The ratio of the A.M and G.M. of two positive numbers a and b, is m: n. Show that 

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

Asked by Abhisek | 2 years ago |  89

1 Answer

Solution :-

Let the two numbers be a and b.

A.M =\( \dfrac{(a+b)}{2}\) and G.M. = \( \sqrt{ab}\)

From the question, we have

\( \dfrac{(a+b)}{2\sqrt{ab}}=\dfrac{m}{n}\)

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

\( (a+b)^2=\dfrac{(4ab\;m^2)}{n^2}\)

\((a+b)^2= \dfrac{2\sqrt{ab}\;m}{n}\)..........(1)

Using the above equation in the identity

\( (a-b)^2 = (a+b)^2- 4ab,\) we obtain

\( (a-b)^2= \dfrac{4abm^2}{n^2}-4ab\)

\( =\dfrac{4ab(m^2-n^2)}{n^2}\)

\( (a-b)= \dfrac{2\sqrt{ab}\sqrt{m^2-n^2}}{n}\).........(2)

Add equation (1) and (2)

2a = \( \dfrac{2\sqrt{ab}}{n}=(m+\sqrt{m^2-n^2})\)

\( a=\dfrac{\sqrt{ab}}{n}=(m+\sqrt{m^2-n^2})\)

Substitute in (1) the value of a .

\( b=\dfrac{2\sqrt{ab}}{n}m-\dfrac{\sqrt{ab}}{n}(m+\sqrt{m^2-n^2})\)

\(\dfrac{\sqrt{ab}}{n}m-\dfrac{\sqrt{ab}}{n}\sqrt{m^2-n^2}\)

\( \dfrac{\sqrt{ab}}{n}(m-\sqrt{m^2-n^2})\)

Therefore,

a : b = \( \dfrac{a}{b}\)

Therefore, it is proved that a:b =

\( (m+\sqrt{m^2-n^2}):(m-\sqrt{m^2-n^2})\)

Answered by Pragya Singh | 2 years ago

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