Let `a_1,a_2,a_3,...` be in A.P. With `a_6=2.` Then the common difference of the A.P. Which maximises the product `a_1a_4a_5` is :

Options

A. `3//2`
B. `6//5`
C. `8//5`
D. `2//3`

Answer

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

### Step 1: Define the terms of the A.P.
Let the first term of the arithmetic progression (A.P.) be \( a \) and the common difference be \( d \). The terms can be expressed as:
- \( a_1 = a \)
- \( a_2 = a + d \)
- \( a_3 = a + 2d \)
- \( a_4 = a + 3d \)
- \( a_5 = a + 4d \)
- \( a_6 = a + 5d \)

Given that \( a_6 = 2 \), we can write:
\[ a + 5d = 2 \]

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

### Step 3: Write the product \( P = a_1 \cdot a_4 \cdot a_5 \)
We need to maximize the product \( P = a_1 \cdot a_4 \cdot a_5 \):
\[ P = a \cdot (a + 3d) \cdot (a + 4d) \]
Substituting \( a = 2 - 5d \):
\[ P = (2 - 5d) \cdot ((2 - 5d) + 3d) \cdot ((2 - 5d) + 4d) \]
This simplifies to:
\[ P = (2 - 5d) \cdot (2 - 2d) \cdot (2 - d) \]

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

### Step 5: Differentiate to find critical points
To maximize \( P \), we differentiate with respect to \( d \) and set it to zero:
\[ \frac{dP}{dd} = -30d^2 + 68d - 32 = 0 \]
Dividing the entire equation by 2 gives:
\[ -15d^2 + 34d - 16 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here \( a = -15, b = 34, c = -16 \):
\[ d = \frac{-34 \pm \sqrt{34^2 - 4 \cdot (-15) \cdot (-16)}}{2 \cdot (-15)} \]
Calculating the discriminant:
\[ 34^2 - 4 \cdot 15 \cdot 16 = 1156 - 960 = 196 \]
Thus:
\[ d = \frac{-34 \pm 14}{-30} \]
Calculating the two possible values:
1. \( d = \frac{-34 + 14}{-30} = \frac{-20}{-30} = \frac{2}{3} \)
2. \( d = \frac{-34 - 14}{-30} = \frac{-48}{-30} = \frac{8}{5} \)

### Step 7: Determine which value maximizes \( P \)
We need to check which of these values gives a maximum product by substituting back into \( P \):
1. For \( d = \frac{2}{3} \)
2. For \( d = \frac{8}{5} \)

After evaluating both, we find that \( d = \frac{8}{5} \) maximizes the product.

### Final Answer:
The common difference \( d \) that maximizes the product \( a_1 a_4 a_5 \) is:
\[ \boxed{\frac{8}{5}} \]