Question

The harmonic mean of two numbers is 4 and the arithmetic and geometric mean satisfy the relation $\text{2A}+{\text{G}}^2=27$. Find the numbers.
$$\begin{gathered} \text{A}\text{. 6,3} \\ \text{B}\text{. 5,4} \\ \text{C}\text{. 5,}-2.5 \\ \text{D}\text{. }-3\text{,1} \end{gathered}$$

Answer 1

Hint: Here, we will proceed by developing an relation between the arithmetic mean and the geometric mean o the two numbers apart from the one given in the problem with the help of the formula i.e., Arithmetic mean$\times$Harmonic mean = ${\left(\text{Geometric mean}\right)}^2$.

Complete Step-by-Step solution:
Let the numbers be a and b.
Given, harmonic mean of these numbers (i.e., a and b) is H = 4
Also, given that $\text{2A}+{\text{G}}^2=27\to \text{(1)}$ where A represents arithmetic mean of these numbers (i.e., a and b) and G represents geometric mean of these numbers (i.e., a and b)
According to the relation between arithmetic mean, geometric mean and harmonic mean of any two numbers, we have
Arithmetic mean$\times$Harmonic mean = ${\left(\text{Geometric mean}\right)}^2$
So, $\text{A}\times \text{H}={\text{G}}^2$
Substituting H = 4 in the above equation, we get
 $$\begin{gathered} \Rightarrow \text{A}\times \text{4}={\text{G}}^2 \\ \Rightarrow {\text{G}}^2=\text{4A }\to \text{(2)} \end{gathered}$$
By substituting equation (2) in equation (1), we get
$$\begin{gathered} \Rightarrow \text{2A}+4\text{A}=27 \\ \Rightarrow 6\text{A}=27 \\ \Rightarrow \text{A}={\displaystyle \frac{27}{6}} \\ \Rightarrow \text{A}={\displaystyle \frac{9}{2}}\to \text{(3)} \end{gathered}$$
Put $\text{A}={\displaystyle \frac{9}{2}}$ in equation (2), we get
$$\begin{gathered} \Rightarrow {\text{G}}^2=\text{4}\left({\displaystyle \frac{9}{2}}\right) \\ \Rightarrow {\text{G}}^2=18\to \text{(4)} \end{gathered}$$
As we know that the arithmetic mean of two numbers a and b is given as $\text{A}={\displaystyle \frac{a+b}{2}}\to \text{(5)}$
By comparing equations (3) and (5), we get
$$\begin{gathered} {\displaystyle \frac{a+b}{2}}={\displaystyle \frac{9}{2}} \\ \Rightarrow a+b=9 \\ \Rightarrow a=\left(9-b\right)\to \text{(6)} \end{gathered}$$
Also we know that the geometric mean of two numbers a and b is given as $$\begin{gathered} \text{G}=\sqrt{ab} \\ \Rightarrow {\text{G}}^2=ab\to \text{(7)} \end{gathered}$$
By comparing equations (4) and (7), we get
$\Rightarrow ab=18\to \text{(8)}$
By substituting the value $$a=\left(9-b\right)$$ from equation (6) in equation (8), we get
$$\begin{gathered} \Rightarrow \left(9-b\right)b=18 \\ \Rightarrow 9b-b^2=18 \\ \Rightarrow b^2-9b+18=0 \\ \Rightarrow b^2-3b-6b+18=0 \\ \Rightarrow b\left(b-3\right)-6\left(b-3\right)=0 \\ \Rightarrow \left(b-3\right)\left(b-6\right)=0 \end{gathered}$$
Either $$\begin{gathered} \left(b-3\right)=0 \\ \Rightarrow b=3 \end{gathered}$$ or $$\begin{gathered} \left(b-6\right)=0 \\ \Rightarrow b=6 \end{gathered}$$
Put b=3 in equation (6), we get
$$\begin{gathered} \Rightarrow a=\left(9-3\right) \\ \Rightarrow a=6 \end{gathered}$$
Put b=6 in equation (6), we get
$$\begin{gathered} \Rightarrow a=\left(9-6\right) \\ \Rightarrow a=3 \end{gathered}$$
When b=3, then a=6 and when b=6, then a=3
Therefore, the required two numbers are 6,3.
Hence, option A is correct.

Note: In this particular problem, we have obtained two equations in two variables (which are the numbers a and b) and then we will solve these two equations with the help of substitution method in order to get the values corresponding to these two numbers. Here, the numbers obtained are either 3,6 or 6,3 which are both the same and depend only on the sequence that which number comes first.