CLASS 9 MATHS · CHAPTER 8 · STUDY NOTES
Predicting What Comes Next: Sequences and Progressions
Understand predicting what comes next: sequences and progressions through short explanations, labelled models and original practice. Read one topic, explain it in your own words, then check.
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A sequence has an order
Separate a term’s position from its value.Positions and values are different
Term value (number)
Position n (positive integer)
- Sequence terms · Only integer positions are valid.
- 1t1=4 · (1, 4)
- 2t2=9 · (2, 9)
- 3t3=14 · (3, 14)
- 4t4=19 · (4, 19)
Only the four shown terms are plotted; the dots do not define a unique continuation without a rule.
Read the plotted values
| Position n | Term value |
|---|---|
| 1 | 4 |
| 2 | 9 |
| 3 | 14 |
| 4 | 19 |
- Sequence
- An ordered list of numbers.
- Term
- A number at a particular position in the list.
In 4, 9, 14, 19, …, t₁ = 4 and t₃ = 14. The small subscript names the position, not a multiplication.
A sequence may stop or continue indefinitely. In this chapter we usually count positions from 1; a stage model can explicitly begin at stage 0.
Take a moment. What do you think?
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Question
In 4, 9, 14, 19, what is t₃?
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Growing dots explain number patterns
Look at what is added at each stage.Count one row at a time
- Triangular number
- The dot count in a triangle with rows 1, 2, …, n: n(n+1)/2.
- Square number
- n² dots arranged in an n by n square.
A triangle grows by adding a whole new row: 1, 3, 6, 10, 15, … . The extra counts are 2, 3, 4, 5, … .
A square grows by adding an L-shaped border. The added counts are 3, 5, 7, … . So 1 + 3 + … + (2n−1) = n². Neither sequence has a constant difference.
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Question
How many dots are in a triangle with five rows?
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Use the position to jump to a term
An explicit rule does not need earlier terms.The rule tₙ = 4n + 1
Term value (number)
Position n (positive integer)
- Sequence terms · Only integer positions are valid.
- 1t1=5 · (1, 5)
- 2t2=9 · (2, 9)
- 3t3=13 · (3, 13)
- 4t4=17 · (4, 17)
- 5t5=21 · (5, 21)
n takes positive integer values. The sequence points follow the stated rule.
Read the plotted values
| Position n | Term value |
|---|---|
| 1 | 5 |
| 2 | 9 |
| 3 | 13 |
| 4 | 17 |
| 5 | 21 |
- Explicit rule
- A formula that gives a term directly from its position.
For tₙ = 4n + 1, the first terms are 5, 9, 13, 17. To find t₂₀, substitute n = 20: 81.
A few terms alone do not guarantee one unique rule. State the assumed rule before extending a pattern.
- 1
Read the rule and the requested position.
- 2
Substitute that integer for n.
- 3
Carry out multiplication before addition.
- 4
Check a small term against the listed sequence.
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Question
Find t₇ for tₙ = 4n + 1.
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A position must be a permitted integer
Solve for n and check whether it is a valid position.11 is a position; 11.25 is between positions
Value (number)
- Number-line marks · Only horizontal position represents the number.
- 110 · (10, 0)
- 211 · (11, 0)
- 311.25 · (11.25, 0)
- 412 · (12, 0)
Positions here are positive integers. 11.25 is not a valid index.
Read the plotted values
| Value | Height |
|---|---|
| 10 | 0 |
| 11 | 0 |
| 11.25 | 0 |
| 12 | 0 |
- Membership test
- Set the term formula equal to a value, solve, then check the allowed position.
For tₙ = 4n + 1, the value 45 gives n = (45−1)/4 = 11, so it belongs.
The value 46 gives n = 11.25. That is not a sequence position; 46 does not belong. A positive real answer is not enough.
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Question
Is 49 a term of tₙ = 4n + 1?
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Build from earlier terms
A recursive rule needs starting values.Double the previous value, then add 1
Term value (number)
Position n (positive integer)
- Sequence terms · Only integer positions are valid.
- 1t1=2 · (1, 2)
- 2t2=5 · (2, 5)
- 3t3=11 · (3, 11)
- 4t4=23 · (4, 23)
- 5t5=47 · (5, 47)
Starting value t₁ = 2; rule tₙ = 2tₙ₋₁ + 1 for n ≥ 2.
Read the plotted values
| Position n | Term value |
|---|---|
| 1 | 2 |
| 2 | 5 |
| 3 | 11 |
| 4 | 23 |
| 5 | 47 |
- Recursive rule
- A rule that gives a term using earlier terms.
With t₁ = 2 and tₙ = 2tₙ₋₁ + 1 for n ≥ 2, the terms are 2, 5, 11, 23, … . Each new value uses the previous one.
The Virahānka–Fibonacci pattern in this book begins 1, 2, 3, 5, 8, … . Add the previous two terms. Other conventions begin 0, 1; state the starting values.
- 1
Write all required starting terms.
- 2
Use the most recent term in the rule.
- 3
Record the result as the next term.
- 4
Repeat; do not replace n with a term value by mistake.
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Question
What comes after 2, 5, 11, 23 using the stated recursive rule?
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An AP adds the same amount
Check every consecutive difference.Equal downward steps
Term value (number)
Position n (positive integer)
- Sequence terms · Only integer positions are valid.
- 1t1=18 · (1, 18)
- 2t2=15 · (2, 15)
- 3t3=12 · (3, 12)
- 4t4=9 · (4, 9)
- 5t5=6 · (5, 6)
- 6t6=3 · (6, 3)
- 7t7=0 · (7, 0)
A decreasing AP has a negative common difference; zero can still be a term.
Read the plotted values
| Position n | Term value |
|---|---|
| 1 | 18 |
| 2 | 15 |
| 3 | 12 |
| 4 | 9 |
| 5 | 6 |
| 6 | 3 |
| 7 | 0 |
- Arithmetic progression (AP)
- A sequence with one constant difference between consecutive terms.
- Common difference d
- Next term minus previous term; it can be positive, zero or negative.
An AP beginning at a has tₙ = a + (n−1)d. There are n−1 jumps from the first term to the nth term.
For 18, 15, 12, 9, …, a = 18 and d = −3. Then t₇ = 18 + 6(−3) = 0. Equal differences put the sequence points on a straight line.
Recursive AP rule
t₁ = a; tₙ = tₙ₋₁ + d for n ≥ 2This gives the same terms as the explicit AP formula.
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Question
Why does the formula use n−1 rather than n?
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