Relation between Arithmetic Means and Geometric Means

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.

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 Geometric Progression



11 and 12 Grade Math

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