In an A.P.,the sixth term `a_(6)=2` .If the product `a_(1)a_(4)a_(5)` is the greatest,then the common difference of the A.P.is equal to

Options

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

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Understanding the given information
We are given that the sixth term of an arithmetic progression (A.P.) is \( a_6 = 2 \). The formula for the \( n \)-th 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: Setting up the equation for the sixth term
Using the formula for the sixth term:
\[ a_6 = a + 5d = 2 \]
This gives us our first equation:
\[ a + 5d = 2 \quad \text{(1)} \]

### Step 3: Expressing the terms \( a_1, a_4, a_5 \)
Now we need to find the product \( a_1 \times a_4 \times a_5 \):
- The first term \( a_1 = a \)
- The fourth term \( a_4 = a + 3d \)
- The fifth term \( a_5 = a + 4d \)

Thus, the product can be expressed as:
\[ P = a \times (a + 3d) \times (a + 4d) \]

### Step 4: Substituting \( a \) from equation (1)
From equation (1), we can express \( a \) in terms of \( d \):
\[ a = 2 - 5d \]
Now substitute this into the product \( P \):
\[ P = (2 - 5d) \times ((2 - 5d) + 3d) \times ((2 - 5d) + 4d) \]
Simplifying the terms:
- \( a + 3d = 2 - 5d + 3d = 2 - 2d \)
- \( a + 4d = 2 - 5d + 4d = 2 - d \)

So, we have:
\[ P = (2 - 5d) \times (2 - 2d) \times (2 - d) \]

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

### Step 6: Finding the maximum product
To find the maximum value of \( P \), we take the derivative \( P' \) and set it to zero:
\[ P' = -32 + 68d - 30d^2 \]
Setting \( P' = 0 \):
\[ -30d^2 + 68d - 32 = 0 \]
This is a quadratic equation in \( d \).

### Step 7: Solving the quadratic equation
Using the quadratic formula \( d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = -30, b = 68, c = -32 \):
\[ d = \frac{-68 \pm \sqrt{68^2 - 4 \times (-30) \times (-32)}}{2 \times (-30)} \]
Calculating the discriminant:
\[ 68^2 = 4624, \quad 4 \times 30 \times 32 = 3840 \]
\[ d = \frac{-68 \pm \sqrt{4624 - 3840}}{-60} = \frac{-68 \pm \sqrt{784}}{-60} = \frac{-68 \pm 28}{-60} \]
Calculating the two possible values for \( d \):
1. \( d = \frac{-40}{-60} = \frac{2}{3} \)
2. \( d = \frac{-96}{-60} = \frac{8}{5} \)

### Step 8: Determining which value gives the maximum product
To determine which value of \( d \) gives the maximum product, we can check the second derivative or evaluate the product at both values.

After checking, we find that \( d = \frac{8}{5} \) gives the maximum product.

### Final Answer
Thus, the common difference \( d \) of the A.P. is:
\[ \boxed{\frac{8}{5}} \]