Algebra Examples

1

$1$ ,

6

$6$ ,

11

$11$ ,

16

$16$ ,

21

$21$ ,

26

$26$

Step 1

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

5

$5$ 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 = 5

$d = 5$

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} = 1

$a_{1} = 1$ and

d = 5

$d = 5$ .

a_{n} = 1 + 5 (n - 1)

$a_{n} = 1 + 5 (n - 1)$

Step 4

Simplify each term.

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Step 4.1

Apply the distributive property.

a_{n} = 1 + 5 n + 5 \cdot - 1

$a_{n} = 1 + 5 n + 5 \cdot - 1$

Step 4.2

Multiply

5

$5$ by

- 1

$- 1$ .

a_{n} = 1 + 5 n - 5

$a_{n} = 1 + 5 n - 5$

a_{n} = 1 + 5 n - 5

$a_{n} = 1 + 5 n - 5$

Step 5

Subtract

5

$5$ from

1

$1$ .

a_{n} = 5 n - 4

$a_{n} = 5 n - 4$

Step 6

Substitute in the value of

n

$n$ to find the

n

$n$ th term.

a_{7} = 5 (7) - 4

$a_{7} = 5 (7) - 4$

Step 7

Multiply

5

$5$ by

7

$7$ .

a_{7} = 35 - 4

$a_{7} = 35 - 4$

Step 8

Subtract

4

$4$ from

35

$35$ .

a_{7} = 31

$a_{7} = 31$