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If the GM and AM between two number are in the ratio n : m, then what is the ratio between the two numbers?1. \(\rm \frac{m+n}{m-n}\)2. \(\rm \frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}\)3. \(\rm \frac{m^2 +n^2}{m^2-n^2}\)4. \(\rm \frac{m^2 -n^2}{m^2+n^2}\) |
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Answer» Correct Answer - Option 2 : \(\rm \frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}\) Concept: Let a and b two numbers. AM = \(\rm (a+b)\over 2\) and GM = \(\rm \sqrt{ab}\) Let, two numbers be a and b, We know, AM = \(\rm (a+b)\over 2\) and GM =\(\rm \sqrt{ab}\) Given that, GM : AM = n : m ⇒ AM : GM = m : n \(⇒ \frac{\rm a+b}{2\rm \sqrt{ab}}= \frac{\rm m}{\rm n}\) \(⇒ \frac{(\rm a+b)^{2}}{4\rm a b}= \frac{\rm m^{2}}{\rm n^{2}}\;\;\;\;\; \ldots \ldots \ldots . . \text { (i) } \\ ⇒\frac{(\rm a+b)^{2}-4 a b}{4\rm a b}=\frac{\rm m^{2}- \rm n^{2}}{ \rm n^{2}} \\ ⇒ \frac{(\rm a-b)^{2}}{4 \rm a b}=\frac{\rm m^{2}-\rm n^{2}}{\rm n^{2}} \ldots \ldots \ldots . . \text { (ii) }\) Since, on dividing eqn (i) and (ii), we get \( \rm \frac{(a+b)^{2}}{(a-b)^{2}}=\frac{m^{2}}{m^{2}-n^{2}}\\ \rm ⇒ \frac{a+b}{a-b}=\frac{m}{\sqrt{m^{2}-n^{2}}} \\ \rm ⇒ \frac{(a+b)+(a-b)}{(a+b)-(a-b)}=\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}\) \(\rm \Rightarrow \frac{2a}{2b}=\frac{a}{b}=\frac{m+\sqrt{m^{2}-n^{2}}}{m-\sqrt{m^{2}-n^{2}}}\) Hence, option (2) is correct. |
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