Question

Find the mean deviation about mean for the series

a , a+d , a+2d , a+3d ......... , a+ nd ?

Answer 1

Mean is just average, right? So it's just
\[\huge \frac{\sum_{k=0}^{n}a+kd}{n+1}\]
Unless I'm missing something?

Answer 2

\[A=\frac{ a+b }{ 2}\]

Answer 3

Now...
\[\huge \sum_{k=0}^{n}a+kd = \sum_{k=1}^{n+1}a+(k-1)d=\sum_{k=1}^{n+1}a-d+kd\]

Answer 4

\[\huge =\sum_{k=1}^{n+1}(a-d) + \sum_{k=1}^{n+1}kd\]

Answer 5

\[\huge \sum_{k=1}^{n+1}(a-d)+ d\sum_{k=1}^{n+1}k\]

Answer 6

Using a rather known formula for the second summation, we get
\[\huge (a-d)(n+1) + \left(\frac{d}{2}\right)(n+1)(n+2)\]

Answer 7

dividing by n+1 to get the mean, we get
\[\huge a - d + \frac{d}{2}(n+2)\]
And that's the average, I reckon :)