If ` a_(n)` be the ` n^(th)` term of an AP and if ` a_(1) = 2` , then the value of the common difference that would make `a_(1) a_(2) a_(3)`
minimum is _________

Answer

To solve the problem, we need to find the common difference \( d \) of an arithmetic progression (AP) that minimizes the product of the first three terms \( a_1, a_2, a_3 \).

### Step-by-Step Solution:

1. **Identify the Terms of the AP**:
Given that the first term \( a_1 = 2 \) and the common difference is \( d \), we can express the first three terms of the AP as:
- \( a_1 = 2 \)
- \( a_2 = a_1 + d = 2 + d \)
- \( a_3 = a_1 + 2d = 2 + 2d \)

2. **Write the Product of the Terms**:
We need to find the product \( P \) of the first three terms:
\[ P = a_1 \cdot a_2 \cdot a_3 = 2 \cdot (2 + d) \cdot (2 + 2d) \]

3. **Simplify the Product**:
Expanding the product:
\[ P = 2 \cdot (2 + d) \cdot (2 + 2d) = 2 \cdot [(2 + d)(2 + 2d)] \]
Now, expand \( (2 + d)(2 + 2d) \):
\[ (2 + d)(2 + 2d) = 4 + 4d + 2d + 2d^2 = 4 + 6d + 2d^2 \]
Therefore,
\[ P = 2(4 + 6d + 2d^2) = 8 + 12d + 4d^2 \]

4. **Minimize the Product**:
To find the value of \( d \) that minimizes \( P \), we can take the derivative of \( P \) with respect to \( d \) and set it to zero:
\[ \frac{dP}{dd} = 12 + 8d \]
Setting the derivative equal to zero:
\[ 12 + 8d = 0 \]
Solving for \( d \):
\[ 8d = -12 \implies d = -\frac{12}{8} = -\frac{3}{2} \]

5. **Conclusion**:
The common difference \( d \) that minimizes the product \( a_1 a_2 a_3 \) is:
\[ d = -\frac{3}{2} \]

### Final Answer:
The value of the common difference that would make \( a_1 a_2 a_3 \) minimum is \( -\frac{3}{2} \).

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