Source: 1976 Euclid Part B Problem 2
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Given that $x$, $y$, and $2$ are in geometric progression, and that $x^{-1}$, $y^{-1}$, and $9x^{-2}$ are in are in arithmetic progression, then find the numerical value of $xy$.
1. Given that \( x \), \( y \), and \( 2 \) are in geometric progression, we can write:
   \[
   \frac{y}{x} = \frac{2}{y} \implies y^2 = 2x \implies x = \frac{y^2}{2}
   \]

2. Given that \( x^{-1} \), \( y^{-1} \), and \( 9x^{-2} \) are in arithmetic progression, we can write:
   \[
   2 \cdot y^{-1} = x^{-1} + 9x^{-2}
   \]
   Substituting \( x = \frac{y^2}{2} \) into the equation, we get:
   \[
   2 \cdot \frac{1}{y} = \frac{2}{y^2} + 9 \left( \frac{2}{y^2} \right)^2
   \]
   Simplifying the right-hand side:
   \[
   2 \cdot \frac{1}{y} = \frac{2}{y^2} + 9 \cdot \frac{4}{y^4} \implies \frac{2}{y} = \frac{2}{y^2} + \frac{36}{y^4}
   \]

3. Multiplying through by \( y^4 \) to clear the denominators:
   \[
   2y^3 = 2y^2 + 36 \implies 2y^3 - 2y^2 - 36 = 0 \implies y^3 - y^2 - 18 = 0
   \]

4. Solving the cubic equation \( y^3 - y^2 - 18 = 0 \):
   \[
   y(y^2 - y - 18) = 0
   \]
   Factoring the quadratic:
   \[
   y^2 - y - 18 = (y - 3)(y + 6)
   \]
   Thus, the solutions are:
   \[
   y = 0, \quad y = 3, \quad y = -6
   \]

5. Since \( y = 0 \) would make \( y^{-1} \) undefined, we discard \( y = 0 \). We are left with \( y = 3 \) and \( y = -6 \).

6. For \( y = 3 \):
   \[
   x = \frac{y^2}{2} = \frac{3^2}{2} = \frac{9}{2}
   \]
   Thus:
   \[
   xy = 3 \cdot \frac{9}{2} = \frac{27}{2}
   \]

7. For \( y = -6 \):
   \[
   x = \frac{y^2}{2} = \frac{(-6)^2}{2} = \frac{36}{2} = 18
   \]
   Thus:
   \[
   xy = -6 \cdot 18 = -108
   \]

8. Since the problem asks for the numerical value of \( xy \), and typically numerical implies real and positive in this context, we consider the positive solution:
   \[
   xy = \frac{27}{2}
   \]

The final answer is \(\boxed{\frac{27}{2}}\).