The geometric mean of \(\left( x_{1}, x_{2}, x_{3}, ... x_{n} \right)\) is X and the geometric mean of \( \left(y_{1}, y_{2}, y_{3}, ...y_{n} \right) \) is Y. Which of the following is/are correct?
1.The geometric mean of \( \left(x_{1}y_{1}, x_{2}y_{2}, x_{3}y_{3}, ... x_{n}y_{n}\right) \) is XY.
2. The geometric mean of \( \left( \large \frac{x_{1}}{y_{1}}, \frac{x_{2}}{y_{2}}, \frac{x_{3}}{y_{3}},.... \frac{x_{n}}{y_{n}} \right)\) is \( \frac{X}{Y} \)
Select the correct and using the code below
Correct Answer: Description for Correct answer:
Let us consider an example\( \Large (x_{1},x_{2},x_{3},..... x_{n}) \)
=(1, 2, 4, 2)
G.M = \( \Large \sqrt[4]{1 X 2 X 4 X 2} = \sqrt[4]{16} = 2 = X\)
and \( \Large (y_{1},y_{2},y_{3},..... y_{n}) \)
=(2,3,4,6) = \( \Large \sqrt[4]{144} = 2(3^{2})^{1/4} \)
\( \Large Y = 2\sqrt{3} \)
Now GM of \( \Large (x_{1}y_{1},x_{2}y_{2},x_{3}y_{3} ..... x_{n}y_{n} ) \)
G.M of (2,6,16,12) = \( \Large \sqrt[4]{2 X 6 X 16 X12} \)
\( \Large =\sqrt{2 X 2 X 3 X 2 X 2 X 2 X 2 X 2 X 2 X 3} \)
\( \Large = 4\sqrt{3} = XY \)
Hence statement 1 is correct
Now,
G.M of \( \Large \left( \frac{x_{1}}{y_{1}}, \frac{x_{2}}{y_{2}}, \frac{x_{3}}{y_{3}} ..... \frac{x_{n}}{y_{n}} \right) \)
G.M of \( \Large \left(\frac{1}{2}, \frac{2}{3}, \frac{4}{4},\frac{2}{6} \right) \)
\( \Large \sqrt[4]{\frac{1}{2} \times \frac{2}{3} \times \frac{4}{4} \times \frac{2}{6}} \)
\( \Large = \sqrt[4]{\frac{1}{3 \times 3}} = \frac{1}{\sqrt{3}} = \frac{x}{y} \)
Hence 2nd statement is also true
Both 1 and 3 are correct
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