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(D) The terms are $x_{i} = 3 	imes (\frac{1}{2})^{i-1}$.We need to find the mean $\bar{x} = \frac{1}{20} \sum_{i=1}^{20} (x_{i} - i)^{2}$.Expanding t

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$142$

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(D) The terms are $x_{i} = 3 	imes (\frac{1}{2})^{i-1}$.We need to find the mean $\bar{x} = \frac{1}{20} \sum_{i=1}^{20} (x_{i} - i)^{2}$.Expanding the sum: $\sum_{i=1}^{20} (x_{i}^{2} - 2ix_{i} + i^{2}) = \sum x_{i}^{2} - 2 \sum ix_{i} + \sum i^{2}$.$1$. $\sum_{i=1}^{20} x_{i}^{2}$ is a $GP$ with first term $a = 9$ and ratio $r = \frac{1}{4}$. Sum $= \frac{9(1 - (1/4)^{20})}{1 - 1/4} = 12(1 - \frac{1}{2^{40}})$.$2$. $\sum_{i=1}^{20} i^{2} = \frac{20(21)(41)}{6} = 2870$.$3$. $\sum_{i=1}^{20} ix_{i} = 3(1) + 3(\frac{1}{2})(2) + 3(\frac{1}{4})(3) + \ldots + 3(\frac{1}{2^{19}})(20)$. This is an $AGP$.Let $S = 3 + 3 + \frac{9}{4} + \ldots$. Using the formula for $AGP$,$S = 12 - \frac{12(22)}{2^{20}} = 12 - \frac{264}{2^{20}}$.Substituting these into the mean formula:$\bar{x} = \frac{1}{20} [12(1 - \frac{1}{2^{40}}) - 2(12 - \frac{264}{2^{20}}) + 2870] = \frac{1}{20} [12 - \frac{12}{2^{40}} - 24 + \frac{528}{2^{20}} + 2870] = \frac{2858}{20} + 	ext{negligible terms} \approx 142.9$.The greatest integer less than or equal to $\bar{x}$ is $142$.

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