Review question

# If $27$, $x$, $y$ are in GP, with sum $21$, what are $x$ and $y$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7258

## Solution

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 $y$.

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 $r = -\frac{2}{3} \quad\text{or}\quad r = -\frac{1}{3}.$

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.