the mean deviation of numbers a a+d a+2d .....a+2nd from their mean is equal to

Dear user,

Given sequence is: a, (a+d), (a+2d)… (a+2nd)

Mean

= total sum of the terms/total number of terms

= a + (a+d) + (a+2d) + …... + (a+2nd)/(2n+1)

The total number of a’s in the question is (2n+1). Thus, they would all sum up to (2n+1)a.

Now adding up d + 2d + 3d + …… + 2nd

This is forming an arithmetic progression

= 2n(d + 2nd)/2

= n(2n+1)d

Thus, mean

= ((2n+1)a + (2n+1)nd)/2n+1

=(2n+1)(a+nd)/(2n+1)

= a+ nd.

Now, finding deviation of

(Deviation = value of Term – Mean value)

1 ^st term: a – (a+ nd) = -nd

2 ^nd term: (a+d) – (a+ nd) = (1-n)d = -(n-1)d

3 ^rd term: (a+ 2d) – (a+ nd) = (2 - n)d = -(n-2)d

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2n ^th term: (a+ (2n-1)d) – (a+ nd) = (n-1)d

(2n+1) ^th term: (a+ 2nd) – (a+ nd) = nd

Now, finding mean deviation

= sum of deviations/ number of terms

= (-nd + (-(n-1)d) +…… + (n-1)d + nd)/2n+1

Now, as we see the 1 ^st term is getting cancelled by (2n+1) ^th term, 2 ^nd term is getting cancelled by 2n ^th term, similarly all the terms will get cancelled except the middle term as the number of terms are odd.

Middle term = number of term/2 + 1  (while dividing by two ignore the fractional part and also don’t                                                                        take roundoff eg. If you get 4.5 consider it as 4 only)

= (2n+1)/2 + 1

= (n+1) ^th term

= a + nd

Now, coming back to the formula of mean deviation.

= (a+nd)/(2n+1)

This is the final answer.

Hope you understood the solution.