If a_0, a_1, ldots, a_{11} are in an arithmet | vedclass

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(C) The mean of the given $12$ terms in arithmetic progression is $\bar{x} = \frac{a_0 + a_{11}}{2}$.Since $a_n = a_0 + nd$,we have $a_{11} = a_0 + 11

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$3|d|$

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(C) The mean of the given $12$ terms in arithmetic progression is $\bar{x} = \frac{a_0 + a_{11}}{2}$.Since $a_n = a_0 + nd$,we have $a_{11} = a_0 + 11d$,so $\bar{x} = a_0 + \frac{11}{2}d$.The terms are $a_0, a_0+d, \ldots, a_0+11d$.The deviations from the mean are $|a_k - \bar{x}| = |a_0 + kd - (a_0 + 5.5d)| = |k - 5.5||d|$.For $k = 0, 1, \ldots, 11$,the values of $|k - 5.5|$ are $5.5, 4.5, 3.5, 2.5, 1.5, 0.5, 0.5, 1.5, 2.5, 3.5, 4.5, 5.5$.The sum of these deviations is $2 	imes (5.5 + 4.5 + 3.5 + 2.5 + 1.5 + 0.5)|d| = 2 	imes 18|d| = 36|d|$.The mean deviation is $\frac{36|d|}{12} = 3|d|$.

${x_i}$	$5$	$7$	$9$	$10$	$12$	$15$
${f_i}$	$8$	$6$	$2$	$2$	$2$	$6$

If $a_0, a_1, \ldots, a_{11}$ are in an arithmetic progression with common difference $d$,then their mean deviation from their arithmetic mean is

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