Consider the two numbers be a and b.
Then, G.M. = \( \sqrt{ab}\).
According to the conditions given in the question,
\( a+b=6\sqrt{ab}\) .........(1)
\( (a+b)^2=36\)
\( (a-b)^2= (a+b)^2-4ab\)
\( =36ab-4ab=32ab\)
\( a-b=\sqrt{32}\sqrt{ab}\)
\( =4\sqrt{2}\sqrt{ab}\) ..........(2)
Add (1) and (2).
\(2a= (6+4\sqrt{2})\sqrt{ab}\)
\( a= (3+2\sqrt{2})\sqrt{ab}\)
Substitute \( a= (3+2\sqrt{2})\sqrt{ab}\) in equation (1) .
\(b= 6\sqrt{ab}-(3+2\sqrt{2})\sqrt{ab}\)
b=\( (3-2\sqrt{2})\sqrt{ab}\)
Divide a by b .
\( \dfrac{a}{b}=\dfrac{(3+2\sqrt{2})\sqrt{ab}}{(3-2\sqrt{2})\sqrt{ab}}\)
= \( \dfrac{(3+2\sqrt{2})}{(3-2\sqrt{2})}\)
Therefore, it is proved that the numbers are in the ratio.
Answered by Abhisek | 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 \( \dfrac{81}{2}\).