**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

Now use the quadratic formula

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.

11th grade math

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