Content

Means

It is a simple matter to find the average of two numbers. For example, the average of 6 and 10 is 8. When we do this, we are really finding a number \(x\) such that \(6,x,10\) forms an arithmetic sequence. In general, if the numbers \(a,x,b\) form an arithmetic sequence, then

\[ x-a = b-x, \quad\text{giving}\quad x = \dfrac{a+b}{2}. \]

This is the average of \(a\) and \(b\), also called the arithmetic mean (AM) of \(a\) and \(b\).

Similarly, we can define the geometric mean (GM) of two positive numbers \(a\) and \(b\) to be the positive number \(x\) such that \(a,x,b\) forms a geometric sequence. In this case, we require

\[ \dfrac{x}{a} = \dfrac{b}{x}, \quad\text{giving}\quad x = \sqrt{ab}. \]

Example

Find the arithmetic and geometric mean of

  1. \(2,\ 18\)
  2. \(3,\ 6\).

Solution

  1. \(AM = \dfrac{2+18}{2} = 10,\quad GM = \sqrt{2\times 18} = 6\).
  2. \(AM = \dfrac{3+6}{2} = \dfrac{9}{2},\quad GM = \sqrt{3\times 6} = \sqrt{18} = 3\sqrt{2}\).

Exercise 11

Suppose \(a\) and \(b\) are positive real numbers. First, take a line segment \(AB\) of length \(a+b\) and mark a point \(X\) on it such that \(AX=a\) (and so \(XB=b\)). Next, draw a semicircle, centred at the midpoint of \(AB\), with \(AB\) as diameter. Finally, raise a perpendicular to \(AB\) at \(X\) to meet the semicircle at \(Y\).

  1. Prove that the length \(XY\) is the geometric mean of \(a\) and \(b\).
  2. By noting that the radius of the semicircle equals the arithmetic mean of \(a\) and \(b\), deduce that \[ \dfrac{a+b}{2} \geq \sqrt{ab}. \] That is, the arithmetic mean of \(a\) and \(b\) is greater than or equal to their geometric mean.

Generalisations of these two means and examples of other types of means are discussed in the Links forward and History and applications sections.


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