Understanding Sequences: Arithmetic, Geometric, and Fibonacci Rules

	$a_{1} > 0$	$a_{1} < 0$
$r > 1$	Increasing $3 \to \times 2 6 \to \times 2 12 \to \times 2 24 \to \times 2 48 \dots$	Decreasing $- 3 \to \times 2 - 6 \to \times 2 - 12 \to \times 2 - 24 \to \times 2 - 48 \dots$
$r = 1$	Constant $3 \to \times 1 3 \to \times 1 3 \to \times 1 3 \to \times 1 3 \dots$	Constant $- 3 \to \times 1 - 3 \to \times 1 - 3 \to \times 1 - 3 \to \times 1 - 3 \dots$
$0 < r < 1$	Decreasing $48 \to \times \frac{1}{2} 24 \to \times \frac{1}{2} 12 \to \times \frac{1}{2} 6 \to \times \frac{1}{2} 3 \dots$	Increasing $- 48 \to \times \frac{1}{2} - 24 \to \times \frac{1}{2} - 12 \to \times \frac{1}{2} - 6 \to \times \frac{1}{2} - 3 \dots$
$r < 0$	Alternating $3 \to \times (- 2) - 6 \to \times (- 2) 12 \to \times (- 2) - 24 \to \times (- 2) 48 \dots$	Alternating $- 3 \to \times (- 2) 6 \to \times (- 2) - 12 \to \times (- 2) 24 \to \times (- 2) - 48 \dots$

a_{1} > 0

a_{1} < 0

r > 1

Increasing

3 \to \times 2 6 \to \times 2 12 \to \times 2 24 \to \times 2 48 \dots

Decreasing

- 3 \to \times 2 - 6 \to \times 2 - 12 \to \times 2 - 24 \to \times 2 - 48 \dots

r = 1

Constant

$3 \to \times 1 3 \to \times 1 3 \to \times 1 3 \to \times 1 3 \dots$

Constant

$- 3 \to \times 1 - 3 \to \times 1 - 3 \to \times 1 - 3 \to \times 1 - 3 \dots$

0 < r < 1

Decreasing

48 \to \times \frac{1}{2} 24 \to \times \frac{1}{2} 12 \to \times \frac{1}{2} 6 \to \times \frac{1}{2} 3 \dots

Increasing

- 48 \to \times \frac{1}{2} - 24 \to \times \frac{1}{2} - 12 \to \times \frac{1}{2} - 6 \to \times \frac{1}{2} - 3 \dots

r < 0

Alternating

$3 \to \times (- 2) - 6 \to \times (- 2) 12 \to \times (- 2) - 24 \to \times (- 2) 48 \dots$

Alternating

$- 3 \to \times (- 2) 6 \to \times (- 2) - 12 \to \times (- 2) 24 \to \times (- 2) - 48 \dots$

Sequence	Classification
$15, 30, 45, 60, \dots$	Arithmetic Sequence

Sequence

Classification

15, 30, 45, 60, \dots

Arithmetic Sequence

Sequence	Classification
$15, 25, 15, 25, \dots$	Neither arithmetic nor geometric

Sequence

Classification

15, 25, 15, 25, \dots

Neither arithmetic nor geometric

Sequence	Classification
$5, 10, 20, 40, \dots$	Geometric Sequence

Sequence

Classification

5, 10, 20, 40, \dots

Geometric Sequence

$n$	$a_{n}$	Rewrite
$1$	$a_{1}$	$a_{1} + 0 \cdot d$
$2$	$a_{2}$	$a_{1} + 1 \cdot d$
$3$	$a_{3}$	$a_{1} + 2 \cdot d$
$4$	$a_{4}$	$a_{1} + 3 \cdot d$
$5$	$a_{5}$	$a_{1} + 4 \cdot d$

n

a_{n}

Rewrite

1

a_{1}

a_{1} + 0 \cdot d

2

a_{2}

a_{1} + 1 \cdot d

3

a_{3}

a_{1} + 2 \cdot d

4

a_{4}

a_{1} + 3 \cdot d

5

a_{5}

a_{1} + 4 \cdot d

Heichi keeps track of the number of pages he reads in a certain amount of time. The hours spent reading and the amount of pages he reads are shown in the table.

Hours, $n$	Number of Pages, $a_{n}$
$1$	$20$
$2$	$35$
$3$	$50$

Write an explicit rule that represents the relationship between the number of pages Heichi reads and the time he spends reading in hours.

We are given a table showing the number of hours spends reading and the corresponding number of pages Heichi reads.

Hours, n	Number of Pages, a_n
1	20
2	35
3	50

We want to write an explicit rule to find the number of pages Heichi reads in n hours. Let's consider just the number of pages for now. They can form an increasing sequence. We can check to see if there is a constant difference between consecutive terms.

This is an arithmetic sequence with first term 20 and common difference 15. Let's use the general formula for the explicit rule of an arithmetic sequence. a_n= a_1+(n-1) d In this rule, a_1 is the first term in a given sequence, d is the common difference from one term to the next, and a_n is the nth term in the sequence. Let's substitute a_1= 20 and d= 15 into the rule and simplify.

This rule gives us the number of pages Heichi can read in n hours.

Ignacio records the distance he runs during various time intervals. The times spent running and the corresponding distance in meters are shown in the table.

Time in minutes, $n$	Distance in meters, $a_{n}$
$3$	$400$
$5$	$640$
$7$	$880$

Write a function that expresses the relationship between the time Ignacio spends running and the distance he covers.

We are given a table showing the times Ignacio spent running and the corresponding distance he ran. We want to write a function to express their relationship.

Time in minutes, n	Distance in meters, a_n
3	400
5	640
7	880

In this table, the values in the time column increase by 2 while the values in the distance row increase by 240. If we were to think of the distances Ignacio runs as a sequence, the third, fifth, and seventh terms would be 400, 640, and 880, respectively. This means that the difference between consecutive terms would be 240÷ 2 =120, and the first term would be 400-2(120)=140.

This is an arithmetic sequence with first term 160 and common difference 120. Now we can use the general formula for the explicit rule of an arithmetic sequence. a_n= a_1+(n-1) d Let's substitute a_1=160 and d=120 into the formula and simplify.

This rule provides the distance, in meters, that Ignacio ran in n minutes.

Describe the value of each term of the sequence as a function of its position.

Position	$3$	$4$	$5$	$6$	$n$
Term	$\frac{1}{3}$	$\frac{1}{9}$	$\frac{1}{2 7}$	$\frac{1}{8 1}$

What is the value of the first term in the sequence?

We want to describe the value of each term as a function of its position. To achieve this, we will attempt to rewrite each given term as a numeric expression that includes its position. Let's take a look at the values.

Position	3	4	5	6	n
Term	1/3	1/9	1/27	1/81

Notice that each term is found by multiplying the previous term by 13 and can be expressed as a power of 13.

Position	3	4	5	6
Term	1/3	1/9	1/27	1/81
Rewrite	1/3 = (1/3)^1	1/3^2= (1/3)^2	1/3^3= (1/3)^3	1/3^4= (1/3)^4

As we can see, the exponents increase by one for each consecutive term. Furthermore, each exponent is 2 less than the position of that term. For example, the exponent of the third term is 1, which is the result of subtracting 2 from its position 3.

Position	3	4	5	6
Term	1/3	1/9	1/27	1/81
Rewrite Exponents	(1/3)^1 = (1/3)^(3- 2)	(1/3)^2 = (1/3)^(4- 2)	(1/3)^3 = (1/3)^(5- 2)	(1/3)^4 = (1/3)^(6- 2)

Considering this pattern, the value of the term in position n can be described as ( 13 )^(n- 2). a_n = ( 1/3)^(n- 2) We can substitute any natural number n to get the corresponding value of term, so this is a function of the sequence.

Let's use the function we found in Part A to find the first term of the sequence. In this formula, we will replace n with 1 and then simplify.

The first term of the sequence is 3.

Sequences

Catch-Up and Review

Describing a Pattern

Sequence

Pattern

Numerical Patterns in Books

Answer

Hint

Solution

Finding the Terms of a Sequence

Special Sequences: Arithmetic Sequences

Arithmetic Sequence

Special Sequences: Geometric Sequences

Geometric Sequence

A Reading Challenge for the Mysterious Library

Hint

Solution

Identifying Sequences

Graphing Sequences

Answer

Hint

Solution

Explicit Rule of Arithmetic Sequences

Proof

Proof

Unlocking the Door

Hint

Solution

Identifying a Specific Term of an Arithmetic Sequence

The Fibonacci Sequence

Sequences

Recommended exercises

	15 Theory slides
	10 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions