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Thread: Geometric Series

  1. #1
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    Geometric Series

    Question 1
    Let A = $\displaystyle {x_1, x_2, ......x_n} $ be a set of positive numbers in geometric progression. Suppose n is odd. Show that the geometric mean and the median of A are the same.


    Isn't the geometric mean = n th root of $\displaystyle \frac{a(1-r^n)}{1-r} $ ? And the median is $\displaystyle \frac{1}{2} ( x_1 + x_n) $ ?

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    Question 2

    Let A = $\displaystyle {x_1, x_2, ......x_n} $ be a set of positive numbers with arithmetic mean, geometric mean and harmonic mean A, G, H respectively.

    In the case n=2, prove that A ≥ G ≥ H.


    Help will be much appreciated. Thank you.
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  2. #2
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    Hello, WWTL@WHL!

    Question 1

    Let $\displaystyle A \;= \;x_1,\,x_2,\,x_3,\,\cdots\,x_n$ be a set of positive numbers in geometric progression.
    Suppose $\displaystyle n$ is odd.
    Show that the geometric mean and the median of $\displaystyle A$ are the same.
    The geometric mean is the $\displaystyle n^{th}$ root of the product of the $\displaystyle n$ terms.

    This is a geometric sequence with first term $\displaystyle a$ and common ratio $\displaystyle r.$
    . . The sequence is: .$\displaystyle a,\,ar,\,ar^2,\,ar^3,\,\cdots ar^{n-1}$

    The product of the terms is: .$\displaystyle (a)(ar)(ar^2)(ar^3)\cdots(ar^{n-1}) \;=\;a^n(r\cdot r^2\cdot r^3 \cdots r^{n-1})$

    . . $\displaystyle = \;a^n\cdot r^{1+2+3+\cdots+(n-1)} \;=\;a^n\cdot r^{\frac{n(n-1)}{2}} $

    The Geometric Mean is: .$\displaystyle \left(a^n\cdot r^{\frac{n(n-1)}{2}}\right)^{\frac{1}{n}} \;=\;\left(a^n\right)^{\frac{1}{n}}\cdot\left(r^{\ frac{n(n-1)}{2}} \right)^{\frac{1}{n}} \;=\;\boxed{ar^{\frac{n-1}{2}}}$


    The median is the "middle" term of the sequence: . .$\displaystyle \boxed{ar^{\frac{n-1}{2}}}$

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  3. #3
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    Excellent. Thank you so much Soroban.

    Any ideas for question 2?

    From wikipedia, I've found that the Harmonic Mean = (geometric mean)^2 divided by the arithmetic mean.

    i.e. $\displaystyle G = \sqrt{AH} $
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