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NDA 2011
Exam
NDA
Year
2011
Subject
Mathematics
Chapter
Sequences and Series

NDA 2011 · Mathematics · Sequences and Series

If the arithmetic and geometric means of two numbers are 10,8...

If the arithmetic and geometric means of two numbers are 10,8 respectively, then one number exceeds the other number by

Options

  1. A88
  2. B1010
  3. C1212
  4. D1616

Correct answer

C. 1212

Step-by-step solution

Let numbers are a and bA⋅M=a+b2⇒10=a+b2⇒a+b=20b\mathrm{A} \cdot \mathrm{M}=\frac{a+b}{2} \Rightarrow 10=\frac{a+b}{2}\Rightarrow \quad a+b=20 ...(i) G.M=ab⇒8=ab⇒ab=64∵(a−b)2=(a+b)2−4ab\mathrm{G} . \mathrm{M}=\sqrt{a b} \Rightarrow 8=\sqrt{a b}\Rightarrow a b=64\because \quad(a-b)^2=(a+b)^2-4 a b=(20)2−4(64)=400−256=144\begin{aligned} & =(20)^2-4(64) \\ & =400-256=144\end{aligned}a−b=12a-b=12 ...(ii) from (i) and (ii) \left.a+b=20a−b=12\begin{array}{rl}a+b & =20 \\ a-b & =12\end{array}\right] \Rightarrow a=16, b=4\therefore \quad 16-4=12\therefore one number exceeds the other number by 12 .
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