156 - 960}}{30} = \frac{34 \pm \sqrt{196}}{30} = \frac{34 \pm 14}{30} \]
This gives us two potential solutions:
\[ d = \frac{48}{30} = \frac{8}{5} \quad \text{and} \quad d = \frac{20}{30} = \frac{2}{3} \]

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

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