The sum of 3 numbers that form an arithmetic sequence is 12. Determine the mean and standard deviation. The answers are $4$ and $\frac{2\sqrt{3}}{3}$ respectively.
My attempt:
$U_1=a-d$
$U_2=a$
$U_3=a+d$
$U_1+U_2+U_3 = 3a = 12$
$a=4$
$Mean = \frac{12}{3}=4$
$(U_1-Mean)^2 = d^2$
$(U_2-Mean)^2 = 0$
$(U_3-Mean)^2 = d^2$
$Total=2d^2$
$\sigma = \sqrt{\frac{2d^2}{3}} = d\frac{\sqrt{6}}{3}$
I can't obtain the difference $d$, which results in not obtaining $\sigma=\frac{2\sqrt{3}}{3}.$