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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 97 of 106
Marks: +1, -0
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+\sqrt{m^{2}-n^{2}}) : (m−m2−n2)(m-\sqrt{m^{2}-n^{2}})
Solution:  
We have a+b2ab\frac{\frac{a+b}{2}}{\sqrt{ab}} = mn\frac{m}{n} ⇒ a+b2ab\frac{a+b}{2\sqrt{ab}} = mn\frac{m}{n}
Applying componendo and dividendo, we get
a+b+2aba+b−2ab\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}} = m+nm−n\frac{m+n}{m-n} ⇒ (a+b)2(a−b)2\frac{(\sqrt{a}+\sqrt{b})^{2}}{(\sqrt{a}-\sqrt{b})^{2}} = m+nm−n\frac{m+n}{m-n} ⇒ a+ba−b\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = m+nm−n\frac{\sqrt{m+n}}{\sqrt{m-n}}
Again applying componendo and dividendo, we get
a+b+a−ba+b−a+b\frac{\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}} = m+n+m−nm+n−m−n\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}} ⇒ ab\frac{\sqrt{a}}{\sqrt{b}} = m+n+m−nm+n−m−n\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}
Squaring on both sides, we get
ab\frac{a}{b} = m+n+m−n+2m2−n2m+n+m−n−2m2−n2\frac{m+n+m-n+2\sqrt{m^{2}-n^{2}}}{m+n+m-n-2\sqrt{m^{2}-n^{2}}}
⇒ ab\frac{a}{b} = 2m+2m2−n22m−2m2−n2\frac{2m+2\sqrt{m^{2}-n^{2}}}{2m-2\sqrt{m^{2}-n^{2}}} ⇒ ab\frac{a}{b} = m+m2−n2m−m2−n2\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}
⇒ a : b = (m+m2−n2)(m+\sqrt{m^{2}-n^{2}}) : (m−m2−n2)(m-\sqrt{m^{2}-n^{2}})
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