Precalculus Examples

$0$ , $4$ , $8$ , $12$ , $16$

Step 1

This is the formula to find the sum of the first $n$ terms of the sequence. To evaluate it, the values of the first and $n$ th terms must be found.

$S_{n} = \frac{n}{2} \cdot (a_{1} + a_{n})$

Step 2

This is an arithmetic sequence since there is a common difference between each term. In this case, adding $4$ to the previous term in the sequence gives the next term. In other words, $a_{n} = a_{1} + d (n - 1)$ .

Arithmetic Sequence: $d = 4$

Step 3

This is the formula of an arithmetic sequence.

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

Step 4

Substitute in the values of $a_{1} = 0$ and $d = 4$ .

$a_{n} = 0 + 4 (n - 1)$

Step 5

Add $0$ and $4 (n - 1)$ .

$a_{n} = 4 (n - 1)$

Step 6

Apply the distributive property.

$a_{n} = 4 n + 4 \cdot - 1$

Step 7

Multiply $4$ by $- 1$ .

$a_{n} = 4 n - 4$

Step 8

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

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

Step 9

Multiply $4$ by $5$ .

$a_{5} = 20 - 4$

Step 10

Subtract $4$ from $20$ .

$a_{5} = 16$

Step 11

Replace the variables with the known values to find $S_{5}$ .

$S_{5} = \frac{5}{2} \cdot (0 + 16)$

Step 12

Add $0$ and $16$ .

$S_{5} = \frac{5}{2} \cdot 16$

Step 13

Cancel the common factor of $2$ .

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Step 13.1

Factor $2$ out of $16$ .

$S_{5} = \frac{5}{2} \cdot (2 (8))$

Step 13.2

Cancel the common factor.

$S_{5} = \frac{5}{2} \cdot (2 \cdot 8)$

Step 13.3

Rewrite the expression.

$S_{5} = 5 \cdot 8$

Step 14

Multiply $5$ by $8$ .

$S_{5} = 40$

Step 15

Convert the fraction to a decimal.

$S_{5} = 40$

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