1. The Big Idea — What Is a Sequence?
A sequence is an ordered list of numbers whose terms are generated
according to a rule. The order matters: changing the order can change the sequence.
3, 7, 11, 15, 19, ...
The first term is usually written as u₁, the second as
u₂, and the nth term as uₙ.
u₁, u₂, u₃, ..., uₙ, ...
Sequence vs Series
These two words are closely related but are not interchangeable.
-
A sequence is a list:
2, 5, 8, 11, ...
-
A series is a sum of terms:
2 + 5 + 8 + 11 + ...
Core distinction:
A sequence tells you what the terms are. A series asks you to add terms together.
How Can a Sequence Be Defined?
A sequence can be described in several ways. In examination questions,
recognising which description you have been given is important.
| Representation |
Example |
What it tells you |
| List of terms |
4, 7, 10, 13, ... |
Shows the pattern directly. |
| Explicit formula |
uₙ = 3n + 1 |
Gives any term directly from n. |
| Recurrence relation |
u₁ = 4, uₙ₊₁ = uₙ + 3 |
Generates each term from the previous term. |
| Summation notation |
Σ(n + 1) |
Describes a sum compactly. |
2. Finding Patterns and the nth Term
The most important idea is that the position of a term is represented by
n. If you can express the term in terms of n, you have found
an explicit formula for the sequence.
uₙ = f(n)
Linear Patterns
Consider:
6, 10, 14, 18, 22, ...
The difference between consecutive terms is constant:
+4, +4, +4, +4
This immediately suggests a linear nth-term formula.
uₙ = 4n + 2
Check:
n = 1 → 4(1) + 2 = 6
n = 2 → 4(2) + 2 = 10
n = 3 → 4(3) + 2 = 14
Good habit:
Never stop after finding a formula. Test at least one or two values of n.
A formula that looks correct can still contain an algebraic mistake.
Not Every Pattern Is Arithmetic
2, 4, 8, 16, 32, ...
The differences are not constant:
+2, +4, +8, +16
But each term is multiplied by 2. This is therefore geometric rather than arithmetic.
3. Arithmetic Sequences
An arithmetic sequence has a constant difference between
consecutive terms. This constant difference is called the
common difference, written as d.
a, a + d, a + 2d, a + 3d, ...
The nth-Term Formula
uₙ = a + (n − 1)d
Why is it n − 1 rather than n?
u₁ = a
u₂ = a + d
u₃ = a + 2d
u₄ = a + 3d
The first term has had zero differences added to it.
Therefore the nth term has had d added exactly n − 1 times.
Worked Example 1 — Basic
Find the nth term of:
5, 9, 13, 17, ...
a = 5
d = 4
uₙ = 5 + (n − 1)(4)
= 5 + 4n − 4
= 4n + 1
Worked Example 2 — Finding a Specific Term
Find the 40th term of the arithmetic sequence:
11, 8, 5, 2, ...
a = 11
d = −3
u₄₀ = 11 + 39(−3)
= 11 − 117
= −106
Worked Example 3 — Finding a and d from Two Terms
An arithmetic sequence has:
u₅ = 17
u₁₂ = 45
Since:
u₅ = a + 4d = 17
u₁₂ = a + 11d = 45
Subtract the equations:
7d = 28
d = 4
Substitute into the first equation:
a + 16 = 17
a = 1
uₙ = 1 + 4(n − 1) = 4n − 3
Higher-level insight:
If two terms of an arithmetic sequence are known, their difference divided
by the difference in their term numbers gives d.
d =
(uₘ − uₙ) / (m − n)
4. Arithmetic Sequence Reasoning
Finding the Position of a Term
Suppose:
uₙ = 7 + 3(n − 1)
Which term is 64?
64 = 7 + 3(n − 1)
64 = 3n + 4
3n = 60
n = 20
Does a Number Occur in the Sequence?
The algebra may produce a non-integer value. If the resulting n is not a
positive integer, the number is not a term of the sequence.
uₙ = 5 + 4(n − 1) = 4n + 1
Test whether 100 occurs:
100 = 4n + 1
4n = 99
n = 24.75
Since 24.75 is not an integer, 100 is not a term.
Arithmetic Mean
If b lies exactly halfway between a and c, then:
b = (a + c) / 2
In an arithmetic progression:
2b = a + c
This is often useful when a missing term must be inserted between known terms.
5. Geometric Sequences
A geometric sequence has a constant ratio between consecutive
non-zero terms. The common ratio is written as r.
a, ar, ar², ar³, ...
uₙ = arⁿ⁻¹
Here:
- a = first term
- r = common ratio
- n = term number
Worked Example 1
3, 6, 12, 24, ...
a = 3
r = 2
uₙ = 3 × 2ⁿ⁻¹
Worked Example 2 — Fractional Ratio
81, 27, 9, 3, ...
r = 27/81 = 1/3
uₙ = 81(1/3)ⁿ⁻¹
Worked Example 3 — Negative Ratio
5, −10, 20, −40, ...
r = −2
uₙ = 5(−2)ⁿ⁻¹
The negative ratio causes the signs to alternate.
Pattern recognition:
A geometric sequence does not have to increase. It can decrease, alternate,
or approach zero.
6. Geometric Sequence Reasoning
Finding r from Two Terms
Suppose:
u₃ = ar²
u₇ = ar⁶
Dividing eliminates a:
u₇ / u₃ = r⁴
Therefore:
r = fourth root of (u₇ / u₃)
Important:
When taking an even root, consider whether both positive and negative values
of r satisfy the original equation. Do not automatically discard the negative root.
Worked Example
A geometric sequence has:
u₂ = 12
u₅ = 96
Since:
u₅ / u₂ = r³
96 / 12 = r³
8 = r³
r = 2
Now:
u₂ = ar = 12
2a = 12
a = 6
uₙ = 6 × 2ⁿ⁻¹
Geometric Mean
If b is the geometric mean of positive numbers a and c, then:
b² = ac
Hence:
b = √(ac)
For example, the positive geometric mean of 4 and 25 is:
√(4 × 25) = √100 = 10
7. Recurrence Relations
A recurrence relation defines a term using one or more earlier terms.
u₁ = 4
uₙ₊₁ = uₙ + 5
Starting from u₁:
u₁ = 4
u₂ = 9
u₃ = 14
u₄ = 19
Arithmetic Recurrence
u₁ = a
uₙ₊₁ = uₙ + d
This recurrence generates the arithmetic sequence:
uₙ = a + (n − 1)d
Geometric Recurrence
u₁ = a
uₙ₊₁ = ruₙ
This generates:
uₙ = arⁿ⁻¹
Iterating a Recurrence
Consider:
u₁ = 2
uₙ₊₁ = 3uₙ − 1
u₂ = 3(2) − 1 = 5
u₃ = 3(5) − 1 = 14
u₄ = 3(14) − 1 = 41
Key idea:
A recurrence is sequential. You normally cannot jump directly to a distant term
unless you have transformed or solved the recurrence.
8. Sigma Notation
The Greek capital letter sigma, Σ, represents summation.
It provides a compact way of writing many terms.
Σ⁵ₙ₌₁ n
means:
1 + 2 + 3 + 4 + 5
Understanding the Components
Σ⁸ₖ₌₃ (2k + 1)
- k is the index.
- 3 is the starting value.
- 8 is the final value.
- 2k + 1 is the term being added.
Therefore:
(2·3 + 1) + (2·4 + 1) + ... + (2·8 + 1)
Changing the Index Letter
The index letter is a dummy variable. For example:
Σ⁵ₙ₌₁ n² = Σ⁵ₖ₌₁ k²
The value does not change because only the name of the index has changed.
Splitting a Sigma
Σ(aₙ + bₙ) = Σaₙ + Σbₙ
Constants can also be taken outside:
Σ(caₙ) = cΣaₙ
9. Standard Summation Results
These results are fundamental when evaluating sums involving powers of n.
Σₙ₌₁ᴺ 1 = N
Σₙ₌₁ᴺ n = N(N + 1)/2
Σₙ₌₁ᴺ n² = N(N + 1)(2N + 1)/6
Σₙ₌₁ᴺ n³ = [N(N + 1)/2]²
Worked Example
Evaluate:
Σₙ₌₁²⁰ (3n + 5)
Split the sum:
3Σₙ₌₁²⁰ n + 5Σₙ₌₁²⁰ 1
= 3[20(21)/2] + 5(20)
= 3(210) + 100
= 730
Worked Example — Quadratic Expression
Σₙ₌₁¹⁰ (n² + 2n − 3)
Σn² + 2Σn − 3Σ1
= [10·11·21/6] + 2[10·11/2] − 3(10)
= 385 + 110 − 30
= 465
Exam method:
If the summand is a polynomial in n, expand or split it into standard sums
before substituting numbers.
10. Arithmetic Series
An arithmetic series is the sum of the terms of an arithmetic sequence.
a + (a+d) + (a+2d) + ... + [a+(n−1)d]
The sum of the first n terms is:
Sₙ = n/2 [2a + (n − 1)d]
If the final term is l:
Sₙ = n/2(a + l)
Why Does the Formula Work?
Write the series forwards:
S = a + (a+d) + (a+2d) + ... + l
Reverse it:
S = l + (l−d) + (l−2d) + ... + a
Adding corresponding terms gives the same value every time:
2S = n(a+l)
Hence:
S = n/2(a+l)
Worked Example
Find the sum of the first 30 terms of:
7, 11, 15, 19, ...
a = 7
d = 4
n = 30
S₃₀ = 30/2 [2(7) + 29(4)]
= 15(14 + 116)
= 15(130)
= 1950
11. Arithmetic Series — Finding Unknowns
Finding n from a Sum
Suppose:
a = 3, d = 4, Sₙ = 465
Use:
465 = n/2 [6 + 4(n−1)]
465 = n/2(4n + 2)
465 = n(2n + 1)
2n² + n − 465 = 0
Factor:
(2n + 31)(n − 15) = 0
n = 15
or
n = −31/2
Since n represents the number of terms:
n = 15
Finding the First Term and Difference
If enough independent conditions are given, form simultaneous equations.
Do not assume that a sequence is determined from one condition alone.
Problem-solving principle:
Count unknowns and count independent equations. This tells you whether the
information is sufficient to determine the sequence uniquely.
12. Geometric Series
A geometric series is the sum of terms of a geometric sequence.
a + ar + ar² + ... + arⁿ⁻¹
For r ≠ 1:
Sₙ = a(1 − rⁿ)/(1 − r)
An equivalent form is:
Sₙ = a(rⁿ − 1)/(r − 1)
Both formulas give the same result.
Deriving the Formula
Start with:
Sₙ = a + ar + ar² + ... + arⁿ⁻¹
Multiply by r:
rSₙ = ar + ar² + ... + arⁿ
Subtract:
Sₙ − rSₙ = a − arⁿ
Sₙ(1−r) = a(1−rⁿ)
Sₙ = a(1−rⁿ)/(1−r)
Worked Example
2 + 6 + 18 + 54 + 162
a = 2
r = 3
n = 5
S₅ = 2(1−3⁵)/(1−3)
= 2(−242)/(−2)
= 242
13. Geometric Series with Negative or Fractional Ratios
Negative Ratio
10 − 5 + 2.5 − 1.25 + ...
a = 10
r = −1/2
The terms alternate in sign because r is negative.
Fractional Ratio Greater Than 1 in Magnitude
If:
|r| > 1
then the magnitude of the terms grows as n increases.
Fractional Ratio Less Than 1 in Magnitude
If:
|r| < 1
then the terms become progressively smaller in magnitude.
This condition becomes especially important when considering an infinite
geometric series.
Do not confuse:
a finite geometric sum can be calculated for any real r, including |r| ≥ 1.
The condition |r| < 1 is specifically required for convergence of an
infinite geometric series.
14. Infinite Geometric Series
Consider:
a + ar + ar² + ar³ + ...
If the magnitude of r is less than 1, the terms approach zero and the
partial sums approach a finite limit.
|r| < 1
In this case:
S∞ = a/(1−r)
Why Does This Work?
Start with the finite formula:
Sₙ = a(1−rⁿ)/(1−r)
If |r| < 1, then:
rⁿ → 0
as n → ∞
Therefore:
S∞ = a(1−0)/(1−r)
= a/(1−r)
Worked Example
12 + 6 + 3 + 1.5 + ...
a = 12
r = 1/2
S∞ = 12/(1−1/2)
= 12/(1/2)
= 24
When Does an Infinite Geometric Series Not Converge?
If:
|r| ≥ 1
the infinite geometric series does not have a finite sum.
| r |
Behaviour |
Finite S∞? |
| 0 < r < 1 |
Positive terms shrink to zero. |
Yes |
| −1 < r < 0 |
Signs alternate and magnitudes shrink. |
Yes |
| r = 1 |
Terms remain constant. |
No |
| r = −1 |
Terms alternate without shrinking. |
No |
| |r| > 1 |
Magnitudes grow. |
No |
15. Infinite Series — Finding an Unknown Ratio
Suppose an infinite geometric series has first term 15 and sum 20.
S∞ = a/(1−r)
20 = 15/(1−r)
20(1−r) = 15
20 − 20r = 15
20r = 5
r = 1/4
The result satisfies |r| < 1, so the infinite sum is valid.
Always perform the convergence check.
Solving algebraically for r is not enough. An infinite geometric sum requires
|r| < 1.
16. Applications of Geometric Series
Repeated Percentage Change
If a quantity is multiplied by a fixed factor each period, the resulting
values form a geometric sequence.
For example, a machine loses 20% of its value each year. Each year it retains:
80% = 0.8
If its initial value is V:
V, 0.8V, 0.8²V, 0.8³V, ...
This is geometric with:
a = V
r = 0.8
If an infinite total of such quantities is meaningful in the model:
S∞ = V/(1−0.8) = 5V
Model carefully:
Percentage decrease must be converted into a multiplier.
A 20% decrease means multiply by 0.8, not subtract 20.
Repeated Bounces
Suppose a ball is dropped from height H and rebounds to 60% of its previous
height each time.
H, 0.6H, 0.6²H, 0.6³H, ...
The bounce heights form a geometric sequence. Questions may ask for the total
upward distance, downward distance, or total distance travelled.
17. Finite vs Infinite Series
| Feature |
Finite Geometric Series |
Infinite Geometric Series |
| Number of terms |
n terms |
Infinitely many |
| Formula |
Sₙ = a(1−rⁿ)/(1−r) |
S∞ = a/(1−r) |
| Condition |
r ≠ 1 for the displayed formula |
|r| < 1 |
| Can |r| ≥ 1? |
Yes |
No finite sum to infinity |
Exam trigger:
The words “sum to infinity” should immediately make you think
about the condition |r| < 1.
18. Connecting Sigma Notation to Series
Sigma notation can describe an arithmetic series directly.
Σⁿₖ₌₁ [a + (k−1)d]
Expanding gives:
a + (a+d) + (a+2d) + ... + [a+(n−1)d]
Therefore:
Σⁿₖ₌₁ [a + (k−1)d]
=
n/2[2a+(n−1)d]
Worked Example
Σ¹⁵ₖ₌₁ (4k − 1)
The first term is:
4(1) − 1 = 3
The common difference is 4.
S₁₅ = 15/2[2(3)+14(4)]
= 15/2(62)
= 465
19. Manipulating Summations
Many difficult-looking sums become straightforward after separating constants
and powers.
Example
Σₖ₌₁²⁰ (5k² − 3k + 7)
Split:
5Σk² − 3Σk + 7Σ1
Use the standard formulas:
5[20·21·41/6]
− 3[20·21/2]
+ 7(20)
= 5(2870)
− 630
+ 140
= 13860
Changing the Limits
Be careful when the lower limit is not 1.
Σ¹⁰ₖ₌₃ k
=
3 + 4 + 5 + ... + 10
One way to evaluate it is:
Σ¹⁰ₖ₌₁ k − Σ²ₖ₌₁ k
= 55 − 3
= 52
Common error:
Σ¹⁰ₖ₌₃ k is not the same as Σ¹⁰ₖ₌₁ k.
Always respect the lower limit.
20. Working Backwards from an nth-Term Formula
Sometimes a sequence is given by a formula rather than its first few terms.
uₙ = n² + 2n
The first four terms are:
u₁ = 1² + 2(1) = 3
u₂ = 2² + 2(2) = 8
u₃ = 3² + 2(3) = 15
u₄ = 4² + 2(4) = 24
3, 8, 15, 24, ...
Notice that the differences are:
5, 7, 9, ...
The first differences are not constant, so this is not arithmetic.
Second Differences
For a quadratic nth-term formula:
uₙ = an² + bn + c
the second differences are constant.
3, 8, 15, 24
First differences:
5, 7, 9
Second differences:
2, 2
Extension / deeper pattern recognition:
Constant second differences indicate a quadratic sequence. This provides a
powerful method for reconstructing a quadratic nth-term formula from data.
21. Quadratic Sequences
Suppose:
4, 9, 16, 25, 36, ...
The first differences are:
5, 7, 9, 11
The second differences are:
2, 2, 2
Therefore try:
uₙ = An² + Bn + C
Since the second difference is 2A:
2A = 2
A = 1
Therefore:
uₙ = n² + Bn + C
Using n = 1:
1 + B + C = 4
B + C = 3
Using n = 2:
4 + 2B + C = 9
2B + C = 5
Subtract:
B = 2
C = 1
uₙ = n² + 2n + 1 = (n+1)²
Structural insight:
Difference tables are not just a calculation trick. They reveal the degree
of the polynomial behind the sequence.
22. Finding Unknown Terms from Conditions
Arithmetic Example
Three consecutive terms of an arithmetic sequence are:
x − 3, 2x + 1, 5x + 5
For consecutive arithmetic terms:
2(middle) = first + third
2(2x+1) = (x−3)+(5x+5)
4x + 2 = 6x + 2
x = 0
The terms are therefore:
−3, 1, 5
Geometric Example
Three consecutive positive geometric terms are:
x, 6, 36/x
The middle term squared equals the product of the outer terms:
6² = x(36/x)
36 = 36
This condition is automatically satisfied for every non-zero x.
Therefore the information given does not uniquely determine x.
Important reasoning lesson:
An equation being true does not mean that it determines the unknown.
Always check whether the conditions provide enough independent information.
23. Mixed Conditions and Simultaneous Equations
More challenging questions may combine a term condition with a sum condition.
Worked Example
An arithmetic sequence has:
u₄ = 11
S₈ = 76
From u₄:
a + 3d = 11
From the sum:
76 = 8/2 [2a + 7d]
76 = 4(2a+7d)
19 = 2a + 7d
From:
a + 3d = 11
multiply by 2:
2a + 6d = 22
Subtract:
d = −3
Therefore:
a − 9 = 11
a = 20
uₙ = 20 − 3(n−1)
Challenge habit:
Translate every piece of information into an equation before manipulating
anything. This prevents hidden assumptions.
24. Relationships Between Terms and Sums
A useful identity for any sequence is:
Sₙ = u₁ + u₂ + ... + uₙ
Therefore:
uₙ = Sₙ − Sₙ₋₁
This means that if a question gives a formula for the sum of the first n terms,
you can recover the nth term by subtracting consecutive sums.
Worked Example
Suppose:
Sₙ = n² + 3n
Then:
uₙ = Sₙ − Sₙ₋₁
= (n²+3n) − [(n−1)² + 3(n−1)]
= n² + 3n − (n² − 2n + 1 + 3n − 3)
= n² + 3n − (n²+n−2)
= 2n + 2
Therefore:
uₙ = 2n + 2
25. Finding a Sequence from Sₙ
The relationship:
uₙ = Sₙ − Sₙ₋₁
becomes particularly useful when the question provides a polynomial expression
for Sₙ.
Example
Sₙ = 2n² − n
First find:
Sₙ₋₁ = 2(n−1)² − (n−1)
= 2(n²−2n+1) − n + 1
= 2n² − 5n + 3
Therefore:
uₙ = (2n²−n) − (2n²−5n+3)
= 4n − 3
Check n = 1:
u₁ = 1
Also:
S₁ = 2(1)² − 1 = 1
The result is consistent.
26. Sequences and Inequalities
Questions may ask for the first term exceeding or falling below a particular value.
The process is to translate the statement into an inequality.
Arithmetic Example
For:
uₙ = 3n + 4
Find the first term greater than 100.
3n + 4 > 100
3n > 96
n > 32
Since n is a positive integer:
n = 33
Geometric Example
Consider:
uₙ = 200(0.8)ⁿ⁻¹
To find when the sequence falls below 50:
200(0.8)ⁿ⁻¹ < 50
(0.8)ⁿ⁻¹ < 0.25
This type of question may require logarithms if an exact term number is requested.
Extension connection:
Taking logarithms can convert an exponential inequality into a linear inequality
in n. Always consider the fact that ln(0.8) is negative when dividing an inequality.
27. Understanding the Behaviour of a Sequence
Arithmetic Sequence
uₙ = a + (n−1)d
- If d > 0, the sequence increases.
- If d < 0, the sequence decreases.
- If d = 0, the sequence is constant.
Geometric Sequence
uₙ = arⁿ⁻¹
The behaviour depends strongly on both a and r.
- 0 < r < 1: magnitude tends to zero.
- r > 1: magnitude generally grows if a ≠ 0.
- r < 0: signs alternate.
- r = 1: constant sequence.
- r = 0: first term may be non-zero, followed by zeros.
Do not use |r| alone to describe whether a sequence increases.
For example, r = −2 gives alternating signs and is not an increasing sequence,
even though |r| > 1.
28. Parameter Questions
A parameter is an unknown constant whose value must be determined from conditions.
Worked Example
The first three terms of an arithmetic sequence are:
k + 2, 2k + 5, 5k + 8
Since the sequence is arithmetic:
2(2k+5) = (k+2) + (5k+8)
4k + 10 = 6k + 10
k = 0
Therefore the sequence is:
2, 5, 8, ...
Parameter + Sum
Harder questions may then use the value of k to determine a sum or another term.
The important thing is to solve the structural condition first, then continue.
Do not substitute blindly.
First identify what mathematical property the wording imposes:
arithmetic → equal differences; geometric → equal ratios; infinite geometric sum
→ |r| < 1.
29. Show-That and Proof-Style Sequence Questions
In a show-that question, the goal is not merely to obtain a numerical answer.
Your algebra must establish the requested expression logically.
Example
Show that the sum of the first n odd positive integers is n².
The odd numbers form an arithmetic sequence:
1, 3, 5, ..., 2n−1
Here:
a = 1
d = 2
l = 2n−1
Therefore:
Sₙ = n/2[1 + (2n−1)]
= n/2(2n)
= n²
Proof habit:
Do not write “obviously” where an algebraic step is required.
A show-that solution should make the chain from the given information to
the required result visible.
30. Advanced Geometric Reasoning
Three Consecutive Geometric Terms
Three positive consecutive geometric terms can be represented as:
a/r, a, ar
This representation is often more efficient than:
a, ar, ar²
because the middle term is exactly a and the terms are symmetrically arranged
around it.
a² = (a/r)(ar)
Example
Three positive consecutive geometric terms have sum 26 and product 216.
Let them be:
a/r, a, ar
Their product is:
(a/r)(a)(ar) = a³
Therefore:
a³ = 216
a = 6
The sum condition becomes:
6/r + 6 + 6r = 26
Divide by 2:
3/r + 3 + 3r = 13
Multiply by r:
3 + 3r + 3r² = 13r
3r² − 10r + 3 = 0
(3r−1)(r−3)=0
r = 1/3
or
r = 3
Both produce the same three terms in reverse order:
18, 6, 2
or
2, 6, 18
Deep insight:
Reversing a geometric sequence replaces r by 1/r.
The same three terms may therefore correspond to two reciprocal ratios
depending on which direction the sequence is listed.
31. Choosing the Correct Tool
A major part of examination success is not memorising formulas independently,
but recognising which structure is present.
| If you see... |
Think... |
| Constant difference |
Arithmetic sequence |
| Constant ratio |
Geometric sequence |
| Sum of first n terms |
Series formula |
| “Sum to infinity” |
S∞ and |r| < 1 |
| Σ notation |
Expand, split, or use standard sums |
| Two known terms |
Use their term positions to eliminate unknowns |
| “First term greater than...” |
Solve an inequality in n |
| Given Sₙ formula |
Use uₙ = Sₙ − Sₙ₋₁ |
| Constant second difference |
Quadratic sequence |
Decision rule:
Identify the mathematical structure before selecting a formula.
Formula recognition should come after structural recognition.
32. Common Traps & Misconceptions
Trap 1 — Using nd Instead of (n−1)d
Correct:
uₙ = a + (n−1)d
The nth term has n−1 jumps from the first term.
Trap 2 — Confusing Difference and Ratio
For:
4, 8, 16, 32, ...
the common difference is not 4. The common ratio is 2.
Trap 3 — Forgetting That n Is an Integer
A sequence term number cannot normally be 7.4 or −3.
Trap 4 — Applying S∞ Without Checking |r| < 1
The formula:
a/(1−r)
is not a universal formula for every infinite geometric expression.
Trap 5 — Sign Errors with Negative r
If r = −2, then:
r² = 4
r³ = −8
r⁴ = 16
Keep brackets around negative ratios when calculating powers.
Trap 6 — Assuming a Pattern Is Unique
A finite list of terms can often be fitted by many different formulas.
In examination questions, the intended structure is normally signalled by
additional information such as “arithmetic”, “geometric”, or a recurrence relation.
Trap 7 — Not Checking the Answer in the Original Conditions
Especially in parameter questions, substitute your result back into the
condition that defined it.
33. Edexcel-Style Exam Strategy
Step 1 — Identify the Object
Ask:
- Is this a sequence or a series?
- Is it arithmetic or geometric?
- Is it finite or infinite?
- Is sigma notation involved?
Step 2 — Extract the Parameters
Arithmetic → a, d, n, l
Geometric → a, r, n
Step 3 — Translate Words into Mathematics
| Wording |
Mathematical translation |
| nth term |
uₙ |
| first n terms |
Sₙ |
| sum to infinity |
S∞ |
| consecutive arithmetic terms |
equal differences |
| consecutive geometric terms |
equal ratios |
| first term exceeding X |
solve uₙ > X |
Step 4 — Check Restrictions
- Is n a positive integer?
- For S∞, is |r| < 1?
- Did a negative ratio require careful brackets?
- Does the result satisfy every condition?
Step 5 — Check the Scale
If ten positive terms are each around 100, a sum of 20 is immediately suspicious.
Estimation can catch arithmetic mistakes before they cost marks.
34. Complete Sequences & Series Formula Atlas
Arithmetic Sequence
uₙ = a + (n−1)d
Arithmetic Series
Sₙ = n/2[2a+(n−1)d]
Sₙ = n/2(a+l)
Geometric Sequence
uₙ = arⁿ⁻¹
Finite Geometric Series
Sₙ = a(1−rⁿ)/(1−r)
r ≠ 1
Infinite Geometric Series
S∞ = a/(1−r)
|r| < 1
Standard Sums
Σₙ₌₁ᴺ 1 = N
Σₙ₌₁ᴺ n = N(N+1)/2
Σₙ₌₁ᴺ n² = N(N+1)(2N+1)/6
Σₙ₌₁ᴺ n³ = [N(N+1)/2]²
Sum-to-Term Relationship
uₙ = Sₙ − Sₙ₋₁
Arithmetic Mean
2b = a+c
Geometric Mean
b² = ac
35. Deep Worked Examples
Example A — Two Conditions in an Arithmetic Sequence
The 7th term of an arithmetic sequence is 25 and the 15th term is 57.
Find the first term and common difference.
a + 6d = 25
a + 14d = 57
8d = 32
d = 4
a + 24 = 25
a = 1
uₙ = 1 + 4(n−1) = 4n−3
Example B — Geometric Sequence from Two Terms
A geometric sequence has u₃ = 20 and u₆ = 160. Find a and r,
assuming r is positive.
ar² = 20
ar⁵ = 160
Divide:
r³ = 160/20 = 8
r = 2
4a = 20
a = 5
uₙ = 5 × 2ⁿ⁻¹
Example C — Infinite Series with a Condition
An infinite geometric series has first term 12 and sum 15.
Find r.
15 = 12/(1−r)
15 − 15r = 12
15r = 3
r = 1/5
Since:
|1/5| < 1
the result is valid.
Example D — Find a Term from a Sum Formula
If:
Sₙ = 3n² + 2n
find uₙ.
Sₙ₋₁ = 3(n−1)² + 2(n−1)
= 3n² − 6n + 3 + 2n − 2
= 3n² − 4n + 1
uₙ = Sₙ − Sₙ₋₁
= (3n²+2n) − (3n²−4n+1)
= 6n−1
36. Very Hard Multi-Step Reasoning
Problem
The first term of a geometric sequence is positive. The sum of the first
three terms is 21 and the sum to infinity is 24. Find the first term and ratio.
Let the first term be a and ratio r.
a(1+r+r²) = 21
Also:
a/(1−r) = 24
Therefore:
a = 24(1−r)
Substitute:
24(1−r)(1+r+r²) = 21
Use:
(1−r)(1+r+r²) = 1−r³
24(1−r³) = 21
1−r³ = 7/8
r³ = 1/8
r = 1/2
Therefore:
a = 24(1−1/2)
= 12
Check:
12 + 6 + 3 = 21
S∞ = 12/(1−1/2) = 24
Why this is powerful:
The problem combines finite and infinite geometric sums.
The key is to express both conditions using the same unknowns and then exploit
the algebraic identity 1+r+r².
37. Real-World Mathematical Models
Salary / Payment Pattern
If a payment increases by a fixed amount each period, use an arithmetic model.
Pₙ = P₁ + (n−1)d
If it increases by a fixed percentage, use a geometric model.
Pₙ = P₁rⁿ⁻¹
Population or Growth Model
A fixed percentage growth per period produces:
P, Pr, Pr², Pr³, ...
Depreciation
A fixed percentage decrease gives a ratio below 1.
r = 1 − percentage decrease
Repeated Physical Processes
Repeated rebounds, repeated reductions, repeated attenuation and similar
processes can generate geometric sequences and series.
Modelling rule:
The mathematics depends on the wording. “Increase by 5 each year” is additive;
“increase by 5% each year” is multiplicative. These produce different sequences.
38. Practice Set — Foundation to Advanced
Question 1 — Arithmetic
Find the nth term of 8, 13, 18, 23, ...
Show Solution
a = 8, d = 5
uₙ = 8 + 5(n−1)
= 5n+3
Question 2 — Geometric
Find the nth term of 4, 12, 36, 108, ...
Show Solution
a = 4, r = 3
uₙ = 4×3ⁿ⁻¹
Question 3 — Arithmetic Sum
Find the sum of the first 25 terms of 2, 6, 10, 14, ...
Show Solution
a = 2, d = 4, n = 25
S₂₅ = 25/2[4+96]
= 1250
Question 4 — Finite Geometric Sum
Find the sum of the first 7 terms of 3, 6, 12, ...
Show Solution
S₇ = 3(1−2⁷)/(1−2)
= 381
Question 5 — Infinite Sum
Find the sum to infinity of 18 + 9 + 4.5 + ...
Show Solution
a = 18, r = 1/2
S∞ = 18/(1−1/2)
= 36
Question 6 — Sigma
Evaluate Σ⁸ₙ₌₁(2n+1).
Show Solution
2Σn + Σ1
= 2(36) + 8
= 80
39. Master Checklist Before the Exam
-
I can distinguish a sequence from a series.
-
I can identify an arithmetic sequence from a constant difference.
-
I can identify a geometric sequence from a constant ratio.
-
I can find the nth term of an arithmetic sequence.
-
I can find the nth term of a geometric sequence.
-
I can find unknown a and d from two arithmetic terms.
-
I can find unknown a and r from two geometric terms.
-
I can evaluate arithmetic sums.
-
I can evaluate finite geometric sums.
-
I can use the infinite geometric formula and check |r| < 1.
-
I can manipulate sigma notation.
-
I know the standard sums for 1, n, n² and n³.
-
I can recover uₙ from Sₙ.
-
I can solve parameter problems.
-
I can handle negative and fractional common ratios.
-
I can solve term-number inequalities.
-
I can recognise quadratic sequences from second differences.
-
I check answers against the original conditions.
Structure → Formula → Algebra → Check
That four-stage process is the core workflow for solving sequences and series
problems reliably.