Question

the algebraic sum of the deviation of a set of N values from their mean is

Answer 1

0 is one of the answer

Answer 2

Well in short the answer is Zero.

In detail
Let's assume that the numbers be a,b,c,d,e.....upto n terms
their mean will be (say M)
$M= \frac{a+b+c+d+e+.........+n}{n}$
$So, [a+b+c+d+e+.........+n] =nM$

Now the algebraic sum of the deviation from their mean is
$S=[a-M]+[b-M]+[c-M]+.....+[n-M]$
$S=[a+b+c+d+e+.........+n]-[M+M+M+M+M+..........]$
$S=nM-nM=0$

Hence prooved

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