TDSM 2.9

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When all terms [math] x_0 = x_1 = \dots = x_{n-1} [/math]

The arthmetic mean is [math]\frac{1}{n} \sum_{i = 0}^{n-1} = \frac{1}{n}\times nx_0 = x_0[/math]

The geometric mean is [math] \sqrt[n]{\prod_{i = 0}^{n-1}} = \sqrt[n]{x_0 ^n} = x_0[/math]

So the arithmetic mean equals the geometric mean when all terms are the same.