Precalculus Examples

3

$3$ ,

11

$11$ ,

19

$19$ ,

27

$27$ ,

35

$35$

Step 1

This is an arithmetic sequence since there is a common difference between each term. In this case, adding

8

$8$ to the previous term in the sequence gives the next term. In other words,

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

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

Arithmetic Sequence:

d = 8

$d = 8$

Step 2

This is the formula of an arithmetic sequence.

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

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

Step 3

Substitute in the values of

a_{1} = 3

$a_{1} = 3$ and

d = 8

$d = 8$ .

a_{n} = 3 + 8 (n - 1)

$a_{n} = 3 + 8 (n - 1)$

Step 4

Simplify each term.

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Step 4.1

Apply the distributive property.

a_{n} = 3 + 8 n + 8 \cdot - 1

$a_{n} = 3 + 8 n + 8 \cdot - 1$

Step 4.2

Multiply

8

$8$ by

- 1

$- 1$ .

a_{n} = 3 + 8 n - 8

$a_{n} = 3 + 8 n - 8$

a_{n} = 3 + 8 n - 8

$a_{n} = 3 + 8 n - 8$

Step 5

Subtract

8

$8$ from

3

$3$ .

a_{n} = 8 n - 5

$a_{n} = 8 n - 5$

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