Find out the value of $d$

Question

If the mean deviation of number $1,\ 1+d,\ 1+2d,\ 1+3d,\ldots,1+100d$ from their mean deviation $255$ then $d$ equals to ?

This was the question asked in AIEEE 2009.

MY EFFORTS: $$\bar{x}=\frac{1^\text{st}\text{ observation} + n^\text{th}\text{ observation}}{2}$$ $$\bar{x}=\frac{1 + (1+100d)}{2}=1+50d$$

$$MEAN\ DEVIATION=\frac{1}{n}\sum x_i-\bar{x}$$ $$MEAN\ DEVIATION=\frac{1}{101}\sum x_i-1-50d$$

Now further i don't know what to do.

Answer 1

Answer:

This should be Mean Absolute Deviation.

Mean = 1+50d

You should calculate the $$\sum_{i=0}^{i=49}x_i=50 + 1225 d$$ $$MEAN ABSOLUTE DEVIATION=2*\left(\frac{1}{101}\sum_{i = 0}^{49}\left|( x_i-(1+50d))\right|\right)$$ $$MEAN ABSOLUTE DEVIATION=2*\frac{1}{101}|\left(\sum_{i = 0}^{49}( x_i)\right)-\left(50*(1+50d)\right)|$$ $$MEAN ABSOLUTE DEVIATION=\left(\frac{1}{101}\right)\left(|100+2450d- 100-5000d|\right) = 255$$ $$2550d = 255*101$$ $$d = \frac{101}{10} = 10.1$$

Thanks

Satish

Answer 2

Hint

It seems that you are bumping on the evaluation of $$\sum_{i=0}^{i=100}x_i=\sum_{i=0}^{i=100}(1+i d)$$ This ia an arithmetic series and the usual formula leads to $$\sum_{i=0}^{i=100}x_i=101 + 5050 d$$

I am sure that you can take from here.