If $27$, $x$, $y$ are in GP, with sum $21$, what are $x$ and $y$?

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As the numbers \(27\), \(x\) and \(y\) are in geometric progression, there is a common ratio \(r\) such that \[\begin{align*} x &= 27 r, \\ y &= xr = 27r^2. \end{align*}\] We are also given that the sum of the three numbers is \(21\), i.e. \[ 27 + x + y = 21. \] By substituting the values for \(x\) and \(y\) into this, we see that \[\begin{align*} 27 + 27r + 27r^2 = 21 &\implies 27r^2 + 27r + 6 = 0 \\ &\implies 9r^2 + 9r + 2 = 0 \\ &\implies (3r + 2)(3r + 1) = 0. \end{align*}\]

Hence, the two possibilities for the common ratio are \[ r = -\frac{2}{3} \quad\text{or}\quad r = -\frac{1}{3}. \]

As the question asked us to find the possible values of \(x\) and \(y\), we are not yet done.

Case 1: \(r = -\frac{2}{3}\)

In this case, \[ x = 27r = 27 \times \left( -\frac{2}{3} \right) = -18 \quad\text{and}\quad y = xr = (-18) \times \left( -\frac{2}{3} \right) = 12. \]

Case 2: \(r = -\frac{1}{3}\)

In this case, \[ x = 27r = 27 \times \left( -\frac{1}{3} \right) = -9 \quad\text{and}\quad y = xr = (-9) \times \left( -\frac{1}{3} \right) = 3. \]