Section 7.2

Sequence

Definition 72 (Arithmetic Sequence)

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

\[ a_{n}=a_{1}+(n-1)d \]

Example 133

Determine \(a_{16}\) and \(a_{n}\) for the arithmetic sequence

\[ 23,20,17,14,\dots \]

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

Definition 73 (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.

Example 134

Consider the arithmetic sequence

\[ 48,44,40,36,\dots \]

Evaluate \(S_{21}\)

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.

Solution:

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

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

Example 135

Evaluate

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

Solution:

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

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

Example 136

Evaluate

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

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

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

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

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