Question

The harmonic mean between two numbers is $\dfrac{21}{5}$ . Their A.M. 'A' and G.M. 'G', satisfy the relation $3A+{{G}^{2}}=36$ . Find the sum of the squares of the numbers.

Answer 1

Hint: Arithmetic Mean (AM): Sum of n numbers divided by n.
For two numbers a and b: $AM=\dfrac{a+b}{2}$ .
Geometric Mean (GM): nth root of the product of n numbers.
For two numbers a and b: $GM=\sqrt{ab}$ .
Harmonic Mean (HM): The reciprocal of the AM of the reciprocal of the numbers.
For two numbers a and b: $HM=\dfrac{1}{\left( \dfrac{\dfrac{1}{a}+\dfrac{1}{b}}{2} \right)}=\dfrac{2ab}{a+b}$ .
 ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ .

Complete step-by-step answer:
Let the two numbers be a and b, so that their $AM=\dfrac{a+b}{2}$ , $GM=\sqrt{ab}$ and $HM=\dfrac{2ab}{a+b}$ .
According to the question:
 $HM=\dfrac{2ab}{a+b}=\dfrac{21}{5}$
⇒ 10ab = 21a + 21b ... (1)
And, $3A+{{G}^{2}}=36$
⇒ $3\left( \dfrac{a+b}{2} \right)+{{\left( \sqrt{ab} \right)}^{2}}=36$
⇒ 3a + 3b + 2ab = 72
⇒ 21a + 21b + 14ab = 504
⇒ 10ab + 14ab = 504 ... [Using equation (1)]
⇒ $ab=\dfrac{504}{24}=21$ ... (2)
Now, squaring both sides of equation (1):
⇒ $100{{(ab)}^{2}}=({{21}^{2}})({{a}^{2}}+{{b}^{2}}+2ab)$
Putting the value of ab = 21 from equation (2), we get:
⇒ $100{{(21)}^{2}}=({{21}^{2}})({{a}^{2}}+{{b}^{2}}+42)$
⇒ $100={{a}^{2}}+{{b}^{2}}+42$
⇒ ${{a}^{2}}+{{b}^{2}}=100-42=58$
Hence, the sum of the squares of the numbers is 58.

Note: AM-GM-HM Inequality: $AM\ge GM\ge HM$ .
Arithmetic Progression (AP): The series of numbers where the difference of any two consecutive terms is the same, is called an Arithmetic Progression.
Geometric Progression (GP): The series of numbers where the ratio of any two consecutive terms is the same, is called a Geometric Progression.
Harmonic Progression (HP): The series of numbers where the reciprocals of the terms are in Arithmetic Progression, is called a Harmonic Progression.