The sixth term of an AP is 2, and its common difference is greater than one. The value of the common difference of the progression so that the product of the first, fourth and fifth terms is greatest is

Options

A. `8//5`
B. `2//3`
C. `5//8`
D. `3//2`

Explanation

To solve the problem, we need to find the common difference \( d \) of an arithmetic progression (AP) such that the product of the first, fourth, and fifth terms is maximized. Given that the sixth term of the AP is 2, we can follow these steps:

### Step 1: Define the terms of the AP
Let \( a \) be the first term and \( d \) be the common difference. The \( n \)-th term of an AP is given by the formula:
\[ a_n = a + (n - 1)d \]
Thus, the terms we need are:
- First term: \( a_1 = a \)
- Fourth term: \( a_4 = a + 3d \)
- Fifth term: \( a_5 = a + 4d \)

### Step 2: Use the information about the sixth term
We know that the sixth term \( a_6 = 2 \). Using the formula for the sixth term:
\[ a_6 = a + 5d = 2 \]
From this, we can express \( a \) in terms of \( d \):
\[ a = 2 - 5d \tag{1} \]

### Step 3: Write the product of the first, fourth, and fifth terms
The product \( P \) of the first, fourth, and fifth terms is:
\[ P = a_1 \cdot a_4 \cdot a_5 = a \cdot (a + 3d) \cdot (a + 4d) \]
Substituting \( a \) from equation (1):
\[ P = (2 - 5d) \cdot ((2 - 5d) + 3d) \cdot ((2 - 5d) + 4d) \]
Simplifying the terms:
\[ P = (2 - 5d) \cdot (2 - 2d) \cdot (2 - d) \]

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

### Step 5: Find the maximum value of \( P \)
To find the maximum value of \( P \), we differentiate \( P \) with respect to \( d \) and set the derivative to zero:
\[ \frac{dP}{dd} = -32 + 68d - 30d^2 \]
Setting the derivative equal to zero:
\[ -30d^2 + 68d - 32 = 0 \]
Multiplying through by -1:
\[ 30d^2 - 68d + 32 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ d = \frac{68 \pm \sqrt{(-68)^2 - 4 \cdot 30 \cdot 32}}{2 \cdot 30} \]
Calculating the discriminant:
\[ d = \frac{68 \pm \sqrt{4624 - 3840}}{60} \]
\[ d = \frac{68 \pm \sqrt{784}}{60} \]
\[ d = \frac{68 \pm 28}{60} \]
Calculating the two possible values:
1. \( d = \frac{96}{60} = \frac{8}{5} \)
2. \( d = \frac{40}{60} = \frac{2}{3} \)

### Step 7: Choose the valid solution
Since we are given that the common difference \( d \) is greater than 1, we select:
\[ d = \frac{8}{5} \]

### Final Answer
The value of the common difference \( d \) that maximizes the product of the first, fourth, and fifth terms is:
\[ \boxed{\frac{8}{5}} \]