Consider three observations $$a, b$$ and $$c$$ such that $$b = a + c$$. If the standard deviation of $$a+2, b+2, c+2$$ is $$d$$, then which of the following is true?

We have three observations $$a, b, c$$ with $$b = a + c$$, and the standard deviation of $$a + 2, b + 2, c + 2$$ is $$d$$.

Since adding a constant to each observation does not change the standard deviation, the standard deviation of $$a, b, c$$ is also $$d$$. The mean of $$a, b, c$$ is $$\bar{x} = \frac{a + b + c}{3} = \frac{a + (a+c) + c}{3} = \frac{2(a+c)}{3} = \frac{2b}{3}$$.

The variance is $$d^2 = \frac{(a - \bar{x})^2 + (b - \bar{x})^2 + (c - \bar{x})^2}{3}$$. Computing each deviation: $$a - \frac{2b}{3} = \frac{3a - 2b}{3} = \frac{3a - 2a - 2c}{3} = \frac{a - 2c}{3}$$, $$b - \frac{2b}{3} = \frac{b}{3} = \frac{a + c}{3}$$, and $$c - \frac{2b}{3} = \frac{3c - 2b}{3} = \frac{c - 2a}{3}$$.

So $$d^2 = \frac{1}{3} \cdot \frac{(a-2c)^2 + (a+c)^2 + (c-2a)^2}{9} = \frac{(a^2 - 4ac + 4c^2) + (a^2 + 2ac + c^2) + (4a^2 - 4ac + c^2)}{27} = \frac{6a^2 - 6ac + 6c^2}{27} = \frac{2(a^2 - ac + c^2)}{9}$$.

Therefore $$9d^2 = 2a^2 - 2ac + 2c^2$$. Now $$3(a^2 + c^2) - 9d^2 = 3a^2 + 3c^2 - 2a^2 + 2ac - 2c^2 = a^2 + 2ac + c^2 = (a + c)^2 = b^2$$.

Hence $$b^2 = 3(a^2 + c^2) - 9d^2$$.