The sixth terms of an AP is 2 and its common difference is greater than 1.
Q. What is the common difference of the AP so that the product of the first, fourth and fifth terms is greatest ?

Options

A. A. `(8)/(5)`
B. B. `9/5`
C. C. 2
D. D. `(11)/(5)`

Answer

To solve the problem step by step, we need to find the common difference \(D\) of the 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 and the common difference \(D\) is greater than 1, we can follow these steps:

### Step 1: Define the terms of the AP
Let the first term of the AP be \(a\) and the common difference be \(D\). The terms can be expressed as follows:
- First term: \(a\)
- Fourth term: \(a + 3D\)
- Fifth term: \(a + 4D\)

### Step 2: Use the information about the sixth term
We know that the sixth term of the AP is given by:
\[ a + 5D = 2 \]
From this, we can express \(a\) in terms of \(D\):
\[ a = 2 - 5D \]

### Step 3: Write the product of the first, fourth, and fifth terms
The product \(P\) of the first, fourth, and fifth terms can be expressed as:
\[ P = a \cdot (a + 3D) \cdot (a + 4D) \]
Substituting the value of \(a\):
\[ P = (2 - 5D) \cdot ((2 - 5D) + 3D) \cdot ((2 - 5D) + 4D) \]
This simplifies to:
\[ P = (2 - 5D) \cdot (2 - 2D) \cdot (2 - D) \]

### Step 4: Expand the product
Now, we will expand the product:
\[ P = (2 - 5D)(2 - 2D)(2 - D) \]
Let’s first expand \((2 - 5D)(2 - 2D)\):
\[ (2 - 5D)(2 - 2D) = 4 - 4D - 10D + 10D^2 = 4 - 14D + 10D^2 \]
Now multiplying with \((2 - D)\):
\[ P = (4 - 14D + 10D^2)(2 - D) \]
Expanding this:
\[ P = 8 - 4D - 28D + 14D^2 + 20D^2 - 10D^3 \]
Combining like terms:
\[ P = 8 - 32D + 34D^2 - 10D^3 \]

### Step 5: Differentiate to find the maximum
To maximize \(P\), we differentiate \(P\) with respect to \(D\):
\[ \frac{dP}{dD} = -32 + 68D - 30D^2 \]
Setting the derivative equal to zero to find critical points:
\[ -30D^2 + 68D - 32 = 0 \]

### 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. \(D = \frac{-40}{-60} = \frac{2}{3}\)
2. \(D = \frac{-96}{-60} = \frac{8}{5}\)

### Step 7: Choose the valid solution
Since we need \(D > 1\), we select:
\[ D = \frac{8}{5} \]

### Conclusion
The common difference of the AP that maximizes the product of the first, fourth, and fifth terms is:
\[ \boxed{\frac{8}{5}} \]