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NDA 2010
Exam
NDA
Year
2010
Subject
Mathematics
Chapter
Sequences and Series

NDA 2010 · Mathematics · Sequences and Series

If the AM and GM between two numbers are in the ratio m: n , then...

If the AM and GM between two numbers are in the ratio m:nm: n, then what is the ratio between the two numbers?

Options

  1. Am+m2−n2m−m2−n2\frac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}
  2. Bm+nm−n\frac{m+n}{m-n}
  3. Cm2−n2m2+n2\frac{m^2-n^2}{m^2+n^2}
  4. Dm2+n2−mnm2+n2+mn\frac{m^2+n^2-m n}{m^2+n^2+m n}

Correct answer

A. m+m2−n2m−m2−n2\frac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}

Step-by-step solution

According to question, a+b2ab=mn⇒a+b+2aba+b−2ab=m+nm−n\begin{aligned} & \frac{a+b}{2 \sqrt{a b}}=\frac{m}{n} \\ & \Rightarrow \frac{a+b+2 \sqrt{a b}}{a+b-2 \sqrt{a b}}=\frac{m+n}{m-n} \end{aligned} (By componendo and devidendo) ⇒(a+b)2(a−b)2=m+nm−n⇒a+ba−b=m+nm−n⇒2a2b=m+n+m−nm+n−m−n\begin{aligned} & \Rightarrow \frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2}=\frac{m+n}{m-n} \\ & \Rightarrow \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{m+n}}{\sqrt{m-n}} \\ & \Rightarrow \frac{2 \sqrt{a}}{2 \sqrt{b}}=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}} \end{aligned} (By componendo and devidendo) ⇒ab=m+m2−n2m−m2−n2\Rightarrow \frac{a}{b}=\frac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}
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