Question

Suppose $x$,$y$, and $z$ form a geometric sequence. if you know that $x+y+z=18$ and $x^2+y^2+z^2=612$, find the value of $y$.

Answer 1

Step 1 : Given information:

It is given that $x+y+z=18$ and $x^2+y^2+z^2=612$.

Step 2 : Framing the equation in terms of the first term $\mathit{a}$ and common ratio $\mathit{r}$:

It's known as the common ratio of the geometric sequence is $r=\frac{x}{y}\mathrm{or}r=\frac{y}{z}$.

So,

$$\begin{gathered} \frac{x}{y}=\frac{y}{z} \\ {y}^2=xz \end{gathered}$$

Substitute $xz$ for $y^2$ in $x^2+y^2+z^2=612$:

$\begin{array}{rcl}{x}^2+xz+z^2& =& 612\\ & \Rightarrow & x^2+2xz-xz+z^2=612\\ & \Rightarrow & {\left(x+z\right)}^2-xz=612\\ & \Rightarrow & {\left(x+z\right)}^2-y^2=612\dots \left(1\right)\end{array}$

Solve $x+y+z=18$ for $x+z$:

$x+z=18-y\dots \left(2\right)$

Step 3: Determine the value of $\mathit{y}$:

Substitute $18-y$ for $x+z$ in equation (1):

$\begin{array}{rcl}{\left(18-y\right)}^2-y^2& =& 612\\ \Rightarrow {18}^2-36y+y^2-y^2& =& 612\\ & \Rightarrow & 324-36y=612\\ \Rightarrow -36y& =& 612-324\\ \Rightarrow y& =& -\frac{288}{36}\\ \Rightarrow & =& -8\end{array}$

Hence, the value of $y$ is $\mathbf{-}\mathbf{8}$.