# Geometric Series

• October 21st 2007, 03:07 PM
WWTL@WHL
Geometric Series
Question 1
Let A = ${x_1, x_2, ......x_n}$ 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 $\frac{a(1-r^n)}{1-r}$ ? And the median is $\frac{1}{2} ( x_1 + x_n)$ ?

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Question 2

Let A = ${x_1, x_2, ......x_n}$ 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.
• October 21st 2007, 06:18 PM
Soroban
Hello, WWTL@WHL!

Quote:

Question 1

Let $A \;= \;x_1,\,x_2,\,x_3,\,\cdots\,x_n$ 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.

The geometric mean is the $n^{th}$ root of the product of the $n$ terms.

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

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

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

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

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

• October 22nd 2007, 12:16 PM
WWTL@WHL
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. $G = \sqrt{AH}$