Arithmetic sequence

Calculus

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Series and Sequences

schedule Jul 1, 2022

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local_offer Arthur

General form

The general form of an arithmetic sequence can be expressed as follows:

$$\{a, a + d, a+2d, ..., a+(n-1)d\}$$

$$1st\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,nth \,\, \text{term}$$

where:

$a$ is the first term
$d$ is the common difference between the terms

Example

Consider the following arithmetic sequence:

$$\{2, 6,10,14,18,22,...\}$$

Here we start with 2 and add 4 to get the next term in the arithmetic sequence.

Therefore in terms of the general form of the arithmetic sequence we can say:

$a$ = 2 (first term)

$d$ = 4 (common difference)

Sum of arithmetic series

To compute the sum $S_n$ of an arithmetic series:

$$S_n = a + (a+d) + \cdots + (a+(n-1)d)$$

We can use the following formula:

$$S_n = \frac{n[a+a_n]}{2}$$

Example

Question. To calculate $S_n$:

$$\{2,6,10,14,18,22\}$$

Solution. Comparing to the general form we can see that $a=2$, $d=4$ and $n=6$. Now plugging these numbers into the formula for sum of arithmetic series:

$$\begin{align*} S_n &= \frac{6[2 + 22]}{2} \\ &= 72 \end{align*}$$

Derivation

We can write the equation for sum of an arithmetic series in the following two ways:

$$\begin{align*} S_n &= {\color{red}a} +{\color{blue}(a+d)}+ \cdots + {\color{green}(a+(n-1)d)} \\ S_n &={\color{red}[a+(n-1)d]} +{\color{blue}[a+(n-2)d]}+\cdots+{\color{green}a} \end{align*}$$

Adding the two equations together:

$$2S_n=[2a+nd-d]+[2a+nd-d]+⋯+[2a+nd-d]$$

Note that there are $n$ terms altogether so the above simplifies to:

$$\begin{align*} 2S_n &= n[2a+nd-d] \\ S_n &= \frac{n[2a+nd-d]}{2} \\ S_n &= \frac{n[a+ {\color{red}a + (n-1)d}]}{2} \end{align*}$$

The key is that the red component can be re-written as:

$$S_n = \frac{n[a+ {\color{red}a_n}]}{2}$$