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).
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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
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