We will discuss here about some of the important relation between Arithmetic Means and Geometric Means.
The following properties are:
Property I: The Arithmetic Means of two positive numbers can never be less than their Geometric Mean.
Proof:
Let A and G be the Arithmetic Means and Geometric Means respectively of two positive numbers m and n.
Then, we have A = m + n/2 and G = ±√mn
Since, m and n are positive numbers, hence it is evident that A > G when G = √mn. Therefore, we are to show A ≥ G when G = √mn.
We have, A  G = m + n/2  √mn = m + n  2√mn/2
A  G = ½[(√m  √n)^2] ≥ 0
Therefore, A  G ≥ 0 or, A ≥ G.
Hence, the Arithmetic Mean of two positive numbers can never be less than their Geometric Means. (Proved).
Property II: If A be the Arithmetic Means and G be the Geometric Means between two positive numbers m and n, then the quadratic equation whose roots are m, n is x^2  2Ax + G^2 = 0.
Proof:
Since, A and G be the Arithmetic Means and Geometric Means respectively of two positive numbers m and n then, we have
A = m + n/2 and G = √mn.
The equation having m, n as its roots is
x^2  x(m + n) + nm = 0
⇒ x^2  2Ax + G^2 = 0, [Since, A = m + n/2 and G = √nm]
Property III: If A be the Arithmetic Means and G be the Geometric Means between two positive numbers, then the numbers are A ± √A^2  G^2.
Proof:
Since, A and G be the Arithmetic Means and Geometric Means respectively then, the equation having its roots as the given numbers is
x^2  2Ax + G^2 = 0
⇒ x = 2A ± √4A^2  4G^2/2
⇒ x = A ± √A^2  G^2
Property IV: If the Arithmetic Mean of two numbers x and y is to their Geometric Mean as p : q, then, x : y = (p + √(p^2  q^2) : (p  √(p^2  q^2).
Solved examples on the properties of Arithmetic and Geometric Means between two given quantities:
1. The Arithmetic and Geometric Means of two positive numbers are 15 and 9 respectively. Find the numbers.
Solution:
Let the two positive numbers be x and y. Then according to the problem,
x + y/2 = 15
or, x + y = 30 .................. (i)
and √xy = 9
or xy = 81
Now, (x  y)^2 = (x + y)^2  4xy = (30)^2  4 * 81 = 576 = (24)^2
Therefore, x  y = ± 24 .................. (ii)
Solving (ii) and (iii), we get,
2x = 54 or 2x = 6
x = 27 or x = 3
When x = 27 then y = 30  x = 30  27 = 3
and when x = 27 then y = 30  x = 30  3 = 27
Therefore, the required numbers are 27 and 3.
2. Find two positive numbers whose Arithmetic Means increased by 2 than Geometric Means and their difference is 12.
Solution:
Let the two numbers be m and n. Then,
m  n = 12 ........................ (i)
It is given that AM  GM = 2
⇒ m + n/2  √mn = 2
⇒ m + n  √mn = 4
⇒ (√m  √n^2 = 4
⇒ √m  √n = ±2 ........................ (ii)
Now, m  n = 12
⇒ (√m + √n)(√m  √n) = 12
⇒ (√m + √n)(±2) = 12 ........................ (iii)
⇒ √m + √n = ± 6, [using (ii)]
Solving (ii) and (iii), we get m = 16, n = 4
Hence, the required numbers are 16 and 4.
3. If 34 and 16 are the Arithmetic Means and Geometric Means of two positive numbers respectively. Find the numbers.
Solution:
Let the two numbers be m and n. Then
Arithmetic Mean = 34
⇒ m + n/2 = 34
⇒ m + n = 68
And
Geometric Mean = 16
√mn = 16
⇒ mn = 256 ............................... (i)
Therefore, (m  n)^2 = (m + n)^2  4mn
⇒ (m – n)^2 = (68)^2  4 × 256 = 3600
⇒ m  n = 60............................... (ii)
On solving (i) and (ii), we get m = 64 and n = 4.
Hence, the required numbers are 64 and 4.
● Geometric Progression
From Relation between Arithmetic Means and Geometric Means to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
