Arithmetic Sequence (with square roots)

Question

$$t_{n} = t_{1} + (n-1) d$$

What is the common difference for the arithmetic sequence $4-\sqrt{5}, 6, 8+ \sqrt{5}$?

How do you find $d$ here?

Answer 1

You just subtract two consecutive terms. What is $6-(4-\sqrt 5)?$

Answer 2

Same way you'd do any other sequence

$d = t_2 - t_1 = t_3 - t_2 = \frac {t_3 - t_1}2 =$

$ 6 - (4 - \sqrt 5) = (8+\sqrt 5) - 6 = \frac {(8+\sqrt 5)-(4-\sqrt 5)}2 =$

$?????$