Precalculus Examples

$0.1$ , $0.2$ , $0.3$ , $0.4$ , $0.5$ , $0.6$ , $0.7$ , $0.8$ , $0.9$

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 $0.1$ to the previous term in the sequence gives the next term. In other words, $a_{n} = a_{1} + d (n - 1)$ .

Arithmetic Sequence: $d = 0.1$

Step 3

This is the formula of an arithmetic sequence.

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

Step 4

Substitute in the values of $a_{1} = 0.1$ and $d = 0.1$ .

$a_{n} = 0.1 + 0.1 (n - 1)$

Step 5

Simplify each term.

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Step 5.1

Apply the distributive property.

$a_{n} = 0.1 + 0.1 n + 0.1 \cdot - 1$

Step 5.2

Multiply $0.1$ by $- 1$ .

$a_{n} = 0.1 + 0.1 n - 0.1$

Step 6

Combine the opposite terms in $0.1 + 0.1 n - 0.1$ .

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Step 6.1

Subtract $0.1$ from $0.1$ .

$a_{n} = 0.1 n + 0$

Step 6.2

Add $0.1 n$ and $0$ .

$a_{n} = 0.1 n$

Step 7

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

$a_{9} = 0.1 (9)$

Step 8

Multiply $0.1$ by $9$ .

$a_{9} = 0.9$

Step 9

Replace the variables with the known values to find $S_{9}$ .

$S_{9} = \frac{9}{2} \cdot (0.1 + 0.9)$

Step 10

Add $0.1$ and $0.9$ .

$S_{9} = \frac{9}{2} \cdot 1$

Step 11

Cancel the common factor of $1$ .

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Step 11.1

Rewrite $2$ as $1 (2)$ .

$S_{9} = \frac{9}{1 (2)} \cdot 1$

Step 11.2

Cancel the common factor.

$S_{9} = \frac{9}{1 \cdot 2} \cdot 1$

Step 11.3

Rewrite the expression.

$S_{9} = \frac{9}{2}$

Step 12

Convert the fraction to a decimal.

$S_{9} = 4.5$

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