A presentation with examples of sequences problems.

2. Jens Martensson
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3. Jens Martensson
Consider the sequence 3 , 6 , 9 , 12 , …
Each term is calculated by adding 3 to the
previous term.
Therefore the next two terms would be
12 + 3 = 𝟏𝟓 and
15 + 3 = 𝟏𝟖
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Example 1
Write the next two terms in the following sequence.

4. Jens Martensson
𝟗 , 𝟐𝟑 , 𝟑𝟕 , , 𝟔𝟓 , , 𝟗𝟑
The fourth and sixth terms are missing.
First, identify the rule (pattern).
To get each term, add 14 to the previous term.
𝑎4 = 𝑎3 + 14
=37 + 14 = 𝟓𝟏 and
𝑎6 = 𝑎5 + 14
=65 + 14 = 𝟕𝟗.
The sequence is 9 , 23, 37 , 𝟓𝟏 , 65 , 𝟕𝟗 , 93
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Example 2
Complete the sequence.

5. Jens Martensson
Example 3
Write the first four terms of the sequence with the rule: the square root of
consecutive natural numbers.
The natural numbers are the counting numbers.
These begin with 1 and continue with 2, 3 and so
on.
The rule would be 𝒂𝒏 = 𝒏
The sequence is 1 , 2 , 3 , 4 , …
Some values can be simplified and we get
𝟏 , 𝟐 , 𝟑 , 𝟐 , …
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6. Jens Martensson
Whole numbers begin at 0 and continue onward.
𝑎1 = 2 × 0 − 3 𝑎2 = 2 × 1 − 3
= 0 − 3 = 2 − 3
= −𝟑 = −𝟏
𝑎3 = 2 × 2 − 3 𝑎4 = 2 × 3 − 3
= 4 − 3 = 6 − 3
= 𝟏 = 𝟑
The first four terms are: −𝟑 , −𝟏 , 𝟏 , 𝟑 , …
The general term is: 𝒂𝒏 = 𝟐 × 𝒏 − 𝟏 − 𝟑
It can be simplified to 2𝑛 − 2 − 3
= 𝟐𝒏 − 𝟓
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Example 4
Write the first four terms of the sequence with the rule: 2 times each consecutive
whole number minus 3.
Write the general term for this sequence.

7. Jens Martensson
Example 5
Determine the rule for each of the sequences below:
a) 1 , 4 , 9 , 16 , …
𝑎1 = 12
𝑎2 = 22
𝑎3 = 32
𝑎4 = 42
To get the terms of this sequence, square each
successive natural number.
Therefore, the rule is 𝒂𝒏 = 𝒏𝟐
b) 10, 20, 30, 40, …
𝑎1 = 1 × 10
𝑎2 = 2 × 10
𝑎3 = 3 × 10
𝑎4 = 4 × 10
To get the terms of this sequence, multiply each
successive natural number by 10.
Therefore the rule is 𝒂𝒏 = 𝒏 × 𝟏𝟎
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8. Jens Martensson
Word Problems
1. While sorting some beads, Amanda
put 16 beads in the first box, 20 beads
in the second box, 24 beads in the
third box, 28 beads in the fourth box,
and 32 beads in the fifth box.
If this pattern continues, how many
beads will Amanda put in the sixth
box?
The sequence is 16 , 20 , 24 , 28 , 32 , …
𝑎1 = 16
𝑎2 = 20
𝑎3 = 24
𝑎4 = 28
𝑎5 = 32
To get each number, add 4 to the previous
number.
Therefore, 𝒂𝟔 = 𝟑𝟔
Type equation here.
Amanda will put 36 beads into the sixth box.
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9. Jens Martensson
Word Problems
2. Lauren is sorting buttons into bowls.
She puts 1 button into the first bowl, 2
buttons into the second, 4 buttons into
the third and 8 buttons into the fourth.
If the pattern continues, how many
buttons would she put into the fifth
bowl?
The sequence is 1 , 2 , 4 , 8 , …
Each value is a power of 2.
𝑎1 = 1 = 20
𝑎2 = 2 = 21
𝑎3 = 4 = 22
𝑎4 = 8 = 23
The general rule is 𝑎𝑛 = 2𝑛−1
Therefore, 𝒂𝟓 = 𝟐𝟒
= 𝟏𝟔
Lauren will put 16 buttons into the fifth bowl.
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