Find the Sum of the Series 3+6+9+12+15

$3 + 6 + 9 + 12 + 15$

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.

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

This is an arithmetic sequence since there is a common difference between each term. In this case, adding $3$ to the previous term in the sequence gives the next term. In other words, $a_{n} = a_{1} + d (n - 1)$ .

Arithmetic Sequence: $d = 3$

This is the formula of an arithmetic sequence.

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

Substitute in the values of $a_{1} = 3$ and $d = 3$ .

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

Simplify each term.

Apply the distributive property.

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

Multiply $3$ by $- 1$ .

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

Combine the opposite terms in $3 + 3 n - 3$ .

Subtract $3$ from $3$ .

$a_{n} = 3 n + 0$

Add $3 n$ and $0$ .

$a_{n} = 3 n$

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

$a_{5} = 3 (5)$

Multiply $3$ by $5$ .

$a_{5} = 15$

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

$S_{5} = \frac{5}{2} \cdot (3 + 15)$

Add $3$ and $15$ .

$S_{5} = \frac{5}{2} \cdot 18$

Cancel the common factor of $2$ .

Factor $2$ out of $18$ .

$S_{5} = \frac{5}{2} \cdot (2 (9))$

Cancel the common factor.

$S_{5} = \frac{5}{2} \cdot (2 \cdot 9)$

Rewrite the expression.

$S_{5} = 5 \cdot 9$

Multiply $5$ by $9$ .

$S_{5} = 45$

Convert the fraction to a decimal.

$S_{5} = 45$

Find the Sum of the Series 3+6+9+12+15

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