Question

The geometric mean of two numbers is $6$ and their arithmetic mean is $6.5$ the numbers are

• A. $(3,12)$
• B. $(4,9)$
• C. $(2,18)$
• D. $(7,6)$

Answer 1

The correct option is B

$(4,9)$

Explanation for the correct option.

Step 1: Obtain the equations from the given conditions.

Assume that, $x$ and $y$ be the required two numbers.

It is given that, the geometric mean of two numbers is equal to $6$.

Therefore,

$$\begin{gathered} \sqrt{xy}=6 \\ \Rightarrow xy=36...\left(1\right) \end{gathered}$$

Also, it is given that the arithmetic mean of the same two numbers is equal to $6.5$.

$$\begin{gathered} \frac{x+y}{2}=6.5 \\ \Rightarrow x+y=13 \\ \Rightarrow x=13-y...\left(2\right) \end{gathered}$$

Step 2: Find the required numbers.

From equation $1$ and equation $2$.

$$\begin{gathered} (13-y)y=36 \\ \Rightarrow 13y-y^2=36 \\ \Rightarrow y^2-13y+36=0 \end{gathered}$$

Find the roots of the above quadratic equation.

$$\begin{gathered} y=\frac{13\pm \sqrt{{13}^2-4·36}}{2} \\ \Rightarrow y=\frac{13\pm \sqrt{169-144}}{2} \\ \Rightarrow y=\frac{13\pm \sqrt{25}}{2} \\ \Rightarrow y=\frac{13\pm 5}{2} \\ \Rightarrow y=\left\{4,9\right\} \end{gathered}$$

When $y=4$,

$$\begin{gathered} x=13-4 \\ \Rightarrow x=9 \end{gathered}$$

When $y=9$,

$$\begin{gathered} x=13-9 \\ \Rightarrow x=4 \end{gathered}$$

Therefore, the required two numbers are $4$ and $9$.

Hence, option $B$ is the correct answer.