# Relation between Geometric and Arithmetic mean

In this section we will discuss relation between geometric and arithmetic mean.
Geometric mean : If a single geometric mean 'G' is inserted between two given numbers 'a' and 'b', then G is known as the geometric mean between 'a' and 'b'.
G.M. = G = $G^{2} = \sqrt{ab}$
Arithmetic mean: If a single arithmetic mean 'A' is inserted between two given numbers 'a' and 'b', then A is known as the arithmetic mean between 'a' and 'b'.
A.M. = A = $\frac{a + b}{2}$

Properties of Arithmetic and Geometric means

Theorem 1: If A and G are arithmetic and geometric mean respectively between two positive numbers a and b, then
A (AM) > G (GM)
Proof : We have,
A.M. = A = $\frac{a + b}{2}$ and G.M. = G = $G^{2} = \sqrt{ab}$

A - G = $\frac{a + b}{2} - \sqrt{ab}$

= $\frac{a + b -2\sqrt{ab} }{2}$

= $\frac{(\sqrt{a} -\sqrt{b})^{2}}{2}$ > 0

∴ A - G > 0
⇒ A > G

Theorem 2: If A and G are arithmetic and geometric mean respectively between two positive numbers a and b, then the quadratic equation having a,b as its roots is
$x^{2} - 2Ax + G^{2}$ = 0
Proof : We have,
A.M. = A = $\frac{a + b}{2}$ and G.M. = G = $G^{2} = \sqrt{ab}$
The given equation has two roots 'a' and 'b' is
$x^{2} - 2Ax + G^{2}$ = 0
$x^{2} - 2\times \frac{a + b}{2}$ + ab = 0
$x^{2} - (a + b)x$ + ab = 0

Theorem 3: If A and G are arithmetic and geometric mean respectively between two positive numbers a and b, then the numbers are A $\pm \sqrt{A^{2} - G^{2}}$
Proof : The given equation has two roots 'a' and 'b' is
$x^{2} - 2Ax + G^{2}$ = 0
x= $\frac {2A \pm \sqrt{A^{2} -G^{2}}}{2}$
∴ x = A $\pm \sqrt{A^{2} - G^{2}}$

## Examples on relation between geometric and arithmetic mean

1) Find two positive numbers whose difference is 12 and whose A.M exceeds the G.M. by 2.
Solution: Let the two numbers be x and y then ,
x - y = 12 -------(i)
A.M - G.M = 2
$\frac{x + y}{2} - \sqrt{xy}$ = 2
x + y - 2$\sqrt{xy}$ = 4
$(\sqrt{x} - \sqrt{y})^{2}$ = 4
$\sqrt{x} - \sqrt{y} = \pm$2 ---------(ii)
Now x - y = 2 $(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})$ = 12
⇒ $(\sqrt{x} + \sqrt{y}) \times \pm$2 = 12
$(\sqrt{x} + \sqrt{y}) = \pm$6 --------(iii)
Solve equation (ii) and (iii) we get
x = 16 and y = 4
So the two numbers are 16 and 4.

2) Find the two numbers whose A.M. is 34 and G.M. is 16.
Solution: Let the two numbers be x and y then ,
AM = 34 = $\frac{x + y}{2}$ ⇒ x + y = 68 ------(i)
GM= 16 = $\sqrt{xy}$ ⇒ xy = 256 -----(ii)
$(x - y)^{2} = (x + y)^{2}$ - 4xy
$(x - y)^{2} = 68^{2} - 4 \times$ 256
$(x - y)^{2}$ = 3600
∴ x- y = 60
and x + y = 68
Adding above two equations we get,
2x = 128
x = 64
∴ y = 4.

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