Precalculus Examples

$15$ , $45$

Step 1

This is an arithmetic sequence since there is a common difference between each term. In this case, adding $30$ to the previous term in the sequence gives the next term. In other words, $a_{n} = a_{1} + d (n - 1)$ .

Arithmetic Sequence: $d = 30$

Step 2

This is the formula of an arithmetic sequence.

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

Step 3

Substitute in the values of $a_{1} = 15$ and $d = 30$ .

$a_{n} = 15 + 30 (n - 1)$

Step 4

Simplify each term.

Tap for more steps...

Step 4.1

Apply the distributive property.

$a_{n} = 15 + 30 n + 30 \cdot - 1$

Step 4.2

Multiply $30$ by $- 1$ .

$a_{n} = 15 + 30 n - 30$

Step 5

Subtract $30$ from $15$ .

$a_{n} = 30 n - 15$

Step 6

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

$a_{3} = 30 (3) - 15$

Step 7

Multiply $30$ by $3$ .

$a_{3} = 90 - 15$

Step 8

Subtract $15$ from $90$ .

$a_{3} = 75$