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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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The ratio of A.M. and G.M. of two positive numbers $a$ and $b$ is $m:n$. Show that $a:b=(m+\sqrt {m^2-n^2}):(m-\sqrt {m^2-n^2})$

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Toolbox:
  • A.M. between $a$ and $b$ is $\large\frac{a+b}{2}$
  • G.M. between $a$ and $b$ is $\sqrt {ab}$
  • Componendo and dividendo : If $\large\frac{a}{b}=\frac{c}{d}$ then $\large\frac{a+b}{a-b}=\frac{c+d}{c-d}$
Given that the ratio of A.M. and G.M. between $a$ and $b$ is $m\::\:n$
But we know that the A.M. between $a$ and $b$ is $\large\frac{a+b}{2}$ and
the G.M. between $a$ and $b$ is $\sqrt {ab}$
$\Rightarrow\:\large\frac{a+b}{2}$$\::\:\sqrt {ab}$$=m\::\:n$
$\Rightarrow\:\large\frac{a+b}{2\sqrt {ab}}$$=\frac{m}{n}$
We know that
Componendo and dividendo : If $\large\frac{a}{b}=\frac{c}{d}$ then $\large\frac{a+b}{a-b}=\frac{c+d}{c-d}$
Step 2
$\therefore\:$ By applying componendo and dividendo we get
$\Rightarrow\:\large\frac{a+b+2\sqrt {ab}}{a+b-2\sqrt {ab}}$$=\frac{m+n}{m-n}$
We know that $a+b+2\sqrt {ab}=(\sqrt a+\sqrt b)^2$ and
We know that $a+b-2\sqrt {ab}=(\sqrt a-\sqrt b)^2$ and
$\Rightarrow\:\large\frac{(\sqrt a+\sqrt b)^2}{(\sqrt a-\sqrt b)^2}$$=\frac{m+n}{m-n}$
Step 3
By taking square root on both the sides we get
$\Rightarrow\:\large\frac{\sqrt a+\sqrt b}{\sqrt a-\sqrt b}$$=\frac{\sqrt {m+n}}{\sqrt {m-n}}$
Step 4
Again applying componendo and dividendo we get
$\Rightarrow\:\large\frac{\sqrt a+\sqrt b+\sqrt a+\sqrt b}{\sqrt a+\sqrt b+\sqrt a-\sqrt b}$$=\frac{\sqrt {m+n}+\sqrt {m-n}}{\sqrt {m+n}-\sqrt {m-n}}$
$\Rightarrow\:\large\frac{\sqrt a}{\sqrt b}=\frac{\sqrt {m+n}+\sqrt {m-n}}{\sqrt {m+n}-\sqrt {m-n}}$
Again squaring both the sides we get
$\Rightarrow\:\large\frac{a}{b}=\bigg(\frac{\sqrt {m+n}+\sqrt {m-n}}{\sqrt {m+n}-\sqrt {m-n}}\bigg)^2$
$\Rightarrow\:\large\frac{a}{b}=\frac{m+n+m-n+2\sqrt {(m+n)(m-n)}}{m+n+m-n-2\sqrt {(m+n)(m-n)}}$
$\Rightarrow\:\large\frac{a}{b}=\frac{2m+2\sqrt {m^2-n^2}}{2m-2\sqrt {m^2-n^2}}$
$\Rightarrow\:\large\frac{a}{b}=\frac{m+\sqrt {m^2-n^2}}{m-\sqrt {m^2-n^2}}$
$\Rightarrow\:a:b=(m+\sqrt {m^2-n^2}):(m-\sqrt {m^2-n^2})$
Hence proved.
answered Apr 1, 2014 by rvidyagovindarajan_1
 

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