Question

If the arithmetic mean of two numbers is

3 times their geometric means. Then

find the ratio of those numbers.​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:Let \: the \:two \: numbers \: be \: x \: and \: y$

So,

• Arithmetic mean between x and y is given by

$\rm :\longmapsto\:A.M. \: = \dfrac{x + y}{2}$

And

• Geometric mean between x and y is given by

$\rm :\longmapsto\:G.M. = \sqrt{xy}$

According to statement,

$\rm :\longmapsto\:A.M = 3G.M.$

$\rm :\longmapsto\:\dfrac{x + y}{2} = 3 \sqrt{xy}$

$\rm :\longmapsto\:\dfrac{x + y}{2 \sqrt{xy} } = \dfrac{3}{1}$

On applying Componendo and Dividendo, we get

$\rm :\longmapsto\:\dfrac{x + y + 2 \sqrt{xy} }{x + y - 2 \sqrt{xy} } = \dfrac{3 + 1}{3 - 1}$

$\rm :\longmapsto\:\dfrac{ {( \sqrt{x} + \sqrt{y} })^{2} }{ {( \sqrt{x} - \sqrt{y}) }^{2} } = \dfrac{4}{2} = 2$

$\rm :\longmapsto\:\dfrac{ \sqrt{x} + \sqrt{y} }{ \sqrt{x} - \sqrt{y} } = \dfrac{ \sqrt{2} }{1}$

Again, apply Componendo and Dividendo, we get

$\rm :\longmapsto\:\dfrac{ \sqrt{x} + \cancel{\sqrt{y}} + \sqrt{x} - \cancel{\sqrt{y}} }{ \cancel{\sqrt{x}} + \sqrt{y} - \cancel{\sqrt{x}} + \sqrt{y} } = \dfrac{ \sqrt{2} + 1 }{ \sqrt{2} - 1}$

$\rm :\longmapsto\:\dfrac{\cancel2 \: \sqrt{x} }{\cancel2 \: \sqrt{y} } = \dfrac{ \sqrt{2} + 1 }{ \sqrt{2} - 1 }$

On squaring both sides, we get

$\rm :\longmapsto\:\dfrac{x}{y} = \dfrac{ {( \sqrt{2} + 1) }^{2} }{ {( \sqrt{2} - 1) }^{2} }$

$\rm :\longmapsto\:\dfrac{x}{y} = \dfrac{2 + 1 + 2 \sqrt{2} }{2 + 1 - 2 \sqrt{2} }$

$\rm :\longmapsto\:\dfrac{x}{y} = \dfrac{3 + 2 \sqrt{2} }{3 - 2 \sqrt{2} }$

$\bf\implies \:x: y = (3 + 2 \sqrt{2}): (3 - 2 \sqrt{2} )$

Additional Information :-

Property I: 

• If the Arithmetic Mean and Geometric Mean of two positive numbers a and b are A and G respectively, then A > G.

Property II: 

• If A be the Arithmetic Mean and G be the Geometric Mean between two positive numbers a and b, then the quadratic equation whose roots are a, b is

x² - 2Ax + G² = 0.

Property III: 

• If A is the Arithmetic Means and G be the Geometric Means between two positive numbers, and then the numbers are A ± √(A² – G²)

Answer 2

Solution :-

we know that, when 2 numbers are a and b ,

• AM = (a + b)/2
• GM = √(a * b) .

given that,

→ 3√(ab) = [(a + b)/2]

squaring both sides ,

→ 9ab = (a + b)² / 4

→ 36ab = a² + b² + 2ab

→ a² + b² + 2ab - 36ab = 0

→ a² + b² - 34ab = 0

now, dividing both sides by b² , we get,

→ (a²/b²) + 1 - (34a/b) = 0

→ (a/b)² - 34(a/b) + 1 = 0

Let (a/b) = x ,

→ x² - 34x + 1 = 0

using, Sridharacharya formula for Solving Quadratic Equation ax² +bx + c = 0 says that,

→ x = [ -b±√(b²-4ac) / 2a ]

we have,

• a = 1
• b = (-34)
• c = 1

then,

→ x = [34 ± √{(-34)² - 4*1*1}] / 2*1

→ x = 34 ± √(1156 - 4) / 2

→ x = 34 ± √(1152)/2

→ x = 34 ± 24√2/2

→ x = 17 ± 12√18

therefore,

→ (a/b) = 17 ± 12√2

→ a : b = 17 + 12√2

→ a : b = 17 - 12√2 .

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