Question

If the ratio of two numbers is (2+√3) : (2-√3), then find the ratio of their arithmetic Mean (AM) and geometric mean (GM).​

Answer 1

ANSWER:

Given:

• Ratio of 2 numbers = (2+√3) : (2-√3)

To Find:

• Ratio of Arithmetic Mean(AM) and Geometric Mean(GM)

Solution:

Let the numbers be x and y.

We are given that,

$\implies x : y = \left(2+\sqrt3\right):\left(2-\sqrt3\right)$

That is,

$\implies \dfrac{x}{y}= \dfrac{2+\sqrt3}{2-\sqrt3}$

Rationalizing the denominator in RHS,

$\implies \dfrac{x}{y}= \dfrac{2+\sqrt3}{2-\sqrt3}\times \dfrac{2+\sqrt3}{2+\sqrt3}$

$\implies \dfrac{x}{y}= \dfrac{(2+\sqrt3)(2+\sqrt3)}{(2-\sqrt3)(2+\sqrt3)}$

So,

$\implies \dfrac{x}{y}= \dfrac{(2+\sqrt3)^2}{(2)^2-(\sqrt3)^2}$

$\implies \dfrac{x}{y}= \dfrac{(2+\sqrt3)^2}{4-3}$

Hence,

$\implies \dfrac{x}{y}= (2+\sqrt3)^2$

Transposing y to RHS,

$\implies x= (2+\sqrt3)^2y - - - -(1)$

Now, we know that, for 2 numbers a and b,

$\hookrightarrow\sf Arithmetic \:Mean(AM) =\dfrac{a+b}{2}$

And,

$\hookrightarrow\sf Geometric \:Mean(GM) =\sqrt{ab}$

So, the ratio of AM and GM is,

$\implies \dfrac{AM}{GM}=\dfrac{\left(\dfrac{a+b}{2}\right)}{\sqrt{ab}}$

$\implies \dfrac{AM}{GM}=\dfrac{a+b}{2\sqrt{ab}}$

Here, the numbers are x and y, so,

$\implies \dfrac{AM}{GM}=\dfrac{x+y}{2\sqrt{xy}}$

Substituting the value of x from (1),

$\implies \dfrac{AM}{GM}=\dfrac{\left((2+\sqrt3)^2y\right)+y}{2\sqrt{\left((2+\sqrt3)^2y\right)y}}$

$\implies \dfrac{AM}{GM}=\dfrac{y\left((2+\sqrt3)^2+1\right)}{2\sqrt{\left((2+\sqrt3)^2y^2\right)}}$

$\implies \dfrac{AM}{GM}=\dfrac{y\!\!\!/\:\left((2+\sqrt3)^2+1\right)}{2(2+\sqrt3)y\!\!\!/}$

$\implies \dfrac{AM}{GM}=\dfrac{(2+\sqrt3)^2+1}{2(2+\sqrt3)}$

Opening the expression, (2+√3)² ,

$\implies \dfrac{AM}{GM}=\dfrac{(2)^2+(\sqrt3)^2+2(2)(\sqrt3)+1}{2(2+\sqrt3)}$

$\implies \dfrac{AM}{GM}=\dfrac{7+4\sqrt3+1}{2(2+\sqrt3)}$

$\implies \dfrac{AM}{GM}=\dfrac{8+4\sqrt3}{2(2+\sqrt3)}$

Taking 2 common,

$\implies \dfrac{AM}{GM}=\dfrac{2\!\!\!/\:(4+2\sqrt3)}{2\!\!\!/\:(2+\sqrt3)}$

$\implies \dfrac{AM}{GM}=\dfrac{4+2\sqrt3}{2+\sqrt3}$

Rationalising the denominator on RHS,

$\implies \dfrac{AM}{GM}=\dfrac{4+2\sqrt3}{2+\sqrt3}\times \dfrac{2-\sqrt3}{2-\sqrt3}$

$\implies \dfrac{AM}{GM}=\dfrac{(4+2\sqrt3)(2-\sqrt3)}{(2+\sqrt3)(2-\sqrt3)}$

$\implies \dfrac{AM}{GM}=\dfrac{8-4\sqrt3+4\sqrt3-6}{(2)^2-(\sqrt3)^2}$

$\implies \dfrac{AM}{GM}=\dfrac{2}{4-3}$

Hence,

$\implies \dfrac{AM}{GM}=\dfrac{2}{1}$

Therefore,

$\implies\bf AM:GM = 2:1$