Question

Three numbers, of which the third is equal to $36$, form a geometric progression. If $36$ is replaced with $20$, then the three numbers form an arithmetic progression. Find these three numbers.

Answer 1

Step-1: Forming relation in three numbers:

Let $x,y,$ and $z$ be the three numbers.

Since the third number of the geometric progression is equal to $36$.

Using the geometric mean concept and it can be written as :

$\begin{array}{rcl}{y}^2& =& xz\\ & \Rightarrow & y^2=x\left(36\right)\left[\text{Substitute}z=36\right]\\ \therefore y^2& =& 36x...\left(3\right)\end{array}$

If $36$ is replaced with $20$, then the three numbers form an arithmetic progression.

Using the arithmetic mean concept and it can be written as :

$\begin{array}{rcl}y& =& \frac{x+z}{2}\\ \therefore y& =& \frac{x+20}{2}...\left(4\right)\left[\text{Substitute}z=20\right]\end{array}$

Step-2: Solving quadratic equation:

Substitute the value of $y$ from equation $\left(4\right)$ in the equation $\left(3\right)$ and solve for $x$ :

$\begin{array}{rcl}{\left(\frac{x+20}{2}\right)}^2& =& 36x\left[\text{Simplifying}\right]\\ & \Rightarrow & \frac{x^2+400+40x}{4}=36x\\ & \Rightarrow & x^2+400+40x=144x\\ & \Rightarrow & x^2-104x+400=0\\ & \Rightarrow & x^2-100x-4x+400=0\\ & \Rightarrow & x\left(x-100\right)-4\left(x-100\right)=0\\ & \Rightarrow & \left(x-4\right)\left(x-100\right)=0\\ \therefore x& =& 4\text{or}100\end{array}$

Step-3: Find the numbers:

Substitutes the value of $x$ in equation $\left(4\right)$ and calculates the value of $y$ :

$\begin{array}{rcl}y& =& \frac{4+20}{2}\text{or}\frac{100+20}{2}\left[\text{Simplifying}\right]\\ & \Rightarrow & y=\frac{24}{2}\text{or}\frac{120}{2}\\ \therefore y& =& 12\text{or}60\end{array}$

Therefore, the three numbers are either $4,12,36\text{or}100,60,20$.

Hence, three numbers are either $\mathbf{4}\mathbf{,}\mathbf{}\mathbf{12}\mathbf{,}\mathbf{}\mathbf{36}\mathbf{}\mathbf{}\mathbf{\text{or}}\mathbf{}\mathbf{}\mathbf{100}\mathbf{,}\mathbf{}\mathbf{60}\mathbf{,}\mathbf{}\mathbf{20}$.