Question

The variance of series $a,a+d,a+2d,.....,a+2nd$ is

• A. $\frac{n(n+1)}{2}{d}^2$
• B. $\frac{n(n+1)}{3}{d}^2$
• C. $\frac{n(n+1)}{6}{d}^2$
• D. $\frac{n(n+1)}{12}{d}^2$

Answer 1

The correct option is B $\frac{n(n+1)}{3}{d}^2$

Given series $a,a+d,a+2d,......,a+2nd$
$\overline{x}=\frac{\frac{(2n+1)}{2}[a+a+2nd]}{(2n+1)}=a+nd$
$\sum (x_i-\overline{x}{)}^2={(a-a-nd)}^2+{(a+d-a-nd)}^2+{(a+2d-a-nd)}^2+.....+{(a+2nd-a-nd)}^2$
$=n^2{d}^2+(1-n{)}^2{d}^2+.......+d^2+0+d^2+.....+n^2{d}^2$
$=2[n^2{d}^2+(n-1{)}^2{d}^2+......+d^2]$
$=2d^2[n^2+(n-1{)}^2+.....+1^2]$
$=2d^2\times \frac{n(n+1)(2n+1)}{6}=\frac{n(n+1)(2n+1)}{3}{d}^2$
$\therefore$ Variance $=\frac{\sum (x_i-\overline{x}{)}^2}{2n+1}=\frac{n(n+1)}{3}{d}^2$