The mean of n items is bar x. If the first te | vedclass

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(C) Let the $n$ items be ${x_1}, {x_2}, \dots, {x_n}$. Then,the original mean is $\bar x = \frac{1}{n} \sum_{i=1}^{n} {x_i}$.Let the new items be ${y_

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$\bar x + \frac{n + 1}{2}$

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(C) Let the $n$ items be ${x_1}, {x_2}, \dots, {x_n}$. Then,the original mean is $\bar x = \frac{1}{n} \sum_{i=1}^{n} {x_i}$.Let the new items be ${y_i} = {x_i} + i$ for $i = 1, 2, \dots, n$.The new mean is $\bar y = \frac{1}{n} \sum_{i=1}^{n} {y_i} = \frac{1}{n} \sum_{i=1}^{n} ({x_i} + i)$.This can be written as $\bar y = \frac{1}{n} \sum_{i=1}^{n} {x_i} + \frac{1}{n} \sum_{i=1}^{n} i$.Since $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$,we have $\bar y = \bar x + \frac{1}{n} \cdot \frac{n(n+1)}{2}$.Therefore,the new mean is $\bar x + \frac{n+1}{2}$.

The mean of $n$ items is $\bar x$. If the first term is increased by $1$,the second by $2$,and so on,then the new mean is:

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