Let `lta_(n)gt` be an arithmetic sequence whose first term is `1` and `ltb_(n)gt` be a geometric sequence whose first term is `2`. If the common ratio of geometric sequence is half the common difference of arithmetic sequence, the minimum value of `(a_(4)b_(1) + a_(3)b_(2) + 2a_(1)b_(3))` is equal to

Options

A. `(25)/(12)`
B. `(-3)/(2)`
C. `(3)/(2)`
D. `(-25)/(12)`

Explanation

To solve the problem, we need to find the minimum value of the expression \( A_4 B_1 + A_3 B_2 + 2 A_1 B_3 \), where \( A_n \) is an arithmetic sequence and \( B_n \) is a geometric sequence.

### Step 1: Define the sequences
1. The first term of the arithmetic sequence \( A_n \) is given as \( A_1 = 1 \).
2. Let the common difference of the arithmetic sequence be \( D \). Thus, the terms can be expressed as:
- \( A_1 = 1 \)
- \( A_2 = 1 + D \)
- \( A_3 = 1 + 2D \)
- \( A_4 = 1 + 3D \)

3. The first term of the geometric sequence \( B_n \) is given as \( B_1 = 2 \).
4. Let the common ratio of the geometric sequence be \( r \). Thus, the terms can be expressed as:
- \( B_1 = 2 \)
- \( B_2 = 2r \)
- \( B_3 = 2r^2 \)

### Step 2: Establish the relationship between \( r \) and \( D \)
According to the problem, the common ratio \( r \) of the geometric sequence is half the common difference \( D \) of the arithmetic sequence:
\[ r = \frac{D}{2} \]

### Step 3: Substitute the terms into the expression
Now, we substitute the terms into the expression \( A_4 B_1 + A_3 B_2 + 2 A_1 B_3 \):
\[ A_4 B_1 = (1 + 3D)(2) = 2 + 6D \]
\[ A_3 B_2 = (1 + 2D)(2r) = (1 + 2D)(2 \cdot \frac{D}{2}) = (1 + 2D)D = D + 2D^2 \]
\[ 2 A_1 B_3 = 2(1)(2r^2) = 4r^2 = 4 \left(\frac{D}{2}\right)^2 = \frac{D^2}{4} \]

### Step 4: Combine the terms
Now, we combine all these terms:
\[ A = 2 + 6D + D + 2D^2 + \frac{D^2}{4} \]
Combining like terms gives:
\[ A = 2 + 7D + 2D^2 + \frac{D^2}{4} \]
To combine \( 2D^2 + \frac{D^2}{4} \), we convert \( 2D^2 \) to have a common denominator:
\[ 2D^2 = \frac{8D^2}{4} \]
Thus,
\[ A = 2 + 7D + \frac{8D^2 + D^2}{4} = 2 + 7D + \frac{9D^2}{4} \]

### Step 5: Find the minimum value of \( A \)
To find the minimum value, we can differentiate \( A \) with respect to \( D \):
\[ \frac{dA}{dD} = 7 + \frac{9D}{2} \]
Setting the derivative to zero to find critical points:
\[ 7 + \frac{9D}{2} = 0 \implies \frac{9D}{2} = -7 \implies D = -\frac{14}{9} \]

### Step 6: Substitute \( D \) back into \( A \)
Now substitute \( D = -\frac{14}{9} \) back into the expression for \( A \):
\[ A = 2 + 7\left(-\frac{14}{9}\right) + \frac{9\left(-\frac{14}{9}\right)^2}{4} \]
Calculating each term:
1. \( 7\left(-\frac{14}{9}\right) = -\frac{98}{9} \)
2. \( \left(-\frac{14}{9}\right)^2 = \frac{196}{81} \)
3. Thus, \( \frac{9 \cdot \frac{196}{81}}{4} = \frac{196}{36} = \frac{49}{9} \)

Combining these:
\[ A = 2 - \frac{98}{9} + \frac{49}{9} = 2 - \frac{49}{9} = \frac{18}{9} - \frac{49}{9} = -\frac{31}{9} \]

### Final Answer
The minimum value of \( A_4 B_1 + A_3 B_2 + 2 A_1 B_3 \) is:
\[ \boxed{-\frac{31}{9}} \]