If \(a_0, a_1, \ldots, a_{11}\) are in an arithmetic progression with common difference \(d\), then their…
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
\(\frac{30}{11}|d|\)
\(2|d|\)
\(3|d|\)
\(12|d|\)
Solution
Since, mean of given data is
\(\begin{aligned}
\bar{x} & =\frac{a_0+a_1+a_2+\ldots+a_{11}}{12}=\frac{a_0+a_{11}}{2} \\
& =\frac{a_1+a_{10}}{2}=\ldots
\end{aligned}\)
Now, deviations from their mean are
\(\begin{aligned}
& \left|\bar{x}-a_0\right|=\frac{\left|a_{11}-a_0\right|}{2}=\frac{11|d|}{2} \\
& \left|\bar{x}-a_1\right|=\frac{\left|a_{10}-a_1\right|}{2}=\frac{9|d|}{2}, \ldots . . . \text { and so on }
\end{aligned}\)
So, sum of deviations
\(\begin{aligned}
& =2\left[\frac{11|d|}{2}+\frac{9|d|}{2}+\frac{7|d|}{2}+\frac{5|d|}{2}+\frac{3|d|}{2}+\frac{|d|}{2}\right] \\
& =36|d|
\end{aligned}\)
\(\therefore\) Mean deviation from their arithmetic mean is
\(\frac{36|d|}{12}=3|d|\)
Hence, option (3) is correct.