Find the Fourth Term 1- square root of 2 , 1 , 1+ square root of 2

$1 - \sqrt{2}$ , $1$ , $1 + \sqrt{2}$

Step 1

This is an arithmetic sequence since there is a common difference between each term. In this case, adding $\sqrt{2}$ to the previous term in the sequence gives the next term. In other words, $a_{n} = a_{1} + d (n - 1)$ .

Arithmetic Sequence: $d = \sqrt{2}$

Step 2

This is the formula of an arithmetic sequence.

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

Step 3

Substitute in the values of $a_{1} = 1 - \sqrt{2}$ and $d = \sqrt{2}$ .

$a_{n} = 1 - \sqrt{2} + \sqrt{2} (n - 1)$

Step 4

Simplify each term.

Apply the distributive property.

$a_{n} = 1 - \sqrt{2} + \sqrt{2} n + \sqrt{2} \cdot - 1$

Move $- 1$ to the left of $\sqrt{2}$ .

$a_{n} = 1 - \sqrt{2} + \sqrt{2} n - 1 \cdot \sqrt{2}$

Rewrite $- 1 \sqrt{2}$ as $- \sqrt{2}$ .

$a_{n} = 1 - \sqrt{2} + \sqrt{2} n - \sqrt{2}$

Step 5

Subtract $\sqrt{2}$ from $- \sqrt{2}$ .

$a_{n} = 1 + \sqrt{2} n - 2 \sqrt{2}$

Step 6

Substitute in the value of $n$ to find the $n$ th term.

$a_{4} = 1 + \sqrt{2} (4) - 2 \sqrt{2}$

Step 7

Move $4$ to the left of $\sqrt{2}$ .

$a_{4} = 1 + 4 \sqrt{2} - 2 \sqrt{2}$

Step 8

Subtract $2 \sqrt{2}$ from $4 \sqrt{2}$ .

$a_{4} = 1 + 2 \sqrt{2}$

Step 9

The result can be shown in multiple forms.

Exact Form:

$a_{4} = 1 + 2 \sqrt{2}$

Decimal Form:

$a_{4} = 3.82842712 \dots$

Find the Fourth Term 1- square root of 2 , 1 , 1+ square root of 2

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