• # question_answer Let G be the geometric mean of two positive numbers a and b, and M be the arithmetic mean of $\frac{1}{a}$and$\frac{1}{b}.$If$\frac{1}{M}:G$is 4 : 5 then a : b can be:   [JEE Main Online Paper ( Held On 12 Apirl  2014 ) A) 1 : 4                                        B) 1 : 2 C) 2 : 3                                        D) 3 : 4

Solution :

$G=\sqrt{ab}$ $M=\frac{\frac{1}{a}+\frac{1}{b}}{2}$                  $M=\frac{a+b}{2ab}$ Given that$\frac{1}{M}:G=4:5$ $\frac{2ab}{(a+b)\sqrt{ab}}=\frac{4}{5}$ $\Rightarrow$$\frac{a+b}{2\sqrt{ab}}=\frac{5}{4}$$\Rightarrow$$\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{5+4}{5-4}$ {Using Componendo & Dividendo} $\Rightarrow$$\frac{({{\sqrt{a)}}^{2}}+{{(\sqrt{b})}^{2}}+2\sqrt{ab}}{{{(\sqrt{a})}^{2}}+{{(\sqrt{b})}^{2}}-2\sqrt{ab}}=\frac{9}{1}$ $\Rightarrow$${{\left( \frac{\sqrt{b}+\sqrt{a}}{\sqrt{b}-\sqrt{a}} \right)}^{2}}=\frac{9}{1}\Rightarrow \frac{\sqrt{b}+\sqrt{a}}{\sqrt{b}-\sqrt{a}}=\frac{3}{1}$ $\Rightarrow$$\Rightarrow \frac{\sqrt{b}+\sqrt{a}+\sqrt{b}-\sqrt{a}}{\sqrt{b}-\sqrt{a}-\sqrt{b}+\sqrt{a}}=\frac{3+1}{3-1}$ {Using Componendo & Dividendo} $\sqrt{\frac{b}{a}}=\frac{4}{2}=2$ $\frac{b}{a}=\frac{4}{1}$ $\frac{a}{b}=\frac{1}{4}\Rightarrow a:b=1:4$

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