Question

The standard deviation of 4 consecutive numbers which are in A.P is $\sqrt{5}$. The common difference (d) of this A.P is

• A. $\pm \sqrt{5}$
• B. $\pm 2\sqrt{5}$
• C. $\pm 2$
  
• D. $\pm \sqrt{2}$

Answer 1

The correct option is C

$\pm 2$

Let the four numbers be a, a+d, a+2d and a+3d
Mean, $\overline{x}=\frac{a+(a+d)+(a+2d)+(a+3d)}{4}=a+1.5d\sigma =\sqrt{\frac{\sum (x_i-\overline{x}{)}^2}{4}}=\sqrt{\frac{(-1.5d{)}^2+(-0.5d{)}^2+(0.5d{)}^2+(1.5d{)}^2}{4}}=\sqrt{5}\Rightarrow 2.25d^2+0.25d^2+0.25d^2+2.25d^2=5\times 4\Rightarrow 5d^2=20\Rightarrow d^2=4\Rightarrow d=\pm 2$