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 | 1 year ago |  70

##### 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)

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 | 1 year ago

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