The sixth term of an A.P. is equal to 2. The value of the common difference of the A.P. which makes the product `a _(1) a_(4)a_(5)` least is given by :

Options

A. `8/5`
B. `4/5`
C. `2/3`
D. None of these

Explanation

To solve the problem, we need to find the common difference \(d\) of an arithmetic progression (A.P.) such that the product \(a_1 \cdot a_4 \cdot a_5\) is minimized, given that the sixth term \(a_6\) is equal to 2.

### Step-by-Step Solution:

1. **Identify 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.

2. **Use the information about the sixth term**:
Given that the sixth term \(a_6 = 2\):
\[ a + 5d = 2 \quad \text{(1)} \]

3. **Express the first, fourth, and fifth terms**:
- The first term \(a_1\) is:
\[ a_1 = a \]
- The fourth term \(a_4\) is:
\[ a_4 = a + 3d \]
- The fifth term \(a_5\) is:
\[ a_5 = a + 4d \]

4. **Write the product \(P = a_1 \cdot a_4 \cdot a_5\)**:
Substituting the expressions for \(a_1\), \(a_4\), and \(a_5\):
\[ P = a \cdot (a + 3d) \cdot (a + 4d) \]

5. **Substitute \(a\) from equation (1)**:
From equation (1), we can express \(a\) in terms of \(d\):
\[ a = 2 - 5d \]
Substituting this into the product \(P\):
\[ P = (2 - 5d) \cdot ((2 - 5d) + 3d) \cdot ((2 - 5d) + 4d) \]
Simplifying:
\[ P = (2 - 5d) \cdot (2 - 2d) \cdot (2 - d) \]

6. **Expand the product**:
First, expand \((2 - 5d)(2 - 2d)\):
\[ (2 - 5d)(2 - 2d) = 4 - 4d - 10d + 10d^2 = 4 - 14d + 10d^2 \]
Now multiply this 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:
\[ P = 8 - 32d + 34d^2 - 10d^3 \]

7. **Find the critical points by differentiating**:
To minimize \(P\), we differentiate with respect to \(d\):
\[ \frac{dP}{dd} = -32 + 68d - 30d^2 \]
Setting the derivative to zero:
\[ -30d^2 + 68d - 32 = 0 \]

8. **Solve the quadratic equation**:
Dividing through by 2:
\[ 15d^2 - 34d + 16 = 0 \]
Using the quadratic formula:
\[ d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{34 \pm \sqrt{(-34)^2 - 4 \cdot 15 \cdot 16}}{2 \cdot 15} \]
\[ d = \frac{34 \pm \sqrt{1156 - 960}}{30} = \frac{34 \pm \sqrt{196}}{30} = \frac{34 \pm 14}{30} \]
This gives:
\[ d = \frac{48}{30} = \frac{8}{5} \quad \text{and} \quad d = \frac{20}{30} = \frac{2}{3} \]

9. **Determine which value minimizes \(P\)**:
We need to check which of these values gives the minimum product \(P\).

After evaluating \(P\) at both values of \(d\), we find that:
- For \(d = \frac{8}{5}\), \(P\) is positive but larger.
- For \(d = \frac{2}{3}\), \(P\) is smaller.

Thus, the value of the common difference \(d\) that minimizes the product \(P = a_1 \cdot a_4 \cdot a_5\) is:
\[ \boxed{\frac{2}{3}} \]