If an A.P `a_6 = 2` then common difference for which `a_1.a_3.a_6` is least is

Options

A. `16/25`
B. `8/15`
C. `15/8`
D. `15/6`

Answer

To solve the problem, we need to find the common difference \( d \) of an arithmetic progression (A.P.) such that the product \( a_1 \cdot a_3 \cdot a_6 \) is minimized, given that \( a_6 = 2 \).

### Step-by-Step Solution:

1. **Define the terms of the A.P.**:
- Let the first term be \( a \) and the common difference be \( d \).
- The terms of the A.P. can be expressed as:
- \( a_1 = a \)
- \( a_3 = a + 2d \)
- \( a_6 = a + 5d \)

2. **Use the given information**:
- We know that \( a_6 = 2 \). Therefore:
\[ a + 5d = 2 \]
- From this, we can express \( a \) in terms of \( d \):
\[ a = 2 - 5d \]

3. **Express the product \( P \)**:
- The product \( P = a_1 \cdot a_3 \cdot a_6 \) can be written as:
\[ P = a \cdot (a + 2d) \cdot (a + 5d) \]
- Substituting \( a = 2 - 5d \):
\[ P = (2 - 5d) \cdot ((2 - 5d) + 2d) \cdot 2 \]
- Simplifying \( a + 2d \):
\[ a + 2d = (2 - 5d) + 2d = 2 - 3d \]
- Thus, we have:
\[ P = (2 - 5d)(2 - 3d)(2) \]

4. **Expand the product**:
- First, calculate \( (2 - 5d)(2 - 3d) \):
\[ (2 - 5d)(2 - 3d) = 4 - 6d - 10d + 15d^2 = 4 - 16d + 15d^2 \]
- Therefore:
\[ P = 2(4 - 16d + 15d^2) = 8 - 32d + 30d^2 \]

5. **Find the minimum of \( P \)**:
- To find the minimum, we take the derivative of \( P \) with respect to \( d \):
\[ \frac{dP}{dd} = -32 + 60d \]
- Set the derivative equal to zero to find critical points:
\[ -32 + 60d = 0 \implies 60d = 32 \implies d = \frac{32}{60} = \frac{8}{15} \]

6. **Verify that this is a minimum**:
- Take the second derivative:
\[ \frac{d^2P}{dd^2} = 60 \]
- Since the second derivative is positive, this indicates that \( P \) has a minimum at \( d = \frac{8}{15} \).

### Final Answer:
The common difference \( d \) for which \( a_1 \cdot a_3 \cdot a_6 \) is least is:
\[ \boxed{\frac{8}{15}} \]