Lesson 1: Sequences

Video Lesson

Competencies (MLC)

At the end of this lesson, you will be able to:

• List patterns which leads to definition of number sequence.
• Explain the concepts of sequence, term of a sequence and its rule.
• Compute any term of a sequence using rule (formula).
• Draw graphs of finite sequences
• Define arithmetic progressions and geometric progressions.
• Determine the terms of arithmetic and geometric sequences.

Key Terms

Brainstorming Questions

20 people live on the first floor of the building, 34 people on the second floor and 48 people on the third floor , and so on. How money people live on the fifth and sixth floor of the building?

Solution

First look at the pattern 20, 34, 48, …… . by simple observation you can guess the

$4^{\text{th}}\text{,}{5}^{\text{th}}\text{and the}{6}^{\text{th}}$

terms of the sequence. the sequence is increased by the common number 14. you simply add 14 on the pre- existed term to get the successive term

20 +14 = 34

34 +14 = 48

48 +14 = 62

62 + 14 = 76

76 +14 = 90 etc

$\text{Therefore, the}{5}^{\text{th}}\text{term is}76\text{and the}{6}^{\text{th}}\text{term is}90\text{.}$

1.1. Introduction to Sequences

Definition:

$\text{A sequence}\left\{a_n\right\}\text{is}$ a special types of function whose domain is the set of positive integers.

$\text{Let}\left\{a_n\right\}\text{be a sequence. Then}$

$$\begin{gathered} \text{i. The functional values}{a}_1,a_2,a_3,\dots ,a_n\text{are called elements of a sequence or terms of a sequence.} \\ {a}_1\text{is the first term,}{a}_2\text{is the second term, and}\left\{a_n\right\}\text{is called the}{n}^{\text{th}}\text{term of the sequence.} \end{gathered}$$

$\text{ii. A sequence}{a}_1,a_2,a_3,\dots ,a_n\text{denoted by}\left\{a_n\right\}\text{.}$

Depending on its last term there are two types of sequence.

• Finite sequence: It is a sequence that has a last term. Its domain is 1,2,3,…,n.
• Infinite sequence: It is a sequence that does not have a last term. Its domain is the set of natural numbers, N.

Example 1:

1. A sequence described by: 2, 3, 4, 5, 6, …, 20 is a finite sequence
2. A sequence described by: 2, 8, 14, 20, 26, …. is an infinite sequence

Note:

A sequence can be described by

• Listing the the terms
• Writing the general term
• Drawing the graph
• Using recurrence relation

Example 2:

$$\begin{gathered} 1.\text{List the first five terms of the sequence whose general term is given below where}n\text{is a positive integer:} \\ {a}_n=2n–1 \\ 2.\text{Write the general formula for the following sequences} \\ \text{a.}\{3,5,7,9,\dots \} \\ \text{b.}\{\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},\dots \} \\ 3.\text{Draw the graph of the sequence}{a}_n=2n+1 \end{gathered}$$

Solution

$$\begin{gathered} 1.\text{a).}{a}_n=2n–1 \\ \text{when}n=1,a_1=2\left(1\right)–1=1 \\ \text{when}n=2,a_2=2\left(2\right)–1=3 \\ \text{when}n=3,a_3=2\left(3\right)–1=5 \\ \text{when}n=4,a_4=2\left(4\right)–1=7 \\ \text{when}n=5,a_5=2\left(5\right)–1=9 \end{gathered}$$

$$\begin{gathered} \text{Therefore the first five terms of the sequence are}1,3,5,7,9 \\ \text{or the sequence can be written as}1,3,5,7,9,11,13,\dots ,2_n–1,\dots \end{gathered}$$

2. (a). {3, 5, 7, 9, ….} this is a sequence with first term 3 and is increased by 2 so that you can add 2 in every pre-existing term to get the next term of the given sequence:

Lets correlate natural numbers with the given sequence

$$\begin{gathered} 3=2\left(1\right)+1 \\ 3+2=5=2\left(2\right)+1 \\ 5+2=7=2\left(3\right)+1 \\ 7+2=9=2\left(4\right)+1 \\ 9+2=11=2\left(5\right)+1 \end{gathered}$$

etc

$$\begin{gathered} \text{by simple observation the general term is determined for any positive integer}n\text{as:} \\ {a}_n=2n+1 \\ \text{b).}\{\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},\dots \} \end{gathered}$$

First observe the pattern of the sequence by integrating each term with the set of positive integers since the domain of a sequence is the set of positive integers

$$\begin{gathered} \frac{1}{2}=\frac{1}{2^1} \\ \frac{1}{4}=\frac{1}{2^2} \\ \frac{1}{8}=\frac{1}{2^3} \\ \frac{1}{16}=\frac{1}{2^4} \end{gathered}$$

then by simple observation we can define the general term using any positive integer

$\text{n, as}{a}_n=\frac{1}{2^n}$

3. Make a table with n and an, as follows.

$\text{Then plot each ordered pair}(n,a_n)\text{.}$

So, the graph becomes

1.1.1. Recursively Defined Sequence

A sequence that is determined by giving the first term or the first few terms by a recursion formula is said to be defined recursively (inductively). The domain of recursive sequence can be the set whole numbers. For example Fibonacci and Mulatu sequences are some examples of recursive sequences.

If all terms of a sequence are the same constant, then it is called a constant sequence. For example, 1,1, 1, 1, …, (n) is a constant sequence.

1.1.2. Fibonacci and Mulatu’s sequence

Fibonacci number: The Fibonacci sequence is extended by adding two consecutive terms to get the next term. The sequence of Fibonacci numbers is defined by the following recurrence relation:

The Fibonacci number is defined as

$$F{n}_{}=\left\{\begin{array}{cc}1& \text{if}n=0\\ 1& \text{if}n=1\\ F_{n–1}+F_{n–2}& \text{if}n\ge 2\end{array}\right\}$$

$$F_n=1,1,2,3,5,8,13,21,\dots$$

Mulatu’s number: The sequence of Mulatu number is defined by the following recurrence relation:

Professor Mulatu introduced a sequence of the form:

$$M{n}_{}=\left\{\begin{array}{cc}4& \text{if}n=0\\ 1& \text{if}n=1\\ M_{n–1}+M_{n–2}& \text{if}n\ge 2\end{array}\right\}$$

$$M_n=4,1,5,6,11,17,28,\dots$$

1.2. Arithmetic Sequence

Definition:

An arithmetic sequence or arithmetic progression is a sequence in which each each term except the first is obtained by adding a fixed number (positive or negative) to the preceding term. The fixed number is called common difference of the sequence.

Derivation of n^th Term in Arithmetic Progression

Final Formula:

Example 3:

1. Given that the first term of an arithmetic sequence is 10, common difference is -4, find the terms from

$$2^{\text{nd}}\text{to}{4}^{\text{th}}\text{term.}$$

$$\begin{gathered} 2.\mathrm{Find}\mathrm{the}\mathrm{general}\mathrm{term}\mathrm{of}\mathrm{the}\mathrm{sequence}{A}_n,\mathrm{where}\mathrm{the}\mathrm{first}\mathrm{term}\mathrm{is}7,\mathrm{common}\mathrm{difference}3. \\ 3.\mathrm{When}\mathrm{the}{7}^{\mathrm{th}}\mathrm{term}\mathrm{is}12\mathrm{and}\mathrm{the}{18}^{\mathrm{th}}\mathrm{term}\mathrm{is}34. \\ a.\mathrm{Find}\mathrm{the}\mathrm{general}\mathrm{term}\mathrm{of}\mathrm{the}\mathrm{sequence}{A}_n. \\ b.\mathrm{Find}{A}_6. \end{gathered}$$

Solution:

$$\begin{gathered} 1.2^{nd}\mathrm{term}=1^{st}\mathrm{term}–4=10–4=6 \\ {3}^{rd}\mathrm{term}=2^{nd}\mathrm{term}–4=6–4=2 \\ {4}^{th}\mathrm{term}=3^{rd}\mathrm{term}–4=2–4=–2 \\ \mathrm{Therefore},\mathrm{the}\mathrm{terms}\mathrm{from}{2}^{nd}\mathrm{to}{4}^{th}\mathrm{term}\mathrm{are}6,2\mathrm{and}–2. \end{gathered}$$

$$\begin{gathered} 2.\mathrm{Given}\mathrm{that}{A}_1=7\mathrm{and}d=3. \\ \mathrm{The}\mathrm{sequence}\mathrm{is}\mathrm{an}\mathrm{arithmetic}\mathrm{sequence}\mathrm{and}\mathrm{its}\mathrm{general}\mathrm{formula}\mathrm{is}{A}_n=A_1+(n–1)d. \\ \mathrm{So} \\ {A}_n=A_1+(n–1)d=7+(n–1)3=7+3n–3=3n+4. \end{gathered}$$

3. a. Applying the arithmetic formula and substituting the existing values yields.

$$\begin{gathered} A_7=A_1+(7–1)d=A_1+6d=12............\left(1\right) \\ {A}_{18}=A_1+(18–1)d=A_1+17d=34............\left(2\right) \end{gathered}$$

Subtracting eq(2) from eq(1),

$$\begin{gathered} –11d=(–22) \\ d=2;and{A}_1+6d=12,implies{A}_1+6\left(2\right)=12, \end{gathered}$$

then

$$\begin{gathered} A_1+12=12implies{A}_1=0 \\ Thereforethegeneraltermbecomes{A}_n=A_1+(n–1)d=0+(n–1)2=0+(n–1)2=2n–2 \\ b.A_n=A_1+(n–1)d=0+(6–1)2=0+5\left(2\right)=10. \end{gathered}$$

Arithmetic Mean Between Two Terms

Definition:

$$\begin{gathered} Letaandbbeanytwonumbers,thenthenumber\frac{a+b}{2}iscalled \\ anA.Mbetweenaandb\left(theaverageofaandb\right) \\ thentermsa,\frac{a+b}{2}andbareinA.Pwithcommondifference\frac{b–a}{2}. \end{gathered}$$

Let

$$\{a,c_1,c_2,c_3,...,c_k,b\}isA.P,$$

then

$$\begin{gathered} c_1,c_2,c_3,...,c_{k}arek–arithmeticmeanbetweenaandb,and \\ b=(k+2)thtermoftheA.P. \end{gathered}$$

Example 4:

$$\begin{gathered} 1.Giventhat1,x,8isanarithmeticsequence,findx \\ 2.Insertfourarithmeticmeanbetween4and29. \end{gathered}$$

Solution:

1.Since it is an arithmetic sequence, the difference between two consecutive terms is constant.

$$x–1=8–ximplies2x=8+1impliesx=\frac{9}{2}$$

$$\begin{gathered} 2.Letthefourarithmeticmeansbe{A}_2,A_3,A_{4}and{A}_5. \\ So,4,A_2,A_3,A_4,A_5,29formarithmeticprogression. \end{gathered}$$

Then,

$$\begin{gathered} A_n=A_1+(n–1)d \\ {A}_1=4, \end{gathered}$$

Thus,

$$\begin{gathered} A_n=4+(n–1)5=5n–1 \\ {A}_2=4+5\times 1=9 \\ implies{A}_3=4+5\times 2=14 \\ {A}_4=4+5\times 3=19 \\ {A}_5=4+5\times 4=24. \end{gathered}$$

1.3. Geometric Sequence

Definition:

A geometric sequence or geometric progression is one in which the ratio between consecutive terms is a non-zero constant. This constant is called the common ration.

Derivation for n^th term of geometric Progression

Final Formula:

$$\begin{gathered} *\; \{G_n\}isgeometricsequenceifr=\frac{G_{n+1}}{G_n},risanelementofthesetofrealnumbers \\ exceptzerowhereriscommonratio. \end{gathered}$$

$$\begin{gathered} \mathrm{*\; I}f\left\{G_n\right\}isageometricsequencewiththefirstterm{G}_{1}andcommonratior, \\ thenthe{n}^{th}termofthesequenceisgivenby: \end{gathered}$$

$$G_n=G_1\times r^{n–1}.$$

Example 5:

$$\begin{gathered} 1.Find{G}_{n}whenthefirsttermis3,commonratiois2. \\ 2.Findthe{4}^{th}termofgeometricsequencewhosefirsttermis1andcommonratiois3. \end{gathered}$$

Solution:

$$\begin{gathered} 1.Given{G}_1=3,r=2.then{G}_n=G_1\times r^{n–1}=3\times 2^{n–1} \\ 2.Given{G}_1=1andr=3,then{G}_n=G_1\times r^{n–1}=1\times 3^{n–1}=3^{n–1}. \\ Therefore,G_4=3^{4–1}=3^3=27. \end{gathered}$$

Geometric Mean between two numbers

Definition:

$$\begin{gathered} Whena,mandbaretermsinageometricsequence,thenmiscalledthegeometricmeanbetween \\ aandb.wherea\ne 0,m\ne 0,b\ne 0. \\ Inageometricsequence,theratiobetweenconsecutivetermsisconstant. \end{gathered}$$

$$\begin{gathered} {\displaystyle r=\frac{m}{a}=\frac{b}{m} \\ ⟹m^2=ab} \end{gathered}$$

Thus

$${\displaystyle m=\pm \sqrt{ab}}$$

• The geometric mean between a and b is

$${\displaystyle GM=\sqrt{ab}}$$

Example 6:

Find a geometric mean between 2 and 8.

Solution:

$$\begin{gathered} \text{Let}a=2\text{and}b=8 \\ \text{and the geometric mean be}m.\text{Then} \end{gathered}$$

$$\begin{gathered} {\displaystyle r=\frac{m}{a}=\frac{b}{m} \\ ⟹m^2=ab=2\times 8=16 \\ ⟹m=\pm \sqrt{16}=\pm 4} \end{gathered}$$

Therefore, the geometric mean between 2 and 8 is 4.