Question

Consider the set $X=\{-5,-3,-1,1,3,5,\mathrm{7...}\}$ consisting of 2012 numbers. Let 'x' be the average of the elements in X and 'y' be twice the average of the first 2012 natural numbers, then what is (x-y) ?

Answer 1

$X=\{-5,-3,-1,1,3,5,\mathrm{7...}\}$ - 2012 numbers
$\therefore S_n=\frac{n}{2}[2a{s}_1+(n-1\left)d\right]$
$=\frac{2012}{2}[2\times -5+(2012-1)\times 2]$
$x=1006\times 2[-5+2011]=2012\times 2006$
$y=2\times \frac{n(n+1)}{2}=\frac{2\times 2012\times 2013}{2}=2012\times 2013$
$(x-y)=2012\times 2006-2012\times 2013$
$=2012\times (-7)=-14084$
$=-14084$.