# Section 7.2


## Sequence

:::{prf:definition} Arithmetic Sequence
:label: arthSeq

An arithmetic sequence is a sequence in which each terms after the first differs from the preceding term by a fixed constant, called the common difference. The common difference is defined as

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

where $n$ is in the domain of the sequence.

In an arithmetic sequence with first term $a_{1}$ and **common difference** $d$, the $n$th term $a_{n}$ is given by the following

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

:::

::::{prf:example}
:label: arthSeqExam1

Determine $a_{16}$ and $a_{n}$ for the arithmetic sequence

$$
23,20,17,14,\dots
$$

:::{dropdown} Solution:

First, we will find the common difference. Since we know this is an
arithmetic sequence we can pick any two consecutive numbers in the
sequence and subtract the former from the later

\begin{align*}
d & =a_{2}-a_{1}\\
 & =20-23=-3
\end{align*}
This means the common difference is $d=-3$ and we know $a_{1}=23$

we then have

\begin{align*}
a_{n} & =23+(n-1)(-3)\\
 & =23-3n+3\\
 & =26-3n
\end{align*}

we can check this with the given sequence 

$$a_{1}=26-3(1)=23$$

and 

$$a_{3}=26-3(3)=17$$

and so on.

Now we can evaluate $a_{16}$

\begin{align*}
a_{16} & =26-3(16)\\
 & =-22
\end{align*}

Therefore, $a_{n}=26-3n$ and $a_{16}=-22$.

:::
::::

## Series

:::{prf:definition} Arithmetic Series

An arithmetic series is the sum of the terms of an arithmetic sequence.

If an arithmetic sequence has first term $a_{1}$ and common different $d$, then the sum $S_{n}$ of the first $n$ terms is given by the following

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

or

$$
S_{n}=\frac{n}{2}\left(2a_{1}+(n-1)d\right).
$$

The first formula is used when the first and last terms are known; otherwise, the second formula is used.
:::

::::{prf:example}
:label: arthSeriesExam1

Consider the arithmetic sequence

$$
48,44,40,36,\dots
$$

Evaluate $S_{21}$

:::{dropdown} Solution:

Using the definition we have

$$
S_{21}=\frac{21}{2}\left(2(48)+(21-1)d\right)
$$

were we need to find $d$. Since this is an arithmetic sequence we can find $d=44-48=-4$. Now we know

\begin{align*}
S_{21} & =\frac{21}{2}\left(2(48)+(21-1)(-4)\right)\\
 & =\frac{21}{2}(96+(-80))\\
 & =\frac{21}{2}(96-80)\\
 & =\frac{21}{2}(16)\\
 & =168
\end{align*}

:::

Evaluate the sum of the first 200 positive integers.

:::{dropdown} Solution:

Since $d=1$ for the list of positive integers we have $a_{1}=1$ and $a_{n}=a_{200}=200$.

\begin{align*}
S_{200} & =\frac{200}{2}\left(a_{1}+a_{200}\right)\\
 & =100(1+200)\\
 & =100(201)\\
 & =20100
\end{align*}

:::
::::

::::{prf:example}
:label: arthSeriesExam2

Evaluate

$$
\sum_{i=1}^{10}\left(4i+8\right)
$$

:::{dropdown} Solution:

We know $a_{1}=4(1)+8=12$ and $a_{10}=4(10)+8=48$. We want to find $S_{10}$ which is

\begin{align*}
S_{10} & =\frac{10}{2}\left(a_{1}+a_{10}\right)\\
 & =5(12+48)\\
 & =5(60)\\
 & =300
\end{align*}

:::
::::

::::{prf:example}
:label: arthSeriesExam3
Evaluate

$$
\sum_{i=5}^{10}(8-2i)
$$

:::{dropdown} Solution:

We have

\begin{align*}
\sum_{i=5}^{10}(8-2i) & =\sum_{i=1}^{10}(8-2i)-\sum_{i=1}^{4}(8-2i)\\
a_{1} & =8-2(1)=6\\
a_{4} & =8-2(4)=0\\
a_{10} & =8-2(10)=-12
\end{align*}

and

\begin{align*}
\sum_{i=1}^{10}(8-2i) & =\frac{10}{2}\left(6+(-12)\right)\\
 & =\frac{10}{2}\cdot(-6)\\
 & =-30
\end{align*}
\begin{align*}
\sum_{i=1}^{4}(8-2i) & =\frac{4}{2}\left(6+0\right)\\
 & =\frac{4}{2}\cdot(6)\\
 & =12
\end{align*}

\begin{align*}
\sum_{i=5}^{10}(8-2i) & =\sum_{i=1}^{10}(8-2i)-\sum_{i=1}^{4}(8-2i)\\
 & =-30-12\\
 & =-42
\end{align*}

:::
::::