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  • The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show that a :b = m plus square root m raised to 2 minus n raised to 2 : m minus square root m raised to 2 minus n raised to 2

 19.  The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show that a: b = \left ( m + \sqrt {m^2 -n^2 }\right ) : \left ( m- \sqrt {m^2 - n^2} \right )
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Answers (1)

best_answer

Let two numbers be a and b.

AM=\frac{a+b}{2}\, \, \, and\, \, \, \, \, GM=\sqrt{ab}

According to the given condition,     

\frac{a+b}{2\sqrt{ab}}=\frac{m}{n}

\Rightarrow \frac{(a+b)^2}{4ab}=\frac{m^2}{n^2}

\Rightarrow (a+b)^2=\frac{4ab.m^2}{n^2}

\Rightarrow (a+b)=\frac{2\sqrt{ab}.m}{n}...................................................................1

(a-b)^2=(a+b)^2-4ab

We get,

             (a-b)^2=\left ( \frac{4abm^2}{n^2} \right )-4ab

           (a-b)^2=\left ( \frac{4abm^2-4abn^2}{n^2} \right )

\Rightarrow (a-b)^2=\left ( \frac{4ab(m^2-n^2)}{n^2} \right )

\Rightarrow (a-b)=\left ( \frac{2\sqrt{ab}\sqrt{(m^2-n^2)}}{n} \right ).....................................................2

From 1 and 2, we get

2a =\left ( \frac{2\sqrt{ab}}{n} \right )\left ( m+\sqrt{(m^2-n^2)} \right )

a =\left ( \frac{\sqrt{ab}}{n} \right )\left ( m+\sqrt{(m^2-n^2)} \right )

Putting the value of a in equation 1, we have

b=\left ( \frac{2.\sqrt{ab}}{n} \right ) m-\left ( \frac{\sqrt{ab}}{n} \right )\left ( m+\sqrt{(m^2-n^2)} \right )

    b=\left ( \frac{\sqrt{ab}}{n} \right ) m-\left ( \frac{\sqrt{ab}}{n} \right )\left ( \sqrt{(m^2-n^2)} \right )

      =\left ( \frac{\sqrt{ab}}{n} \right )\left (m- \sqrt{(m^2-n^2)} \right )

\therefore a:b=\frac{a}{b}=\frac{\left ( \frac{\sqrt{ab}}{n} \right )\left (m+ \sqrt{(m^2-n^2)} \right )}{\left ( \frac{\sqrt{ab}}{n} \right )\left (m- \sqrt{(m^2-n^2)} \right )}

                       =\frac{\left (m+ \sqrt{(m^2-n^2)} \right )}{\left (m- \sqrt{(m^2-n^2)} \right )}

a:b=\left (m+ \sqrt{(m^2-n^2)} \right ) : \left (m- \sqrt{(m^2-n^2)} \right )

Posted by

seema garhwal

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