Solveeit Logo
Login

Question

Mathematics Question on Sequences and Series

The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show that a:b=(m+√m2−n2):(m−√m2−n2).a : b =\sqrt{ (m + √m^2 - n^2) }: (m -\sqrt{ √m^2 - n^2)}.

Answer

Let the two numbers be a and b.
A.M=a+b2andG.M.=√abA.M = \frac{a + b }{ 2} and G.M. = √ab
According to the given condition,
a+b2√ab=mn\frac{a + b }{ 2} √ab =\frac{ m }{ n}
⇒(a+b)24(ab)=m2n2⇒ \frac{(a + b)^2 }{4(ab)} =\frac{ m^2 }{ n^2}
⇒(a+b)2=4abm2n2⇒ (a + b)^2 = \frac{4abm^2}{ n^2}
⇒(a+b)=2ab mn...(1)⇒ (a + b)= \frac{2\sqrt{ab\,m} }{ n} ...(1)
Using this in the identity (a-b) 2 = (a + b) 2- 4ab, we obtain
(a−b)2=4ab m2n2−4ab=4ab(m2−n2)n2(a - b)^2 = \frac{4ab\,m2 }{ n^2} - 4ab = \frac{4ab(m^2 - n2) }{ n^2}
⇒(a−b)=2√ab√m2−n2n...(2)⇒ (a - b) =\frac{ 2 √ab √m^2 - n^2}{ n} ...(2)
Adding (1) and (2), we obtain
2a=2√abn(m+m2−n2)2a = \frac{2 √ab}{ n} (m + \sqrt{m2 - n2})
⇒a=√abn(m+m2−n2)⇒ a = \frac{√ab }{ n} (m + \sqrt{m^2 - n^2})
Substituting the value of a in (1), we obtain
b=2√abnm−√abn(m+m2−n2)b =\frac{ 2 √ab }{ n} m -\frac{ √ab }{ n} (m + \sqrt{m^2 - n^2)}
=√abnm−√abnm2−n2= \frac{√ab}{n} m -\frac{ √ab}{ n} \sqrt{m^2 - n^2}
=√abn(m−m2−n2)= \frac{√ab }{n} (m - \sqrt{m^2 - n^2)}

∴a:b=ab=abn(m+m2−n2)abn(m−m2−n2)=(m+m2−n2)(m−m2−n2)∴ a : b =\frac{ a }{ b} = \frac{\frac{\sqrt{ab}}{n}(m+\sqrt{m^2-n^2)}}{\frac{\sqrt{ab}}{n}(m-\sqrt{m^2-n^2})}=\frac{(m+\sqrt{m^2-n^2})}{(m-\sqrt{m^2-n^2})}

Thus, a:b=(m+m2−n2):(m−m2−n2).a : b = (m + \sqrt{m^2 - n^2}) : (m - \sqrt{m^2 - n^2)}.