Question

If the harmonic mean between two numbers is $\dfrac{16}{5}$, their arithmetic mean is A, their geometric mean is G and $2A+{{G}^{2}}=26$, then the numbers are:
(a)6, 8
(b)4, 8
(c)2, 8
(d)1, 8

Answer 1

Hint: Here, we have to apply the relationship between arithmetic mean, AM, geometric mean, GM and harmonic mean, HM. That is, $AM\times HM=G{{M}^{2}}$. The value we obtain through this should be substituted in $2A+{{G}^{2}}=26$. From this we will get the value of A. Then find the value of G. Now, by the definition of arithmetic mean and geometric mean, i.e. $AM=\dfrac{a+b}{2}$ and $GM=\sqrt{ab}$ and using the value of A and G we can find the value of two numbers a and b.

Complete step-by-step answer:
Here, we are given that the harmonic mean between two numbers is $\dfrac{16}{5}$, the arithmetic mean is A and geometric mean is G. We are also given that $2A+{{G}^{2}}=26$.
Now, we have to find the two numbers.
We know that the three Pythagorean means are arithmetic mean, geometric mean and harmonic mean.
Now, we can define these three means.
We have two numbers a and b then the arithmetic mean, AM of a and b is defined as:
$AM=\dfrac{a+b}{2}$
Now, for the two numbers a and b the geometric mean, GM is defined as:
$GM=\sqrt{ab}$
We have the harmonic mean, HM of two numbers a and b is defined as:
$HM=\dfrac{2}{\dfrac{1}{a}+\dfrac{1}{b}}$
We also have a relationship between AM, GM and HM and is given by,
$AM\times HM=G{{M}^{2}}$ ….. (1)
Here, we have
 $\begin{align}
  & HM=\dfrac{16}{5} \\
 & AM=A \\
 & GM=G \\
\end{align}$
Now, by substituting these values in equation (1),
$\Rightarrow A\times \dfrac{16}{5}={{G}^{2}}$
$\Rightarrow \dfrac{16}{5}A={{G}^{2}}$ ….. (2)
We also have that:
$2A+{{G}^{2}}=26$ …… (3)
Next, by substituting equation (2) in equation (3), we get the equation:
$2A+\dfrac{16}{5}A=26$
Now, by taking the LCM,
$\begin{align}
  & \Rightarrow \dfrac{5\times 2A+1\times 16A}{5}=26 \\
 & \Rightarrow \dfrac{10A+16A}{4}=26 \\
 & \Rightarrow \dfrac{26A}{5}=26 \\
\end{align}$
Now, by cross multiplication,
$\Rightarrow A=\dfrac{5\times 26}{26}$
Again, by cancellation, we get:
$A=5$
Now, by substituting the value of $A=5$ in equation (2) we get:
$\dfrac{16}{5}\times 5={{G}^{2}}$
Next, by cancellation we obtain:
$16={{G}^{2}}$
We have:
$\begin{align}
  & AM=\dfrac{a+b}{2} \\
 & \Rightarrow 5=\dfrac{a+b}{2} \\
\end{align}$
Now, by cross multiplication,
$\Rightarrow 10=a+b$
We also know that:
$\begin{align}
  & GM=\sqrt{ab} \\
 & \Rightarrow G=\sqrt{ab} \\
\end{align}$
By squaring,
$\begin{align}
  & \Rightarrow {{G}^{2}}=ab \\
 & \Rightarrow 16=ab \\
\end{align}$
We got,
 $\begin{align}
  & a+b=10 \\
 & ab=16 \\
\end{align}$
Now, we can write:
$b=10-a$
Next, substitute, this $b=10-a$ in $ab=16$ we get:
$\begin{align}
  & a(10-a)=16 \\
 & \Rightarrow 10a-{{a}^{2}}=16 \\
\end{align}$
Now, by taking 16 to the left side we get 16,
$\Rightarrow 10a-{{a}^{2}}-16=0$
Now, multiplying by -1 throughout the equation, and rearranging we get:
$\Rightarrow {{a}^{2}}-10a+16=0$
In the next step let us split -10a as -2a - 8a,
$\Rightarrow {{a}^{2}}-2a-8a+16=0$
For the first two terms a is a common factor, so take it outside. Similarly, for the last two terms we can take -8 outside,
$\Rightarrow a(a-2)-8(a-2)=0$
Next, we have a – 2 as the common factor, se we can take a – 2 outside,
$\Rightarrow (a-2)(a-8)=0$
So, we can say that,
$a-2=0$ or $a-8=0$
Now, consider $a-2=0$, by taking -2 to the right side, it becomes 2,
$\Rightarrow a=2$
Similarly, for $a-8=0$, take -8 to the right side, it becomes 8,
$\Rightarrow a=8$
Now, substitute, a = 2 in $a+b=10$,
$\Rightarrow 2+b=10$
Now, take 2 to the right side,
$\begin{align}
  & \Rightarrow b=10-2 \\
 & \Rightarrow b=8 \\
\end{align}$
Next, for a = 8 in $a+b=10$,
$\Rightarrow 2+b=10$
Now, take 8 to the right side,
$\begin{align}
  & \Rightarrow b=10-8 \\
 & \Rightarrow b=2 \\
\end{align}$
Hence, the numbers are 2 and 8.
Therefore, the correct answer for this question is option (c).

Note: The harmonic mean has the least value compared to the arithmetic and geometric mean. The arithmetic mean has the maximum value compared to the geometric and harmonic mean. That is,
AM > GM > HM.