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 | 1 year agoConstruct a quadratic in x such that A.M. of its roots is A and G.M. is G.
Find the geometric means of the following pairs of numbers:
(i) 2 and 8
(ii) a3b and ab3
(iii) –8 and –2
Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).