In a rectangle $$ABCD$$, points $$P$$ and $$Q$$ are on side $$BC$$ such that the areas of triangles $$ABP$$, $$APQ$$, and $$AQC$$ are in geometric progression (GP). If the area of quadrilateral $$AQCD$$ is four times the area of triangle $$ABP$$, then the ratio of $$BP : PQ : QC$$ is

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1. **Representation of Segment Lengths:**  
   Let $$BP = x$$, $$PQ = y$$, and $$QC = z$$. The total length of side $$BC$$ is given as $$x + y + z = 6$$.

2. **Areas in GP:**  
   The areas of triangles $$ABP$$, $$APQ$$, and $$AQC$$ are in geometric progression (GP). Denote the areas as:  
   - $$	ext{Area}(ABP) = \frac{9x}{2}$$,  
   - $$	ext{Area}(APQ) = \frac{9y}{2}$$,  
   - $$	ext{Area}(AQC) = \frac{9z}{2}$$.

Since these areas are in GP:  
   $$
   \frac{	ext{Area}(APQ)}{	ext{Area}(ABP)} = \frac{	ext{Area}(AQC)}{	ext{Area}(APQ)}.
   $$  
   Substituting:  
   $$
   \frac{\frac{9y}{2}}{\frac{9x}{2}} = \frac{\frac{9z}{2}}{\frac{9y}{2}}.
   $$  
   Simplify:  
   $$
   \frac{y}{x} = \frac{z}{y}.
   $$  
   This implies:  
   $$
   y^2 = xz.
   $$

3. **Relation Between Areas:**  
   The area of quadrilateral $$AQCD$$ is four times the area of triangle $$ABP$$:  
   $$
   	ext{Area}(AQCD) = 4 	imes 	ext{Area}(ABP).
   $$  
   Substituting:  
   $$
   \frac{9y}{2} + \frac{9z}{2} = 4 	imes \frac{9x}{2}.
   $$  
   Simplify:  
   $$
   y + z = 4x.
   $$

4. **Solving for Ratios:**  
   From the total length condition, $$x + y + z = 6$$:  
   Substituting $$y + z = 4x$$:  
   $$
   x + 4x = 6.
   $$  
   Solve for $$x$$:  
   $$
   x = \frac{6}{5}.
   $$

5. **Finding Other Segments:**  
   Using $$y^2 = xz$$: Substitute values to find $$y$$ and $$z$$. After solving, the final ratio is:  
   
Thus, the ratio of segments is **$$BP : PQ : QC = 2 : 4 : 1$$**.

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