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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Sequences and Series

NDA 2018 · Mathematics · Sequences and Series

If x_ i > 0, y_ i > 0(i=1,2,3 n) are the values of two variable X...

If xi>0,yi>0(i=1,2,3\mathrm{x}_{\mathrm{i}} > 0, \mathrm{y}_{\mathrm{i}} > 0(\mathrm{i}=1,2,3\qquad n) are the values of two variable X and Y with geometric mean P and Q respectively, then the geometric mean of XY\frac{X}{Y} is

Options

  1. APQ\frac{\mathrm{P}}{\mathrm{Q}}
  2. Bantilog⁡(PQ)\operatorname{antilog} \left(\frac{\mathrm{P}}{\mathrm{Q}}\right)
  3. Cn(log⁡P−log⁡Q)\mathrm{n}(\log \mathrm{P}-\log \mathrm{Q})
  4. Dn(log⁡P+log⁡Q)\mathrm{n}(\log \mathrm{P}+\log \mathrm{Q})

Correct answer

B. antilog⁡(PQ)\operatorname{antilog} \left(\frac{\mathrm{P}}{\mathrm{Q}}\right)

Step-by-step solution

(x1⋅x2……xn)1/n=P(y1⋅y2………yn)1/n=Q(x1⋅x2…………xny1⋅y2………⋅yn)1/n=(x1⋅x2…………xn)1/n(y1⋅y2…………yn)1/n=PQ\begin{aligned} & \left(x_1 \cdot x_2 \ldots \ldots x_n\right)^{1 / n}=\mathrm{P} \\ & \left(y_1 \cdot y_2 \ldots \ldots \ldots y_n\right)^{1 / n}=\mathrm{Q} \\ & \left(\frac{x_1 \cdot x_2 \ldots \ldots \ldots \ldots x_n}{y_1 \cdot y_2 \ldots \ldots \ldots \cdot y_n}\right)^{1 / n}=\frac{\left(x_1 \cdot x_2 \ldots \ldots \ldots \ldots x_n\right)^{1 / n}}{\left(y_1 \cdot y_2 \ldots \ldots \ldots \ldots y_n\right)^{1 / n}}=\frac{\mathrm{P}}{\mathrm{Q}}\end{aligned}
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