In a geometric progression (G.P.), the ratio between consecutive terms is constant. Given the terms \( \frac{1}{16}, x, \frac{1}{4}, y, \ldots \), we can denote the common ratio as \( r \).
From the first three terms:
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The ratio between the first and the second term is: \[ r = \frac{x}{\frac{1}{16}} \quad \Rightarrow \quad x = \frac{1}{16} r \]
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The ratio between the second and the third term is: \[ r = \frac{\frac{1}{4}}{x} \quad \Rightarrow \quad \frac{1}{4} = x \cdot r \]
Substituting the expression for \( x \) into the second equation: \[ \frac{1}{4} = \left( \frac{1}{16} r \right) r \] \[ \frac{1}{4} = \frac{1}{16} r^2 \] Multiplying both sides by 16 gives: \[ 4 = r^2 \quad \Rightarrow \quad r = 2 \quad \text{(since we take the positive root for a G.P.)} \]
Now substituting \( r = 2 \) back to find \( x \): \[ x = \frac{1}{16} \cdot 2 = \frac{1}{8} \]
Now, to find \( y \), we use the ratio between the third and the fourth terms: \[ y = \frac{1}{4} \cdot r = \frac{1}{4} \cdot 2 = \frac{1}{2} \]
Finally, we compute the product of \( x \) and \( y \): \[ x \cdot y = \frac{1}{8} \cdot \frac{1}{2} = \frac{1}{16} \]
Therefore, the product of \( x \) and \( y \) is: \[ \boxed{\frac{1}{16}} \]
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