Sum of the First $n$ Terms of an Arithmetic Series

If a series is arithmetic the sum of the first $n$ terms, denoted $S_{n}$ , there are ways to find its sum without actually adding all of the terms.

To find the sum of the first $n$ terms of an arithmetic series use the formula, $n$ terms of an arithmetic sequence use the formula,
$S_{n} = \frac{n (a_{1} + a_{n})}{2}$ ,
where $n$ is the number of terms, $a_{1}$ is the first term and $a_{n}$ is the last term.

The series $3 + 6 + 9 + 12 + \dots + 30$ can be expressed as sigma notation $\sum_{n = 1}^{10} 3 n$ . This expression is read as the sum of $3 n$ as $n$ goes from $1$ to $10$

Example 1:

Find the sum of the first $20$ terms of the arithmetic series if $a_{1} = 5$ and $a_{20} = 62$ .

$\begin{array}{l} S_{20} = \frac{20 (5 + 62)}{2} \\ S_{20} = 670 \end{array}$

Example 2:

Find the sum of the first $40$ terms of the arithmetic sequence
$2, 5, 8, 11, 14, \dots$

First find the $40$ ^th term:

$\begin{array}{l} a_{40} = a_{1} + (n - 1) d \\ = 2 + 39 (3) = 119 \end{array}$

Then find the sum:

$\begin{array}{l} S_{n} = \frac{n (a_{1} + a_{n})}{2} \\ S_{40} = \frac{40 (2 + 119)}{2} = 2420 \end{array}$

Example 3:

Find the sum:

$\sum_{k = 1}^{50} (3 k + 2)$

First find $a_{1}$ and $a_{50}$ :

$\begin{array}{l} a_{1} = 3 (1) + 2 = 5 \\ a_{20} = 3 (50) + 2 = 152 \end{array}$

Then find the sum:

$\begin{array}{l} S_{k} = \frac{k (a_{1} + a_{k})}{2} \\ S_{50} = \frac{50 (5 + 152)}{2} = 3925 \end{array}$

Sum of the First n Terms of an Arithmetic Series

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Sum of the First $n$ Terms of an Arithmetic Series