For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Dominic Chukwuemeka

Welcome to Sequences

I greet you this day,
First: read the notes. Second: view the videos. Third: solve the questions/solved examples. Fourth: check your solutions with my thoroughly-explained solutions. Fifth: check your answers with the calculators as applicable.
I wrote the codes for these calculators using JavaScript, a client-side scripting language. Please use the latest Internet browsers. The calculators should work.
Comments, ideas, areas of improvement, questions, and constructive criticisms are welcome. You may contact me.

If you are my student, please do not contact me here. Contact me via the school's system. Thank you for visiting!!!

Samuel Dominic Chukwuemeka (Samdom For Peace) B.Eng., A.A.T, M.Ed., M.S

Stories


First Story

It's $2019$
A family of the Dad, Mom, Daughter, Son
The son was not at home.

Daughter: Good morning Dad
Dad: Good morning little angel.
How are you?
Daughter: I am okay...but not really okay
Dad: What is the problem this time?
Daughter: I celebrate my birthday every year but my brother does not
It's been $3$ years since we had his birthday
Why is that? ...speaking like an American lol
And I'm sure he has not joined Jehovah's witnesses
Dad: Where did you get your "smart mouth" from?
Daughter: From you, of course
Dad: No, you got your intelligence from me...
but you got your smart mouth from your Mom.
Daughter: Dad!!!
I am going to ask Mom why we do not celebrate Jude's birthday...
and I'm going to tell her what you said

She runs to her Mom

Daughter: Good morning Mom
Mom: Good morning my princess.
You are wonderful this morning.
Daughter: Hmmm...are you sure...cause I'm not
Mom: Come on, my dear; what's the problem?
Anyway, what should you say when you receive a complement?
Daughter: Thank you Mom.
What about Jude's birthday?
I celebrate my birthday every year.
But, what about my brother, Jude?
It's been many years since we celebrated his birthday.
I want us to celebrate his birthday every year.
Mom: My dear, your brother was born in a leap year
His birthday is on $29th$ February
Each leap year has $29$ days rather than $28$ days
A leap year comes up every four years
Your brother's birthday is like a sequence...every four years....
Bring it to Mathematics
It is a Sequence
An Arithmetic Sequence
where the common difference is $4$
When was the last time we had Jude's birthday?
Daughter: In $2016$
Mom: So, the next time would be...
Daughter: next year, $2020$...
Mom: and ...
Daughter counts ...1, 2, 3, 4
Daughter: $2024, 2028, 2032,...$
But, why?
What is a leap year?
And why does each leap year have $29$ days?
And why should my brother's birthday fall on such a day?
You mean ...leap year ...year that ... "leaps like a frog"?
Mom: Interesting... well go and find out
I have no idea
Daughter: Well, you should know!
Mom: Excuse me...what a smart mouth?
Daughter: And by the way, Dad said I got my smart mouth from you
Mom: Did he really say that?
The daughter nodded
Daughter: Yes, he did
Mom: Do not mind your Dad. He is an as....

Dad has been listening to the interaction.
He interrupts...

Dad: Honey, what time do we need to go to the store?
Daughter: Lol...He called you honey...lol... that is what he does when he gets in trouble

She runs to the computer
The Dad goes to the Mom

Dad: Did you sign out of your email?
Mom: No, I did not.
Let us find a way to distract her
So, I can go and sign out.
We definitely do not want her to ask questions again from my email

Daughter was listening to the conversation but they did not realize

Daughter: I am not going to contact anyone this time
I just want to find out what both of you should have known
I want to find out the information about a leap year
Why do we have leap years?
Why does it have to be only February that is affected by leap years?
I just want my brother to celebrate birthday every year just like I do.
Mom and Dad: You are correct. We should know. Let's join you to find out.

Second Story

Objectives


Students will:

(1.) Discuss sequences.

(2.) Identify the types of sequences.

(3.) Determine the $n$th term of an arithmetic sequence.

(4.) Determine the $n$th term of a geometric sequence.

(5.) Determine the sum of the $n$ terms of an arithmetic sequence.

(6.) Determine the sum of the $n$ terms of a geometric sequence.

(7.) Determine the sum to infinity of an "applicable" geometric sequence.

(8.) Solve applied problems involving arithmetic sequences.

(9.) Solve applied problems involving geometric sequences.

(10.) Determine the $nth$ term of a "general" sequence.

(11.) Determine the $nth$ term of recursive sequences.

Arithmetic Sequence


An Arithmetic Sequence is a sequence whose consecutive terms are obtained by the addition of a constant value.
OR
An Arithmetic Sequence is a sequence whose difference between consecutive terms is a constant value.
It is also known as an Arithmetic Progression.
The next (succeeding) term of the sequence is found by adding a constant to the previous (preceeding) term of that sequence.
That constant is known as the common difference.
So, the common difference is the constant that is added to a term of an Arithmetic Sequence in order to give the next term of that sequence.

Example 1: In the sequence: $3, 10, 17, 24, ...$
First term, $a = 3$
Second term = $10$
Third term = $17$
Fourth term = $24$

Teacher: how did we obtain the:
second term from the first term?
third term from the second term?
fourth term from the third term?


$ 2nd\:\: term - 1st\:\: term = 10 - 3 = 7 \\[3ex] 3rd\:\: term - 2nd\:\: term = 17 - 10 = 7 \\[3ex] 4th\:\: term - 3rd\:\: term = 24 - 17 = 7 \\[3ex] $ So, we have a common difference ("common" because it is a constant value, "difference" because it is the result of a subtraction procedure).
So, the common difference, $d = 7$

Teacher: Does it make sense?
What would be the fifth term?


Example 2: In the sequence: $-3, -10, -17, -24, ...$
First term, $a = -3$
Second term = $-10$
Third term = $-17$
Fourth term = $-24$

Teacher: how did we obtain the:
second term from the first term?
third term from the second term?
fourth term from the third term?


$ 2nd\:\: term - 1st\:\: term = -10 - (-3) = -10 + 3 = -7 \\[3ex] 3rd\:\: term - 2nd\:\: term = -17 - (-10) = -17 + 10 = -7 \\[3ex] 4th\:\: term - 3rd\:\: term = -24 - (-17) = -24 + 17 = -7 \\[3ex] $ Here, the common difference, $d = -7$

Teacher: What would be the fifth term?
What would be the seventieth term?
Student: That would be a lot of work
Is there an easier way to find it?
Teacher: yes of course! ☺☺☺


Back to Example 1 (as it would be easier to demonstrate what we want to do)
We already know that:
first term = $a$
common difference = $d$

So, we have:

$ 1st\:\: term = a = 3 \\[3ex] 2nd\:\: term = 3 + 7 = 10 = a + d \\[3ex] 3rd\:\: term = 10 + 7 = 17 = a + d + d = a + 2d \\[3ex] 4th\:\: term = 17 + 7 = 24 = a + d + d + d = a + 3d \\[3ex] $ Teacher: In that sense and using Algebra only (forget Arithmetic (the results/numbers) now), because we want to come up with a formula
What are the:
$5th, 6th, 7th, 10th, nth$ terms?


$ 5th\:\: term = a + 4d \\[3ex] 6th\:\: term = a + 5d = a + (6 - 1)d \\[3ex] 7th\:\: term = a + 6d = a + (7 - 1)d \\[3ex] 10th\:\: term = a + 9d = a + (10 - 1)d \\[3ex] nth\:\: term = a + (n - 1)d \\[3ex] nth\:\: term = a + d(n - 1) \\[3ex] $ So, the $nth$ term of an Arithmetic Sequence is: $AS_n = a + d(n - 1)$

Teacher: Do you now understand how that formula was derived?
So, back to Example 2.
What is the seventieth term?


Back to Example 2:

$ AS_n = a + d(n - 1) \\[3ex] a = -3 \\[3ex] n = 70 \\[3ex] d = -7 \\[3ex] AS_{70} = -3 + -7(70 - 1) \\[3ex] = -3 + -7(69) \\[3ex] = -3 - 483 \\[3ex] = -486 $

Now, that we have figured out the derivation of the formula for the $n$th term of an Arithmetic Sequence, let us find out how the formula for the sum of the first $n$ terms of an Arithmetic Sequence is derived.
I shall begin with my method. I think it is much easier. Besides, you can easily recall it if your teacher does not give you the formula.

Samuel Chukwuemeka's Method (SamDom For Peace Method) for Finding the Sum of the First $n$ Terms of an Arithmetic Sequence
Published May 28, 2019 (05/28/2019)
My method connects Combinatorics with Algebra
Let us begin this way:
Sum of the First:
$ two\:\:terms = a + (a + d) = a + a + d = 2a + d \\[3ex] three\:\:terms = a + (a + d) + (a + 2d) = a + a + d + a + 2d = 3a + 3d \\[3ex] four\:\:terms = a + (a + d) + (a + 2d) + (a + 3d) = a + a + d + a + 2d + a + 3d = 4a + 6d $

Let us save some time ☺☺☺
The sum of the first five terms is the sum of the first four terms and the fifth term
The sum of the first six terms is the sum of the first five terms and the sixth term
The sum of the first seven terms is the sum of the first six terms and the seventh term

Sum of the First:
$ five\:\:terms = (4a + d) + (a + 4d) = 4a + d + a + 4d = 5a + 5d \\[3ex] six\:\:terms = (5a + 5d) + (a + 5d) = 5a + 5d + a + 5d = 6a + 10d \\[3ex] seven\:\:terms = (6a + 10d) + (a + 6d) = 6a + 10d + a + 6d = 7a + 16d $

In Summary
Sum of the First:
$ two\:\:terms = 2a + d \\[3ex] three\:\:terms = 3a + 3d \\[3ex] four\:\:terms = 4a + 6d \\[3ex] five\:\:terms = 5a + 5d \\[3ex] six\:\:terms = 6a + 10d \\[3ex] seven\:\:terms = 7a + 16d $

Let us deviate to Pascal's Triangle and the Binomial Expansion
Let us write the first eight steps of the Pasca's Triangle
$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~1 \\[5ex] ~~~~~~~~~~~~~~~~~~~~~~~1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~1 \\[3ex] ~~~~~~~~~~~~~~~~~~1~~~~~~~~~~~~~~~~~~~~~~~~~~~~2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~1 \\[3ex] ~~~~~~~~~~~~~1~~~~~~~~~~~~~~~~~~~~3~~~~~~~~~~~~~~~~~~~~~~~~~~~3~~~~~~~~~~~~~~~~~~~~~~~~~~~1 \\[3ex] ~~~~~~~~1~~~~~~~~~~~~~~~~~~~~~4~~~~~~~~~~~~~~~~~~~~~~~6~~~~~~~~~~~~~~~~~~~~~~~~~~~4~~~~~~~~~~~~~~~~1 \\[3ex] ~~~~1~~~~~~~~~~~~~5~~~~~~~~~~~~~~~~~~~~~~~10~~~~~~~~~~~~~~~~~~~~~~~~~~10~~~~~~~~~~~~~~~~~~~~~5~~~~~~~~~~~~~1 \\[3ex] $

Let us compare the coefficients of $a$ and $d$ with the coefficients in the Pascal's Triangle
Forget the $1's$ in the Pascal's Triangle
Forget the first two steps in the Pascal's Triangle
Why? The first step in the Pascal's Triangle are the coefficients of the expansion $(x + y)^0$
The second step in the Pascal's Triangle are the coefficients of the expansion $(x + y)^1$
So, we do not need those steps.
Let us begin with the third step in the Pascal's Triangle
The third step in the Pascal's Triangle are the coefficients of the expansion $(x + y)^2$
Compare the coefficients of $a$ and $d$ in the Sum of the first two terms of an Arithmetic Sequence with the second and third coefficients of the expansion $(x + y)^2$ in the Pascal's Triangle
Compare the coefficients of $a$ and $d$ in the Sum of the first three terms of an Arithmetic Sequence with the second and third coefficients of the expansion $(x + y)^3$ in the Pascal's Triangle
Compare the coefficients of $a$ and $d$ in the Sum of the first four terms of an Arithmetic Sequence with the second and third coefficients of the expansion $(x + y)^4$ in the Pascal's Triangle
What do you notice?

The coefficients are the same

Recall that the coefficients in the Pascal's Triangle: (as explained in the video link)
In the expansion of $(x + y)^1$ is $C(1, 0)$ and $C(1,1)$
In the expansion of $(x + y)^2$ is $C(2, 0)$, $C(2, 1)$, and $C(2, 2)$
In the expansion of $(x + y)^3$ is $C(3, 0)$, $C(3, 1)$, $C(3, 2)$, and $C(3, 3)$
In the expansion of $(x + y)^4$ is $C(4, 0)$, $C(4, 1)$, $C(4, 2)$, $C(4, 3)$, and $C(4, 4)$
$ C(3, 1) = 3 \\[3ex] C(4, 1) = 4 \\[3ex] C(n, 1) = n \\[3ex] Why?\:\: How? \\[3ex] Based\:\: on\:\:the\:\:Combinations\:\:Formula \\[3ex] n! = n * (n - 1) * (n - 2) * (n - 3) * ... * 1 \\[3ex] n! = n * (n - 1)! \\[3ex] n! = n * (n - 1) * (n - 2)! \\[3ex] C(n, r) = \dfrac{n!}{(n - r)! r!} \\[5ex] C(n, 1) = \dfrac{n!}{(n - 1)! 1!} \\[5ex] = \dfrac{n * (n - 1)!}{(n - 1)! 1} \\[5ex] C(n, 1) = n \\[5ex] C(n, 2) = \dfrac{n!}{(n - 2)! 2!} \\[5ex] = \dfrac{n * (n - 1) * (n - 2)!}{(n - 2)! 2 * 1} \\[5ex] C(n, 2) = \dfrac{n(n - 1)}{2} \\[5ex] $

This implies that:
Sum of the First:
$ two\:\:terms = 2a + d = C(2, 1)a + C(2, 2)d \\[3ex] three\:\:terms = 3a + 3d = C(3, 1)a + C(3, 2)d \\[3ex] four\:\:terms = 4a + 6d = C(4, 1)a + C(4, 2)d \\[3ex] five\:\:terms = 5a + 5d = C(5, 1)a + C(5, 2)d \\[3ex] six\:\:terms = 6a + 10d = C(6, 1)a + C(6, 2)d \\[3ex] seven\:\:terms = 7a + 16d = C(7, 1)a + C(7, 2)d \\[3ex] n\:\:terms = C(n, 1)a + C(n, 2)d = na + \dfrac{n(n-1)}{2}d \\[5ex] \therefore SAS_n = na + \dfrac{n(n-1)}{2}d $

Student: But, this was not the formula that you wrote in the "Formulas" section
Teacher: Well, you can simplify this formula to be that one
Student: Is this formula correct?
Teacher: Very correct
It can be simplified to that one.
Let us simplify it to that one

$ SAS_n = na + \dfrac{n(n-1)}{2}d \\[5ex] SAS_n = \dfrac{2na + n(n-1)d}{2} \\[5ex] SAS_n = \dfrac{n[2a + (n - 1)d]}{2} \\[5ex] SAS_n = \dfrac{n}{2}[2a + d(n - 1)] $

Symbols and Formulas - Arithmetic Sequence


To learn how to solve for a variable in terms of other variables, please visit: Solved Examples - Literal Equations

Formulas

$ (1.)\:\: AS_n = a + d(n - 1) \\[3ex] (2.)\:\: p = a + d(n - 1) \\[3ex] (3.)\:\: SAS_n = \dfrac{n}{2}(a + p) \\[5ex] (4.)\:\: SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[5ex] (5.)\:\: n = \dfrac{p - a + d}{d} \\[5ex] (6.)\:\: n = \dfrac{-(2a - d) \pm \sqrt{(2a - d)^2 + 8d*SAS_n}}{2d} \\[5ex] $

Calculators for Arithmetic Sequence


Please NOTE: Only integers and decimals are allowed.
Fractions are not allowed for this calculator.
  • Given: first term, common difference, number of terms
    To Find: other details

  • Given: first term, last term, number of terms
    To Find: other details

  • Given: first term, second term, last term
    To Find: other details

  • Given: first term, common difference, sum of $n$ terms
    To Find: other details

  • Given: first term, last term, sum of $n$ terms
    To Find: other details


  • Given: any two terms
    To Find: the common difference, first term

The th term is

The th term is

  • Given: any two terms

    To Find: any $nth$ term, sum of the first any $n$ terms

The th term is


The th term is




The sum of the first terms is

Geometric Sequence


Symbols and Formulas - Geometric Sequence


To learn how to solve for a variable in terms of other variables, please visit: Solved Examples - Literal Equations

Formulas

$ (1.)\:\: GS_n = ar^{n - 1} \\[3ex] (2.)\:\: SGS_n = \dfrac{a(r^{n} - 1)}{r - 1} \:\:for\:\: r \gt 1 \\[5ex] (3.)\:\: SGS_n = \dfrac{a(1 - r^{n})}{1 - r} \:\:for\:\: r \lt 1 \\[5ex] (4.)\:\: n = \dfrac{\log{\left[\dfrac{SGS_n(r - 1)}{a} + 1\right]}}{\log r} $

Calculators - Geometric Sequence


Please NOTE: Only integers and decimals are allowed.
Fractions are not allowed for this calculator.
  • Given: first term, second term, number of terms
    To Find: other details

  • Given: first term, last term, number of terms
    To Find: other details
  • Given: first term, second term, last term
    To Find: other details
  • Given: first term, last term, number of terms
    To Find: other details
  • Given: any two terms
    To Find: the common ratio, first term

The th term is

The th term is

  • Given: any two terms
    To Find: any $nth$ term, sum of the first any $n$ terms

The th term is

The th term is



The sum of the first terms is

References


Chukwuemeka, S.D (2016, April 30). Samuel Chukwuemeka Tutorials - Math, Science, and Technology. Retrieved from https://www.samuelchukwuemeka.com
Bittinger, M. L., Beecher, J. A., Ellenbogen, D. J., & Penna, J. A. (2017). Algebra and Trigonometry: Graphs and Models ($6^{th}$ ed.). Boston: Pearson.
Rosen, K. H. (2013). Discrete mathematics and its applications ($8^{th}$ ed.). New York: McGraw-Hill.
Sullivan, M., & Sullivan, M. (2017). Algebra & Trigonometry ($7^{th}$ ed.). Boston: Pearson.
CrackACT. (n.d.). Retrieved from http://www.crackact.com/act-downloads/
VAST LEARNERS. (n.d). Retrieved from https://www.vastlearners.com/free-jamb-past-questions/