The numbers \(27\), \(x\), \(y\) are in geometric progression. If the sum of these three numbers is \(21\) calculate the possible values of \(x\) and \(y\).

*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 \[ r = -\frac{2}{3} \quad\text{or}\quad r = -\frac{1}{3}. \]

Alternatively, one could use the quadratic formula to find the possible values of \(r\).

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

It’s possible to do the question without mentioning \(r\).

Three numbers \(a,b\) and \(c\) are in geometric progression if and only if \(ac=b^2\).

So here we have \(27y=x^2\), and also \(27+x+y=21\), or \(y = -x-6\).

So we have \(27(-x-6)=x^2\), and solving this quadratic equation gives us the same answers as above.