This is “Arithmetic Sequences and Series”, section 9.2 from the book Advanced Algebra (v. 1.0).
For more information on the source of this book, or why it is available for free, please see the project's home page. You can browse or download additional books there. You may also download a PDF copy of this book (41 MB) or just this chapter (1 MB), suitable for printing or most e-readers, or a .zip file containing this book's HTML files (for use in a web browser offline).
An arithmetic sequenceA sequence of numbers where each successive number is the sum of the previous number and some constant d., or arithmetic progressionUsed when referring to an arithmetic sequence., is a sequence of numbers where each successive number is the sum of the previous number and some constant d.
$${a}_{n}={a}_{n-1}+d\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{Arithmetic\text{\hspace{0.17em}}Sequence}$$
And because ${a}_{n}-{a}_{n-1}=d$, the constant d is called the common differenceThe constant d that is obtained from subtracting any two successive terms of an arithmetic sequence; ${a}_{n}-{a}_{n-1}=d.$. For example, the sequence of positive odd integers is an arithmetic sequence,
$$1,3,5,7,9,\dots $$
Here ${a}_{1}=1$ and the difference between any two successive terms is 2. We can construct the general term ${a}_{n}={a}_{n-1}+2$ where,
$$\begin{array}{ccc}\hfill {a}_{1}& =& 1\hfill \\ \hfill {a}_{2}& =& {a}_{1}+2=1+2=3\hfill \\ \hfill {a}_{3}& =& {a}_{2}+2=3+2=5\hfill \\ \hfill {a}_{4}& =& {a}_{3}+2=5+2=7\hfill \\ \hfill {a}_{5}& =& {a}_{4}+2=7+2=9\hfill \\ & \vdots & \hfill \end{array}$$
In general, given the first term ${a}_{1}$ of an arithmetic sequence and its common difference d, we can write the following:
$$\begin{array}{ccc}\hfill {a}_{2}& =& {a}_{1}+d\hfill \\ \hfill {a}_{3}& =& {a}_{2}+d=\left({a}_{1}+d\right)+d={a}_{1}+2d\hfill \\ \hfill {a}_{4}& =& {a}_{3}+d=\left({a}_{1}+2d\right)+d={a}_{1}+3d\hfill \\ \hfill {a}_{5}& =& {a}_{4}+d=\left({a}_{1}+3d\right)+d={a}_{1}+4d\hfill \\ & \vdots & \hfill \end{array}$$
From this we see that any arithmetic sequence can be written in terms of its first element, common difference, and index as follows:
$${a}_{n}={a}_{1}+\left(n-1\right)d\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{Arithmetic\text{\hspace{0.17em}}Sequence}$$
In fact, any general term that is linear in n defines an arithmetic sequence.
Find an equation for the general term of the given arithmetic sequence and use it to calculate its 100^{th} term: $7,10,13,16,19,\dots $
Solution:
Begin by finding the common difference,
$$d=10-7=3$$
Note that the difference between any two successive terms is 3. The sequence is indeed an arithmetic progression where ${a}_{1}=7$ and $d=3.$
$$\begin{array}{ccc}\hfill {a}_{n}& =& {a}_{1}+\left(n-1\right)d\hfill \\ & =& 7+\left(n-1\right)\cdot 3\hfill \\ & =& 7+3n-3\hfill \\ & =& 3n+4\hfill \end{array}$$
Therefore, we can write the general term ${a}_{n}=3n+4.$ Take a minute to verify that this equation describes the given sequence. Use this equation to find the 100^{th} term:
$${a}_{100}=3\left(100\right)+4=304$$
Answer: ${a}_{n}=3n+4$; ${a}_{100}=304$
The common difference of an arithmetic sequence may be negative.
Find an equation for the general term of the given arithmetic sequence and use it to calculate its 75^{th} term: $6,4,2,0,-2,\dots $
Solution:
Begin by finding the common difference,
$$d=4-6=-2$$
Next find the formula for the general term, here ${a}_{1}=6$ and $d=-2.$
$$\begin{array}{ccc}\hfill {a}_{n}& =& {a}_{1}+\left(n-1\right)d\hfill \\ & =& 6+\left(n-1\right)\cdot \left(-2\right)\hfill \\ & =& 6-2n+2\hfill \\ & =& 8-2n\hfill \end{array}$$
Therefore, ${a}_{n}=8-2n$ and the 75^{th} term can be calculated as follows:
$$\begin{array}{ccc}\hfill {a}_{75}& =& 8-2\left(75\right)\hfill \\ & =& 8-150\hfill \\ & =& -142\hfill \end{array}$$
Answer: ${a}_{n}=8-2n$; ${a}_{100}=-142$
The terms between given terms of an arithmetic sequence are called arithmetic meansThe terms between given terms of an arithmetic sequence..
Find all terms in between ${a}_{1}=-8$ and ${a}_{7}=10$ of an arithmetic sequence. In other words, find all arithmetic means between the 1^{st} and 7^{th} terms.
Solution:
Begin by finding the common difference d. In this case, we are given the first and seventh term:
$$\begin{array}{cccc}\hfill {a}_{n}& =& {a}_{1}+\left(n-1\right)d\hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}{Use\text{\hspace{0.17em}}n=7.}\\ & & & \\ \hfill {a}_{7}& =& {a}_{1}+\left(7-1\right)d\hfill & \\ \hfill {a}_{7}& =& {a}_{1}+6d\hfill & \end{array}$$
Substitute ${a}_{1}=-8$ and ${a}_{7}=10$ into the above equation and then solve for the common difference d.
$$\begin{array}{ccc}\hfill 10& =& -8+6d\hfill \\ \hfill 18& =& 6d\hfill \\ \hfill 3& =& d\hfill \end{array}$$
Next, use the first term ${a}_{1}=-8$ and the common difference $d=3$ to find an equation for the nth term of the sequence.
$$\begin{array}{ccc}\hfill {a}_{n}& =& -8+\left(n-1\right)\cdot 3\hfill \\ & =& -8+3n-3\hfill \\ & =& -11+3n\hfill \end{array}$$
With ${a}_{n}=3n-11$, where n is a positive integer, find the missing terms.
$$\begin{array}{c}{a}_{1}=3\left(1\right)-11=3-11=-8\phantom{\rule{10.1em}{0ex}}\\ \begin{array}{c}\phantom{\rule{0.5em}{0ex}}{a}_{2}=3\left(2\right)-11=6-11=-5\\ \phantom{\rule{0.6em}{0ex}}{a}_{3}=3\left(3\right)-11=9-11=-2\\ {a}_{4}=3\left(4\right)-11=12-11=1\\ {a}_{5}=3\left(5\right)-11=15-11=4\\ \phantom{\rule{0.75em}{0ex}}{a}_{6}=3\left(6\right)-11=18-11=7\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}\}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{arithmetic\text{\hspace{0.17em}}means}\\ {a}_{7}=3\left(7\right)-11=21-11=10\phantom{\rule{10.1em}{0ex}}\end{array}$$
Answer: −5, −2, 1, 4, 7
In some cases, the first term of an arithmetic sequence may not be given.
Find the general term of an arithmetic sequence where ${a}_{3}=-1$ and ${a}_{10}=48.$
Solution:
To determine a formula for the general term we need ${a}_{1}$ and $d.$ A linear system with these as variables can be formed using the given information and ${a}_{n}={a}_{1}+\left(n-1\right)d$:
$$\begin{array}{cccc}\{\begin{array}{c}{a}_{3}={a}_{1}+\left(3-1\right)d\\ {a}_{10}={a}_{1}+\left(10-1\right)d\end{array}& \underset{\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{0.17em}}}{\Rightarrow}& \{\begin{array}{c}-1={a}_{1}+2d\\ 48={a}_{1}+9d\end{array}& \begin{array}{c}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{Use\text{\hspace{0.17em}}{a}_{3}=-1.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}{Use\text{\hspace{0.17em}}{a}_{10}=48.}\end{array}\end{array}$$
Eliminate ${a}_{1}$ by multiplying the first equation by −1 and add the result to the second equation.
$$\begin{array}{l}\{\begin{array}{ccc}-1& =& {a}_{1}+2d\\ 48& =& {a}_{1}+9d\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{c}\text{\hspace{0.17em}}\underset{\text{\hspace{0.17em}}\text{\hspace{0.17em}}}{\stackrel{{\times \left(-1\right)}\text{\hspace{0.17em}}}{\Rightarrow}}\\ \end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{\xaf}{\begin{array}{c}\\ +\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}\{\begin{array}{ccc}1& =& -{a}_{1}-2d\\ 48& =\text{\hspace{0.17em}}& \text{\hspace{0.17em}}{a}_{1}+\text{\hspace{0.17em}}\text{\hspace{0.17em}}9d\end{array}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{ccc}49& =& 7d\\ 7& =& d\end{array}\text{\hspace{0.17em}}\text{\hspace{0.05em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$$
Substitute $d=7$ into $-1={a}_{1}+2d$ to find ${a}_{1}.$
$$\begin{array}{ccc}\hfill -1& =& {a}_{1}+2\left(7\right)\hfill \\ \hfill -1& =& {a}_{1}+14\hfill \\ \hfill -15& =& {a}_{1}\hfill \end{array}$$
Next, use the first term ${a}_{1}=-15$ and the common difference $d=7$ to find a formula for the general term.
$$\begin{array}{ccc}\hfill {a}_{n}& =& {a}_{1}+\left(n-1\right)d\hfill \\ & =& -15+\left(n-1\right)\cdot 7\hfill \\ & =& -15+7n-7\hfill \\ & =& -22+7n\hfill \end{array}$$
Answer: ${a}_{n}=7n-22$
Try this! Find an equation for the general term of the given arithmetic sequence and use it to calculate its 100^{th} term: $\frac{3}{2},2,\frac{5}{2},3,\frac{7}{2},\dots $
Answer: ${a}_{n}=\frac{1}{2}n+1$; ${a}_{100}=51$
An arithmetic seriesThe sum of the terms of an arithmetic sequence. is the sum of the terms of an arithmetic sequence. For example, the sum of the first 5 terms of the sequence defined by ${a}_{n}=2n-1$ follows:
$$\begin{array}{ccc}\hfill {S}_{5}& =& {\displaystyle \underset{n=1}{\overset{5}{\Sigma}}\left(2n-1\right)}\hfill \\ & =& \left[2\left({1}\right)-1\right]+\left[2\left({2}\right)-1\right]+\left[2\left({3}\right)-1\right]+\left[2\left({4}\right)-1\right]+\left[2\left({5}\right)-1\right]\hfill \\ & =& 1+3+5+7+9\hfill \\ & =& 25\hfill \end{array}$$
Adding 5 positive odd integers, as we have done above, is managable. However, consider adding the first 100 positive odd integers. This would be very tedious. Therefore, we next develop a formula that can be used to calculate the sum of the first n terms, denoted ${S}_{n}$, of any arithmetic sequence. In general,
$${S}_{n}={a}_{1}+\left({a}_{1}+d\right)+\left({a}_{1}+2d\right)+\dots +{a}_{n}$$
Writing this series in reverse we have,
$${S}_{n}={a}_{n}+\left({a}_{n}-d\right)+\left({a}_{n}-2d\right)+\dots +{a}_{1}$$
And adding these two equations together, the terms involving d add to zero and we obtain n factors of ${a}_{1}+{a}_{n}$:
$$\begin{array}{ccc}\hfill 2{S}_{n}& =& \left({a}_{1}+{a}_{n}\right)+\left({a}_{1}+{a}_{n}\right)+\dots +\left({a}_{n}+{a}_{1}\right)\hfill \\ \hfill 2{S}_{n}& =& n\left({a}_{1}+{a}_{n}\right)\hfill \end{array}$$
Dividing both sides by 2 leads us the formula for the nth partial sum of an arithmetic sequenceThe sum of the first n terms of an arithmetic sequence given by the formula: ${S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}.$:
$${S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}$$
Use this formula to calculate the sum of the first 100 terms of the sequence defined by ${a}_{n}=2n-1.$ Here ${a}_{1}=1$ and ${a}_{100}=199.$
$$\begin{array}{ccc}\hfill {S}_{100}& =& \frac{100\left({a}_{1}+{a}_{100}\right)}{2}\hfill \\ & =& \frac{100\left(1+199\right)}{2}\hfill \\ & =& \mathrm{10,000}\hfill \end{array}$$
Find the sum of the first 50 terms of the given sequence: 4, 9, 14, 19, 24, …
Solution:
Determine whether or not there is a common difference between the given terms.
$$d=9-4=5$$
Note that the difference between any two successive terms is 5. The sequence is indeed an arithmetic progression and we can write
$$\begin{array}{ccc}\hfill {a}_{n}& =& {a}_{1}+\left(n-1\right)d\hfill \\ & =& 4+\left(n-1\right)\cdot 5\hfill \\ & =& 4+5n-5\hfill \\ & =& 5n-1\hfill \end{array}$$
Therefore, the general term is ${a}_{n}=5n-1.$ To calculate the 50^{th} partial sum of this sequence we need the 1^{st} and the 50^{th} terms:
$$\begin{array}{ccc}\hfill {a}_{1}& =& 4\hfill \\ \hfill {a}_{50}& =& 5\left(50\right)-1=249\hfill \end{array}$$
Next use the formula to determine the 50^{th} partial sum of the given arithmetic sequence.
$$\begin{array}{ccc}\hfill {S}_{n}& =& \frac{n({a}_{1}+{a}_{n})}{2}\hfill \\ \hfill {S}_{50}& =& \frac{50.({a}_{1}+{a}_{50})}{2}\hfill \\ & =& \frac{50(4+249)}{2}\hfill \\ & =& 25(253)\hfill \\ & =& \mathrm{6,325}\hfill \end{array}$$
Answer: ${S}_{50}=\mathrm{6,325}$
Evaluate: $\underset{n=1}{\overset{35}{\Sigma}}\left(10-4n\right)$.
Solution:
In this case, we are asked to find the sum of the first 35 terms of an arithmetic sequence with general term ${a}_{n}=10-4n.$ Use this to determine the 1^{st} and the 35^{th} term.
$$\begin{array}{ccc}\hfill {a}_{1}& =& 10-4\left(1\right)=6\hfill \\ \hfill {a}_{35}& =& 10-4\left(35\right)=-130\hfill \end{array}$$
Next use the formula to determine the 35^{th} partial sum.
$$\begin{array}{ccc}\hfill {S}_{n}& =& \frac{n\left({a}_{1}+{a}_{n}\right)}{2}\hfill \\ \hfill {S}_{35}& =& \frac{35\cdot \left({a}_{1}+{a}_{35}\right)}{2}\hfill \\ & =& \frac{35\left[6+\left(-130\right)\right]}{2}\hfill \\ & =& \frac{35\left(-124\right)}{2}\hfill \\ & =& -\mathrm{2,170}\hfill \end{array}$$
Answer: −2,170
The first row of seating in an outdoor amphitheater contains 26 seats, the second row contains 28 seats, the third row contains 30 seats, and so on. If there are 18 rows, what is the total seating capacity of the theater?
Figure 9.2
Roman Theater (Wikipedia)
Solution:
Begin by finding a formula that gives the number of seats in any row. Here the number of seats in each row forms a sequence:
$$26,28,30,\dots $$
Note that the difference between any two successive terms is 2. The sequence is an arithmetic progression where ${a}_{1}=26$ and $d=2.$
$$\begin{array}{ccc}\hfill {a}_{n}& =& {a}_{1}+\left(n-1\right)d\hfill \\ & =& 26+\left(n-1\right)\cdot 2\hfill \\ & =& 26+2n-2\hfill \\ & =& 2n+24\hfill \end{array}$$
Therefore, the number of seats in each row is given by ${a}_{n}=2n+24.$ To calculate the total seating capacity of the 18 rows we need to calculate the 18^{th} partial sum. To do this we need the 1^{st} and the 18^{th} terms:
$$\begin{array}{ccc}\hfill {a}_{1}& =& 26\hfill \\ \hfill {a}_{18}& =& 2\left(18\right)+24=60\hfill \end{array}$$
Use this to calculate the 18^{th} partial sum as follows:
$$\begin{array}{ccc}{S}_{n}\hfill & =\hfill & \frac{n\left({a}_{1}+{a}_{n}\right)}{2}\hfill \\ \hfill {S}_{18}& =& \frac{18\cdot \left({a}_{1}+{a}_{18}\right)}{2}\hfill \\ & =& \frac{18\left(26+60\right)}{2}\hfill \\ & =& 9\left(86\right)\hfill \\ & =& 774\hfill \end{array}$$
Answer: There are 774 seats total.
Try this! Find the sum of the first 60 terms of the given sequence: 5, 0, −5, −10, −15, …
Answer: ${S}_{60}=-\mathrm{8,550}$
Write the first 5 terms of the arithmetic sequence given its first term and common difference. Find a formula for its general term.
${a}_{1}=5$; $d=3$
${a}_{1}=12$; $d=2$
${a}_{1}=15$; $d=-5$
${a}_{1}=7$; $d=-4$
${a}_{1}=\frac{1}{2}$; $d=1$
${a}_{1}=\frac{2}{3}$; $d=\frac{1}{3}$
${a}_{1}=1$; $d=-\frac{1}{2}$
${a}_{1}=-\frac{5}{4}$; $d=\frac{1}{4}$
${a}_{1}=1.8$; $d=0.6$
${a}_{1}=-4.3$; $d=2.1$
Given the arithmetic sequence, find a formula for the general term and use it to determine the 100^{th} term.
3, 9, 15, 21, 27,…
3, 8, 13, 18, 23,…
−3, −7, −11, −15, −19,…
−6, −14, −22, −30, −38,…
−5, −10, −15, −20, −25,…
2, 4, 6, 8, 10,…
$\frac{1}{2}$, $\frac{5}{2}$, $\frac{9}{2}$, $\frac{13}{2}$, $\frac{17}{2}$,…
$-\frac{1}{3}$, $\frac{2}{3}$, $\frac{5}{3}$, $\frac{8}{3}$, $\frac{11}{3}$,…
$\frac{1}{3}$, 0, $-\frac{1}{3}$, $-\frac{2}{3}$, −1,…
$\frac{1}{4}$, $-\frac{1}{2}$, $-\frac{5}{4}$, −2, $-\frac{11}{4}$,…
0.8, 2, 3.2, 4.4, 5.6,…
4.4, 7.5, 10.6, 13.7, 16.8,…
Find the 50^{th} positive odd integer.
Find the 50^{th} positive even integer.
Find the 40^{th} term in the sequence that consists of every other positive odd integer: 1, 5, 9, 13,…
Find the 40^{th} term in the sequence that consists of every other positive even integer: 2, 6, 10, 14,…
What number is the term 355 in the arithmetic sequence −15, −5, 5, 15, 25,…?
What number is the term −172 in the arithmetic sequence 4, −4, −12, −20, −28,…?
Given the arithmetic sequence defined by the recurrence relation ${a}_{n}={a}_{n-1}+5$ where ${a}_{1}=2$ and $n>1$, find an equation that gives the general term in terms of ${a}_{1}$ and the common difference d.
Given the arithmetic sequence defined by the recurrence relation ${a}_{n}={a}_{n-1}-9$ where ${a}_{1}=4$ and $n>1$, find an equation that gives the general term in terms of ${a}_{1}$ and the common difference d.
Given the terms of an arithmetic sequence, find a formula for the general term.
${a}_{1}=6$ and ${a}_{7}=42$
${a}_{1}=-\frac{1}{2}$ and ${a}_{12}=-6$
${a}_{1}=-19$ and ${a}_{26}=56$
${a}_{1}=-9$ and ${a}_{31}=141$
${a}_{1}=\frac{1}{6}$ and ${a}_{10}=\frac{37}{6}$
${a}_{1}=\frac{5}{4}$ and ${a}_{11}=\frac{65}{4}$
${a}_{3}=6$ and ${a}_{26}=-40$
${a}_{3}=16$ and ${a}_{15}=76$
${a}_{4}=-8$ and ${a}_{23}=30$
${a}_{5}=-7$ and ${a}_{37}=-135$
${a}_{4}=-\frac{23}{10}$ and ${a}_{21}=-\frac{25}{2}$
${a}_{3}=\frac{1}{8}$ and ${a}_{12}=-\frac{11}{2}$
${a}_{5}=13.2$ and ${a}_{26}=61.5$
${a}_{4}=-1.2$ and ${a}_{13}=12.3$
Find all arithmetic means between the given terms.
${a}_{1}=-3$ and ${a}_{6}=17$
${a}_{1}=5$ and ${a}_{5}=-7$
${a}_{2}=4$ and ${a}_{8}=7$
${a}_{5}=\frac{1}{2}$ and ${a}_{9}=-\frac{7}{2}$
${a}_{5}=15$ and ${a}_{7}=21$
${a}_{6}=4$ and ${a}_{11}=-1$
Calculate the indicated sum given the formula for the general term.
${a}_{n}=3n+5$; ${S}_{100}$
${a}_{n}=5n-11$; ${S}_{100}$
${a}_{n}=\frac{1}{2}-n$; ${S}_{70}$
${a}_{n}=1-\frac{3}{2}n$; ${S}_{120}$
${a}_{n}=\frac{1}{2}n-\frac{3}{4}$; ${S}_{20}$
${a}_{n}=n-\frac{3}{5}$; ${S}_{150}$
${a}_{n}=45-5n$; ${S}_{65}$
${a}_{n}=2n-48$; ${S}_{95}$
${a}_{n}=4.4-1.6n$; ${S}_{75}$
${a}_{n}=6.5n-3.3$; ${S}_{67}$
Evaluate.
Find the sum of the first 200 positive integers.
Find the sum of the first 400 positive integers.
The general term for the sequence of positive odd integers is given by ${a}_{n}=2n-1$ and the general term for the sequence of positive even integers is given by ${a}_{n}=2n.$ Find the following.
The sum of the first 50 positive odd integers.
The sum of the first 200 positive odd integers.
The sum of the first 50 positive even integers.
The sum of the first 200 positive even integers.
The sum of the first k positive odd integers.
The sum of the first k positive even integers.
The first row of seating in a small theater consists of 8 seats. Each row thereafter consists of 3 more seats than the previous row. If there are 12 rows, how many total seats are in the theater?
The first row of seating in an outdoor amphitheater contains 42 seats, the second row contains 44 seats, the third row contains 46 seats, and so on. If there are 22 rows, what is the total seating capacity of the theater?
If a triangular stack of bricks has 37 bricks on the bottom row, 34 bricks on the second row and so on with one brick on top. How many bricks are in the stack?
Each successive row of a triangular stack of bricks has one less brick until there is only one brick on top. How many rows does the stack have if there are 210 total bricks?
A 10-year salary contract offers $65,000 for the first year with a $3,200 increase each additional year. Determine the total salary obligation over the 10 year period.
A clock tower strikes its bell the number of times indicated by the hour. At one o’clock it strikes once, at two o’clock it strikes twice and so on. How many times does the clock tower strike its bell in a day?
Is the Fibonacci sequence an arithmetic sequence? Explain.
Use the formula for the nth partial sum of an arithmetic sequence ${S}_{n}=\frac{n\left({a}_{1}+{a}_{n}\right)}{2}$ and the formula for the general term ${a}_{n}={a}_{1}+\left(n-1\right)d$ to derive a new formula for the nth partial sum ${S}_{n}=\frac{n}{2}\left[2{a}_{1}+\left(n-1\right)d\right].$ Under what circumstances would this formula be useful? Explain using an example of your own making.
Discuss methods for calculating sums where the index does not start at 1. For example, $\underset{n=15}{\overset{35}{\Sigma}}\left(3n+4\right)}=\mathrm{1,659}.$
A famous story involves Carl Friedrich Gauss misbehaving at school. As punishment, his teacher assigned him the task of adding the first 100 integers. The legend is that young Gauss answered correctly within seconds. What is the answer and how do you think he was able to find the sum so quickly?
5, 8, 11, 14, 17; ${a}_{n}=3n+2$
15, 10, 5, 0, −5; ${a}_{n}=20-5n$
$\frac{1}{2}$, $\frac{3}{2}$, $\frac{5}{2}$, $\frac{7}{2}$, $\frac{9}{2}$; ${a}_{n}=n-\frac{1}{2}$
1, $\frac{1}{2}$, 0, $-\frac{1}{2}$, −1; ${a}_{n}=\frac{3}{2}-\frac{1}{2}n$
1.8, 2.4, 3, 3.6, 4.2; ${a}_{n}=0.6n+1.2$
${a}_{n}=6n-3$; ${a}_{100}=597$
${a}_{n}=1-4n$; ${a}_{100}=-399$
${a}_{n}=-5n$; ${a}_{100}=-500$
${a}_{n}=2n-\frac{3}{2}$; ${a}_{100}=\frac{397}{2}$
${a}_{n}=\frac{2}{3}-\frac{1}{3}n$; ${a}_{100}=-\frac{98}{3}$
${a}_{n}=1.2n-0.4$; ${a}_{100}=119.6$
99
157
38
${a}_{n}=5n-3$
${a}_{n}=6n$
${a}_{n}=3n-22$
${a}_{n}=\frac{2}{3}n-\frac{1}{2}$
${a}_{n}=12-2n$
${a}_{n}=2n-16$
${a}_{n}=\frac{1}{10}-\frac{3}{5}n$
${a}_{n}=2.3n+1.7$
1, 5, 9, 13
$\frac{9}{2}$, 5, $\frac{11}{2}$, 6, $\frac{13}{2}$
18
15,650
−2,450
90
−7,800
−4,230
38,640
124,750
−18,550
−765
10,578
20,100
2,500
2,550
${k}^{2}$
294 seats
247 bricks
$794,000
Answer may vary
Answer may vary