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NDA 2021
Exam
NDA
Year
2021
Subject
Mathematics
Chapter
Sequences and Series

NDA 2021 · Mathematics · Sequences and Series

Let x be the HM and y be the GM of two positive numbers m and n ....

Let xx be the HM and yy be the GM of two positive numbers mm and nn. If 5x=4y5 x=4 y, then which one of the following is correct?

Options

  1. A5m=4n5 m=4 n
  2. B2m=n2 m=n
  3. C4m=5n4 m=5 n
  4. Dm=4nm=4 n

Correct answer

D. m=4nm=4 n

Step-by-step solution

x=2m⋅nm+n and  y=m⋅n5x=4y\begin{aligned} & x=\frac{2 m \cdot n}{m+n} \quad \text{\text{ and } } y=\sqrt{m \cdot n} \\ & 5 x=4 y\end{aligned} A.M. of m and n=m+n2 As 5⋅(2m⋅n)m+n=4⋅m⋅nn=\frac{m+n}{2}\text{ As } 5 \cdot \frac{(2 m \cdot n)}{m+n}=4 \cdot \sqrt{m \cdot n}5m⋅n2m⋅n=m+n52m⋅n=(m+n)\begin{aligned} & \frac{5 m \cdot n}{2 \sqrt{m \cdot n}}=m+n \\ & \frac{5}{2} \sqrt{m \cdot n}=(m+n)\end{aligned}52=mn+nm\frac{5}{2}=\sqrt{\frac{m}{n}}+\sqrt{\frac{n}{m}} Let mn=K,\sqrt{\frac{m}{n}}=K, \quad then 52=K+1K2(K2+1)=5K\frac{5}{2}=K⁺\frac{1}{K}2\left(K^2+1\right)=5 K2K2−5K+2=0(K−2)(2K−1)=0\begin{aligned} & 2 K^2-5 K+2=0 \\ & (K-2)(2 K-1)=0\end{aligned}$\begin{array}{rlrl}\therefore & K & =2 \text{\text{ or } } \frac{1}{2} & \\ & \sqrt{\frac{m}{n}}=2 & \Rightarrow m=4 n \\ & \text{or, } \sqrt{\frac{m}{n}}=\frac{1}{2} & n=4 m\end{array}
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