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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 62 of 106
Marks: +1, -0
The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3+22):(3−22)(3+2\sqrt{2}) : (3-2\sqrt{2}).
Solution:  
Let a and b be the two numbers such that a + b = 6 ab\sqrt{ab}
⇒ a + b = 3 (2ab(2\sqrt{ab}) ⇒ a+b2ab\frac{a+b}{2\sqrt{ab}} = 31\frac{3}{1}
Applying componendo and dividendo, we have, a+b+2aba+b−2ab\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}} = 3+13−1\frac{3+1}{3-1}
⇒ (a+b)2(a−b)2\frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2} = 42\frac{4}{2} ⇒ a+ba−b\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = 21\frac{\sqrt{2}}{1}
Again applying componendo & dividendo, we have
(a+b)+(a−b)(a+b)−(a−b)\frac{(\sqrt{a}+\sqrt{b})+(\sqrt{a}-\sqrt{b})}{(\sqrt{a}+\sqrt{b})-(\sqrt{a}-\sqrt{b})} = 2+12−1\frac{\sqrt{2}+1}{\sqrt{2}-1} ⇒ 2a2b\frac{2\sqrt{a}}{2\sqrt{b}} = 2+12−1\frac{\sqrt{2}+1}{\sqrt{2}-1} ⇒ ab\frac{\sqrt{a}}{\sqrt{b}} = 2+12−1\frac{\sqrt{2}+1}{\sqrt{2}-1}
Squaring both sides, we get
ab\frac{a}{b} = (2+1)2(2−1)2\frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)^2} = 2+1+222+1−22\frac{2+1+2\sqrt{2}}{2+1-2\sqrt{2}} ⇒ ab\frac{a}{b} = 3+223−22\frac{3+2\sqrt{2}}{3-2\sqrt{2}}
Hence, proved.
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