The numbers ` x_(1) , x_(2) , x_(3) …..` form an infinite decreasing GP . If x = 1 , then the common ratio of the progression
for which the expression ` 6x_(5) - 16x_(4) - 3x_(3) + 12x_(2) ` is maximum , is

Options

A. `(1)/(3)`
B. `(2)/(3)`
C. `(3)/(4)`
D. `(1)/(2)`

Answer

To solve the problem, we need to find the common ratio \( r \) of an infinite decreasing geometric progression (GP) such that the expression \( 6x_5 - 16x_4 - 3x_3 + 12x_2 \) is maximized, given that \( x_1 = 1 \).

### Step-by-step Solution:

1. **Define the terms of the GP**:
- Since \( x_1 = 1 \) and the common ratio \( r \) is between 0 and 1 (because the GP is decreasing), we can express the terms as:
\[ x_1 = 1, \quad x_2 = r, \quad x_3 = r^2, \quad x_4 = r^3, \quad x_5 = r^4 \]

2. **Substitute the terms into the expression**:
- We substitute \( x_2, x_3, x_4, \) and \( x_5 \) into the expression:
\[ 6x_5 - 16x_4 - 3x_3 + 12x_2 = 6r^4 - 16r^3 - 3r^2 + 12r \]

3. **Define the function**:
- Let \( f(r) = 6r^4 - 16r^3 - 3r^2 + 12r \).

4. **Find the derivative**:
- To maximize the function, we need to find its derivative and set it to zero:
\[ f'(r) = 24r^3 - 48r^2 - 6r + 12 \]

5. **Set the derivative to zero**:
- Setting the derivative equal to zero:
\[ 24r^3 - 48r^2 - 6r + 12 = 0 \]

6. **Simplify the equation**:
- Divide the entire equation by 6:
\[ 4r^3 - 8r^2 - r + 2 = 0 \]

7. **Factor the polynomial**:
- We can factor this polynomial:
\[ (r - 2)(4r^2 + 8r - 1) = 0 \]
- This gives us one root \( r = 2 \).

8. **Solve the quadratic equation**:
- Now we solve \( 4r^2 + 8r - 1 = 0 \) using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} \]
\[ = \frac{-8 \pm \sqrt{64 + 16}}{8} = \frac{-8 \pm \sqrt{80}}{8} = \frac{-8 \pm 4\sqrt{5}}{8} = \frac{-2 \pm \sqrt{5}}{2} \]

9. **Evaluate the roots**:
- This gives us two potential roots:
\[ r_1 = \frac{-2 + \sqrt{5}}{2}, \quad r_2 = \frac{-2 - \sqrt{5}}{2} \]
- Since \( r \) must be between 0 and 1, we only consider \( r_1 \):
\[ r = \frac{-2 + \sqrt{5}}{2} \]

10. **Check the range**:
- We need to check if \( r_1 \) is in the range \( (0, 1) \):
- Since \( \sqrt{5} \approx 2.236 \), we have:
\[ r \approx \frac{-2 + 2.236}{2} \approx \frac{0.236}{2} \approx 0.118 \]
- This value is indeed between 0 and 1.

### Final Answer:
The common ratio \( r \) for which the expression is maximized is:
\[ \boxed{\frac{-2 + \sqrt{5}}{2}} \]