Let x_1, x_2, ldots, x_{100} be in an arithme | vedclass

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(C) Given the mean of $100$ terms is $200$,the sum is $S_{100} = 100 	imes 200 = 20000$. Using the sum formula $S_n = \frac{n}{2}(2a + (n-1)d)$,we ha

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$10049.50$

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(C) Given the mean of $100$ terms is $200$,the sum is $S_{100} = 100 	imes 200 = 20000$. Using the sum formula $S_n = \frac{n}{2}(2a + (n-1)d)$,we have $\frac{100}{2}(2(2) + 99d) = 20000$. $50(4 + 99d) = 20000$ $\Rightarrow 4 + 99d = 400$ $\Rightarrow 99d = 396$ $\Rightarrow d = 4$. The $i$-th term is $x_i = a + (i-1)d = 2 + (i-1)4 = 4i - 2$. Given $y_i = i(x_i - i) = i(4i - 2 - i) = i(3i - 2) = 3i^2 - 2i$. The mean of $y_i$ is $\frac{1}{100} \sum_{i=1}^{100} (3i^2 - 2i)$. $= \frac{1}{100} \left[ 3 \sum_{i=1}^{100} i^2 - 2 \sum_{i=1}^{100} i ight]$. $= \frac{1}{100} \left[ 3 \frac{100(101)(201)}{6} - 2 \frac{100(101)}{2} ight]$. $= \frac{1}{100} \left[ \frac{100(101)(201)}{2} - 100(101) ight] = \frac{101(201)}{2} - 101 = 101(100.5 - 1) = 101 	imes 99.5 = 10049.50$.

Let $x_1, x_2, \ldots, x_{100}$ be in an arithmetic progression,with $x_1 = 2$ and their mean equal to $200$. If $y_i = i(x_i - i)$ for $1 \leq i \leq 100$,then the mean of $y_1, y_2, \ldots, y_{100}$ is

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