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

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

*
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.

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 on arithmetic sequence.

(9.) Solve applied problems on geometric sequence.

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

(11.) Determine the $n$th term of a quadratic sequence.

(12.) Determine the $n$th term of a triangular sequence.

(13.) Determine the $nth$ term of a Fibonacci sequence.

(14.) Determine the $nth$ terms of recursive sequences.

sequences, series, arithmetic progression, geometric progression, arithmetic sequence, geometric sequence, first term, common difference, common ratio, number of terms, general term, nth term, sum of terms, last term, sum to infinity, quadratic sequence, triangular sequence, Fibonacci sequence, recursive sequence, general sequence, arithmetic mean, geometric mean, convergent series, divergent series, first difference, second difference, linear expression, quadratic expression, linear function, quadratic function, linear equation, quadratic equation

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.**

It is also known as a **Linear Sequence.**

Because it is also known as a **Linear Sequence**, one can also define it as a sequence in which
the highest exponent of the number of terms in the general term ($nth$ term) is one.

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, $d = 7$

What would be the fifth term?

First term, $a = -3$

Second term = $-10$

Third term = $-17$

Fourth term = $-24$

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, $d = -7$

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

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] $

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: $AS_n = a + d(n - 1)$

So, back to Example 2.

What is the seventieth term?

Back to

$ 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.

Guess what? It was even asked in the National Senior Certificate (NSC) exam.

**NSC** Derive a formula for the sum of the first $n$ terms of an arithmetic sequence if the
first term of the sequence is $a$ and the common difference is $d$.

I shall begin with my method. I think it is much easier.

Besides, I think it is easier to recall.

**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)] $

**Another Method for Finding the Sum of the First $n$ Terms of an Arithmetic Sequence**

Assume we have five terms of an Arithmetic Sequence: $7, 12, 17, 22, 27$

$
Recall \\[3ex]
a = 7 \\[3ex]
d = 12 - 7 = 5 \\[3ex]
2nd\:\:term = AS_2 = a + d = 7 + 5 = 12 \\[3ex]
3rd\:\:term = AS_3 = a + 2d = 7 + 2(5) = 7 + 10 = 17 \\[3ex]
4th\:\:term = AS_4 = a + 3d = 7 + 3(5) = 7 + 15 = 22 \\[3ex]
5th\:\:term = AS_5 = a + 4d = 7 + 4(5) = 7 + 20 = 27 \\[3ex]
nth\:\:term = AS_n = a + (n - 1)d = a + d(n - 1) \\[3ex]
Going\:\:backwards \\[3ex]
4th\:\:term = AS_5 - d = 27 - 5 = 22 \\[3ex]
3rd\:\:term = AS_5 - 2d = 27 - 2(5) = 27 - 10 = 17 \\[3ex]
2nd\:\:term = AS_5 - 3d = 27 - 3(5) = 27 - 15 = 12 \\[3ex]
$
This means that if we have $n$ terms (instead of five terms),

term preceding the $nth$ term = $AS_n - d$

term preceding the term preceeding the $nth$ term = $AS_n - 2d$

term preceding the term preceding the term preceeding the $nth$ term = $AS_n - 3d$

and so on and so forth...

What did we just write?

$
Sum \\[3ex]
For\:\:n\:\:terms \\[3ex]
SAS_n = 1st\:\:term + 2nd\:\:term + 3rd\:\:term + ... + up\:\:to\:\:nth\:\:term \\[3ex]
SAS_n = a + (a + d) + (a + 2d) + (a + 3d) + ... + AS_n...eqn.(1) \\[3ex]
Also...going\:\:backwards \\[3ex]
SAS_n = AS_n + (AS_n - d) + (AS_n - 2d) + (AS_n - 3d) + ... + a...eqn.(2) \\[3ex]
eqn.(1) + eqn(2)\:\:gives \\[3ex]
SAS_n + SAS_n = (a + AS_n) + (a + d + AS_n - d) + (a + 2d + AS_n - 2d) + (a + 3d + AS_n - 3d) + ... + (AS_n + a) \\[3ex]
2SAS_n = (a + AS_n) + (a + AS_n) + (a + AS_n) + (a + AS_n) + ... + (a + AS_n) \\[3ex]
2SAS_n = n(a + AS_n) \\[3ex]
SAS_n = \dfrac{n}{2}(a + AS_n)...Formula\:\:Number\:4 \\[5ex]
But\:\: AS_n = a + d(n - 1) \\[3ex]
\rightarrow SAS_n = \dfrac{n}{2}[a + a + d(n - 1)] \\[5ex]
SAS_n = \dfrac{n}{2}[2a + d(n - 1)]...Formula\:\:Number\:5
$

## Symbols and Meanings

- $AS_n$ = $n$th term of an Arithmetic Sequence
- $a$ = first term
- $p$ = last term
- $d$ = common difference
- $n$ = number of terms
- $SAS_n$ = sum of the $n$ terms of an Arithmetic Sequence

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

$ (1.)\:\: AS_n = a + d(n - 1) \\[5ex] (2.)\:\: AS_n = vn + w \:\:where\:\: v = d \:\:and\:\: w = a - d \\[5ex] (3.)\:\: p = a + d(n - 1) \\[5ex] (4.)\:\: SAS_n = \dfrac{n}{2}(a + AS_n) \\[7ex] (5.)\:\: SAS_n = \dfrac{n}{2}[2a + d(n - 1)] \\[7ex] (6.)\:\: n = \dfrac{2 * SAS_n}{a + p} \\[7ex] (7.)\:\: n = \dfrac{p - a + d}{d} \\[7ex] (8.)\:\: n = \dfrac{-(2a - d) \pm \sqrt{(2a - d)^2 + 8d*SAS_n}}{2d} $

A **Geometric Sequence** is a sequence whose consecutive terms are obtained by the multiplication of a
constant value.

OR

A **Geometric Sequence** is a sequence whose quotient between consecutive terms is a constant value.

It is also known as an **Geometric Progression.**

It is also known as an **Exponential Sequence.**

The next (succeeding) term of the sequence is found by multiplying a constant to the previous (preceeding) term
of that sequence.

That constant is known as the common ratio OR common quotient.

So, the **common ratio** OR **common quotient** is the constant that is multiplied to a term of a Geometric Sequence in order
to give the next term of that sequence.

__Example 1:__ In the sequence: $3, 21, 147, 1029, ...$

First term, $a = 3$

Second term = $21$

Third term = $147$

Fourth term = $1029$

*
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 \div 1st\:\: term = 21 \div 3 = 7 \\[3ex] 3rd\:\: term \div 2nd\:\: term = 147 \div 21 = 7 \\[3ex] 4th\:\: term \div 3rd\:\: term = 1029 \div 147 = 7 \\[3ex] $ So, we have a common ratio ("common" because it is a constant value, "ratio" because it is the result of a division procedure).

So, the common ratio, $r = 7$

What would be the fifth term?

First term, $a = -3$

Second term = $21$

Third term = $-147$

Fourth term = $1029$

second term from the first term?

third term from the second term?

fourth term from the third term?

$ 2nd\:\: term \div 1st\:\: term = 21 \div (-3) = -7 \\[3ex] 3rd\:\: term \div 2nd\:\: term = -147 \div 21 = -7 \\[3ex] 4th\:\: term \div 3rd\:\: term = 1029 \div (-147) = -7 \\[3ex] $ Here, the common ratio or the common quotient, $r = -7$

What would be the tenth term?

Student: I guess there would be an easier way to find it as well.

Teacher: yes of course! ☺☺☺

Back to

We already know that:

first term = $a$

common ratio = $r$

So, we have:

$ 1st\:\: term = a = 3 \\[3ex] 2nd\:\: term = 3 * 7 = 21 = a * r \\[3ex] 3rd\:\: term = 21 * 7 = 147 = a * r * r = a * r^2 = ar^2 \\[3ex] 4th\:\: term = 147 * 7 = 1029 = a * r * r * r = a * r^3 = ar^3 \\[3ex] $

What are the:

$5th, 6th, 7th, 10th, nth$ terms?

$ 5th\:\: term = ar^4 \\[3ex] 6th\:\: term = ar^5 = a * r^{6 - 1} \\[3ex] 7th\:\: term = ar^6 = a * r^{7 - 1} \\[3ex] 10th\:\: term = ar^9 = a * r^{10 - 1} \\[3ex] nth\:\: term = a * r^{n - 1} \\[3ex] nth\:\: term = ar^{n - 1} \\[3ex] $ So, the $nth$ term of an Arithmetic Sequence is: $GS_n = ar^{n - 1}$

So, back to Example 2.

What is the tenth term?

Back to

$ GS_n = ar^{n - 1} \\[3ex] a = -3 \\[3ex] n = 10 \\[3ex] r = -7 \\[3ex] GS_{10} = -3 * -7^{10 - 1} \\[3ex] = -3 * -7^{9} \\[3ex] = -3 * -40353607 \\[3ex] = 121060821 $

Now, that we have figured out the derivation of the formula for the $n$th term of a Geometric Sequence, let us find out how the formula for the sum of the first $n$ terms of a Geometric Sequence is derived.

## Symbols and Meanings

- $GS_n$ = $n$th term of a Geometric Sequence
- $a$ = first term
- $p$ = last term
- $r$ = common ratio
- $n$ = number of terms
- $SGS_n$ = sum of the $n$ terms of a Geometric Sequence
- $S_{\infty}$ = sum to infinity of a Geometric Sequence

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

$ (1.)\:\: GS_n = ar^{n - 1} \\[5ex] (2.)\:\: SGS_n = \dfrac{a(r^{n} - 1)}{r - 1} \:\:for\:\: r \gt 1 \\[7ex] (3.)\:\: SGS_n = \dfrac{a(1 - r^{n})}{1 - r} \:\:for\:\: r \lt 1 \\[7ex] (4.)\:\: n = \dfrac{\log{\left[\dfrac{SGS_n(r - 1)}{a} + 1\right]}}{\log r} \\[7ex] (5.)\:\: If\:\:r \lt 1,\:\:the\:\:series\:\:converges\:\:and\:\: S_{\infty} = \dfrac{a}{1 - r} \\[7ex] (6.)\:\: If\:\:r \gt 1,\:\:the\:\:series\:\:diverges \\[5ex] (7.)\:\: If\:\:r = 1,\:\:S_{\infty}\:\:DNE \\[5ex] (8.)\:\: r = \dfrac{S_{\infty} - a}{S_{\infty}} \\[7ex] (9.)\:\: a = S_{\infty}(1 - r) $

A **Quadratic Sequence** is defined as a sequence in which the highest exponent of the number
of terms in the general term ($nth$ term) is two.

Let us compare and contrast the Linear Sequence and the Quadratic Sequence.

Linear Sequence (Arithmetic Sequence) | Quadratic Sequence |
---|---|

$AS_n = a + d(n -1)$ $AS_n = vn + w$ where $v = d$ and $w = a - d$ |
$QS_n = an^2 + bn + c$ |

The highest exponent of $n$ in the $nth$ term is $1$ | The highest exponent of $n$ in the $nth$ term is $2$ |

Compare to a Linear Expression; Linear Function; Linear Equation | Compare to a Quadratic Expression; Quadratic Function; Quadratic Equation |

The first difference or the difference between a consecutive term and the term is equal In other words, a linear sequence has a common "first" difference between a consecutive term and the term. For example: $ 4, 7, 10, 13,... \\[3ex] \underline{First\:\:Difference} \\[3ex] 7 - 4 = \color{red}{3} \\[3ex] 10 - 7 = \color{red}{3} \\[3ex] 13 - 10 = \color{red}{3} $ |
The second difference or the difference of the difference of consecutive terms is equal In other words, a quadratic sequence has a common "second" difference between consecutive terms For example: $ 8, 14, 22, 32,... \\[3ex] \underline{First\:\:Difference} \\[3ex] 14 - 8 = 6 \\[3ex] 22 - 14 = 8 \\[3ex] 32 - 22 = 10 \\[3ex] \underline{Second\:\:Difference} \\[3ex] 8 - 6 = \color{red}{2} \\[3ex] 10 - 8 = \color{red}{2} $ |

** Given:** A Sequence of Numbers

At least the first four terms of the sequence is required.

This is important in order to ensure that the second difference is the same.

Let the Sequence = $1st$, $2nd$, $3rd$, $4th$, ...

$ \underline{First\:\:Difference} \\[3ex] 2nd - 1st = 2nd - 1st \\[3ex] 3rd - 2nd = 3rd - 2nd \\[3ex] 4th - 3rd = 4th - 3rd \\[3ex] \underline{Second\:\:Difference} \\[3ex] 3rd - 2nd - (2nd - 1st) \\[3ex] = 3rd - 2nd - 2nd + 1st \\[3ex] = 3rd - 2(2nd) + 1st \\[3ex] = 1st + 3rd - 2(2nd) \\[3ex] Also: \\[3ex] 4th - 3rd - (3rd - 2nd) \\[3ex] = 4th - 3rd - 3rd + 2nd \\[3ex] = 4th - 2(3rd) + 2nd \\[3ex] = 2nd + 4th - 2(3rd) \\[3ex] The\:\:Second\:\:Difference\:\:must\:\:be\:\:the\:\:same \\[3ex] \implies 1st + 3rd - 2(2nd) = 2nd + 4th - 2(3rd) \\[3ex] $ Therefore, for a Sequence to be Quadratic;

The $1st + 3rd - 2(2nd)$ must be equal to $2nd + 4th - 2(3rd)$

** Given:** A Quadratic Sequence

Only the first three terms of the sequence is required.

Let the Quadratic Sequence = $1st$, $2nd$, $3rd$

$ nth\:\:term = general\:\:term = QS_n \\[3ex] QS_n = an^2 + bn + c \\[3ex] QS_1 = First\:\:term = 1st \implies n = 1 \\[3ex] QS_1 = a(1)^2 + b(1) + c \\[3ex] QS_1 = a(1) + b + c \\[3ex] 1st = a + b + c \\[3ex] a + b + c = 1st...eqn.(1) \\[3ex] QS_2 = Second\:\:term = 2nd \implies n = 2 \\[3ex] QS_2 = a(2)^2 + b(2) + c \\[3ex] QS_2 = a(4) + 2b + c \\[3ex] 2nd = 4a + 2b + c \\[3ex] 4a + 2b + c = 2nd...eqn.(2) \\[3ex] QS_3 = Third\:\:term = 3rd \implies n = 3 \\[3ex] QS_3 = a(3)^2 + b(3) + c \\[3ex] QS_3 = a(9) + 3b + c \\[3ex] 3rd = 9a + 3b + c \\[3ex] 9a + 3b + c = 3rd...eqn.(3) \\[3ex] We\:\:need\:\:to\:\:find\:\:a \\[3ex] First:\:\:let\:\:us\:\:eliminate\:\: c \\[3ex] eqn.(2) - eqn.(1) \implies (4a - a) + (2b - b) + (c - c) = 2nd - 1st \\[3ex] 3a + b = 2nd - 1st...eqn.(4) \\[3ex] eqn.(3) - eqn.(1) \implies (9a - a) + (3b - b) + (c - c) = 3rd - 1st \\[3ex] 8a + 2b = 3rd - 1st...eqn.(5) \\[3ex] Next:\:\:let\:\:us\:\:eliminate\:\: c \\[3ex] 2 * eqn(4) \implies 2(3a + b) = 2(2nd - 1st) \\[3ex] 6a + 2b = 2(2nd) - 2(1st)...eqn(6) \\[3ex] eqn.(5) - eqn.(6) \implies (8a - 6a) + (2b - 2b) = 3rd - 1st - [2(2nd) - 2(1st)] \\[3ex] 2a = 3rd - 1st - 2(2nd) + 2(1st) \\[3ex] 2a = -1st + 2(1st) + 3rd - 2(2nd) \\[3ex] 2a = 1st + 3rd - 2(2nd) \\[3ex] \therefore a = \dfrac{1st + 3rd - 2(2nd)}{2} \\[5ex] We\:\:have\:\:found\:\:a \\[3ex] Let\:\:us\:\:find\:\:b \\[3ex] From\:\:eqn.(4) \\[3ex] 3a + b = 2nd - 1st...eqn.(4) \\[3ex] b = 2nd - 1st - 3a \\[3ex] 3a = 3\left[\dfrac{1st + 3rd - 2(2nd)}{2}\right] \\[5ex] 3a = \dfrac{3}{2}\left[1st + 3rd - 2(2nd)\right] \\[5ex] 3a = \dfrac{3(1st)}{2} + \dfrac{3(3rd)}{2} - 3(2nd) \\[5ex] \rightarrow b = 2nd - 1st - \left[\dfrac{3(1st)}{2} + \dfrac{3(3rd)}{2} - 3(2nd)\right] \\[5ex] b = 2nd - 1st - \dfrac{3(1st)}{2} - \dfrac{3(3rd)}{2} + 3(2nd) \\[5ex] b = 2nd + 3(2nd) - 1st - \dfrac{3(1st)}{2} - \dfrac{3(3rd)}{2} \\[5ex] b = 4(2nd) - \dfrac{2(1st)}{2} - \dfrac{3(1st)}{2} - \dfrac{3(3rd)}{2} \\[5ex] b = \dfrac{8(2nd)}{2} - \dfrac{5(1st)}{2} - \dfrac{3(3rd)}{2} \\[5ex] \therefore b = \dfrac{8(2nd) - 5(1st) - 3(3rd)}{2} \\[5ex] We\:\:have\:\:found\:\:b \\[3ex] Let\:\:us\:\:find\:\:c \\[3ex] From\:\:eqn.(1) \\[3ex] a + b + c = 1st...eqn.(1) \\[3ex] c = 1st - b - a \\[3ex] c = 1st - \left[\dfrac{8(2nd) - 5(1st) - 3(3rd)}{2}\right] - \left[\dfrac{1st + 3rd - 2(2nd)}{2}\right] \\[5ex] c = \dfrac{2(1st)}{2} - \left[\dfrac{8(2nd) - 5(1st) - 3(3rd)}{2}\right] - \left[\dfrac{1st + 3rd - 2(2nd)}{2}\right] \\[5ex] c = \dfrac{2(1st) - 8(2nd) + 5(1st) + 3(3rd) - 1st - 3rd + 2(2nd)}{2} \\[5ex] c = \dfrac{2(1st) + 5(1st) - 1st + 2(2nd) - 8(2nd) + 3(3rd) - 3rd}{2} \\[5ex] c = \dfrac{6(1st) - 6(2nd) + 2(3rd)}{2} \\[5ex] c = \dfrac{6(1st)}{2} - \dfrac{6(2nd)}{2} + \dfrac{2(3rd)}{2} \\[5ex] \therefore c = 3(1st) - 3(2nd) + 3rd \\[3ex] We\:\:have\:\:found\:\:c \\[3ex] Let\:\:us\:\:write\:\:the\:\:nth\:\:term \\[3ex] QS_n = an^2 + bn + c \\[3ex] Substitute\:\:the\:\:values\:\:of\:\:a, b, c \\[3ex] \therefore QS_n = \dfrac{1st + 3rd - 2(2nd)}{2} * n^2 + \dfrac{8(2nd) - 5(1st) - 3(3rd)}{2} * n + 3(1st) - 3(2nd) + 3rd $

## Symbols and Meanings

- $QS_n$ = $n$th term of a Quadratic Sequence = $an^2 + bn + c$
- $1st$ = first term
- $2nd$ = second term
- $3rd$ = third term
- $4th$ = fourth term
- $n$ = number of terms
- $a$ = coefficient of $n^2$
- $b$ = coefficient of $n$

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

$ QS = 1st,\:\:\:\:2nd,\:\:\:3rd,\:\:\:4th,... \\[5ex] QS_n = an^2 + bn + c \\[5ex] (1.)\:\: a = \dfrac{1st + 3rd - 2(2nd)}{2} \\[7ex] (2.)\:\: b = \dfrac{8(2nd) - 5(1st) - 3(3rd)}{2} \\[7ex] (3.)\:\: c = 3(1st) - 3(2nd) + 3rd \\[5ex] (4.)\:\: \therefore QS_n = \dfrac{1st + 3rd - 2(2nd)}{2} * n^2 + \dfrac{8(2nd) - 5(1st) - 3(3rd)}{2} * n + 3(1st) - 3(2nd) + 3rd \\[7ex] The\:\:Left\:\:Hand\:\:Side\:\:will\:\:always\:\:be\:\:equal\:\:this\:\:Right\:\:Hand\:\:Side \\[5ex] (5.)\:\: a + b + c = 1st \\[5ex] (6.)\:\: 4a + 2b + c = 2nd \\[5ex] (7.)\:\: 9a + 3b + c = 3rd \\[5ex] (8.)\:\: 3a + b = 2nd - 1st \\[5ex] (9.)\:\: 8a + 2b = 3rd - 1st $

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