Let $G_{1}$ and $G_{2}$ be the geometric means of two series $x_{1}, x_{2}, . . . . . ., x_{n}$ and $y_{1}, y_{2} . . . . . y_{n}$ respectively. If G isthe geometric mean of series $x_{i} / y_{i}, i = 1, 2, . . ., n$ , then G is equal to

G_{1} - G_{2}

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l o g G_{1} / l o g_{2}

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l o g (G_{1} / G_{2})

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G_{1} / G_{2}

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Solution

The correct option is A $G_{1} / G_{2}$
$G = {(\frac{x_{1}}{y_{1}} \times \frac{x_{2}}{y_{2}} \times \frac{x_{3}}{y_{3}} \times . . \times \frac{x_{n}}{y_{n}})}^{\frac{1}{n}} = [{(x_{1} x_{2} x_{3} . . . . x_{n})}^{\frac{1}{n}}] / [{(y_{1} y_{2} y_{3} . . . . y_{n})}^{\frac{1}{n}}] = \frac{G_{1}}{G_{2}}$
Ans: D

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Similar questions

G_{1}, G_{2}, . . . . . . . G_{n}

If n geometric means between a and b be $G_{1}, G_{2}, . . . . G_{n}$ and a geometric mean of a and b be G, then the true relation is

x_{1}, x_{2}, x_{3}, . . . . . . . x_{n}

G_{1}

y_{1}, y_{2}, y_{3}, . . . . . y_{n}

G_{2}

\frac{x_{1}}{y_{1}}, \frac{x_{2}}{y_{2}}, \frac{x_{3}}{y_{3}}, . . . . . . \frac{x_{n}}{y_{n}}

G_{1} a n d G_{2}

x_{1}, x_{2}, . . . . . x_{n} a n d y_{1}, y_{2} . . . . ., y_{n}

x_{i} / y_{i}

i = 1, 2, . . . ., n

G_{1}, G_{2}

\frac{G_{1}^{2}}{G_{2}} + \frac{G_{2}^{2}}{G_{1}}

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