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?

You are overthinking this. The common difference is ... the difference between any two consecutive terms in the sequence! So subtract either of the two consecutive pairs and you will get the common difference. (Check your work by subtracting both pairs!) — hardmath, Dec 20 '18 at 06:30

score 0 · Accepted Answer · answered Dec 20 '18 at 00:33

0

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

answered Dec 20 '18 at 00:33

Ross Millikan

score 0 · Answer 2 · answered Dec 20 '18 at 00:35

0

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 =$

$?????$

answered Dec 20 '18 at 00:35

fleablood

2 Answers2