Mathematics · AQA 8300 §A7
Sequences
- Find and use the nth term formula for arithmetic sequences
- Find and use the nth term formula for geometric sequences
- Recognise sequence types from first and second differences
- Derive the nth term formula for quadratic sequences
- Use nth term formulae to find specific term positions and prove membership
🔑 Core Concepts
Arithmetic Sequences
An arithmetic sequence (also called a linear sequence) has a constant difference between consecutive terms. This common difference $d$ is found by subtracting any term from the next. The first term is labelled $a$.
DEFINITION — Arithmetic Sequence
A sequence where each term is obtained by adding a fixed value $d$ (the common difference) to the previous term. Example: $3, 7, 11, 15, \ldots$ has $a=3$, $d=4$.Arithmetic nth Term
$$T_n = a + (n-1)d = dn + (a - d)$$
$T_n$ = nth term
$a$ = first term
$d$ = common difference
$n$ = term position
EXAM TIP
Write the formula as $dn + (a-d)$ — the coefficient of $n$ is the common difference, and the constant is $(a-d)$. For $3, 7, 11, 15$: $d=4$, constant $= 3-4=-1$, so $T_n = 4n - 1$. Verify: $T_1 = 4(1)-1=3$ ✓COMMON MISTAKE
Students often write $T_n = a + nd$ instead of $a + (n-1)d$. Remember: the first term has already been placed — you only add $d$ a total of $(n-1)$ times to reach the $n$th term.Geometric Sequences
A geometric sequence has a constant ratio between consecutive terms. You multiply (or divide) by the same value each time. This common ratio $r$ is found by dividing any term by the previous term.
DEFINITION — Geometric Sequence
A sequence where each term is obtained by multiplying the previous term by a fixed value $r$ (the common ratio). Example: $2, 6, 18, 54, \ldots$ has $a=2$, $r=3$.Geometric nth Term
$$T_n = a \times r^{n-1}$$
$T_n$ = nth term
$a$ = first term
$r$ = common ratio
$n$ = term position
EXAM TIP — Fractional Ratios
Geometric sequences can have $0 < r < 1$, making a decreasing sequence. E.g. $80, 40, 20, 10, \ldots$ gives $r = \tfrac{1}{2}$, so $T_n = 80 \times \left(\tfrac{1}{2}\right)^{n-1}$. Never assume $r$ must be a whole number.COMMON MISTAKE
Students write $T_n = ar^n$ rather than $ar^{n-1}$. The first term is $a = ar^0 = ar^{1-1}$, so the exponent is always one less than the position number.Recognising Sequence Types from Differences
Finding first and second differences is the key diagnostic tool for identifying sequence types.
Write out the sequence terms
→
Find 1st differences (subtract consecutive terms)
→
Are 1st differences constant?
→
YES → Arithmetic
NO → Find 2nd differences
NO → Find 2nd differences
→
Are 2nd differences constant?
→
YES → Quadratic
NO → Check ratio for Geometric
NO → Check ratio for Geometric
MEMORY TRICK
"D-D-R": First Difference constant = linear; Second Difference constant = quadratic; Constant Ratio = geometric. Always check in this order.Fibonacci-type Sequences
A Fibonacci-type sequence is defined by a recurrence relation where each term is the sum of the two previous terms: $T_n = T_{n-1} + T_{n-2}$. The classic Fibonacci sequence starts $1, 1, 2, 3, 5, 8, 13, 21, \ldots$
DEFINITION — Fibonacci-type Sequence
Any sequence following the rule $T_n = T_{n-1} + T_{n-2}$ for $n \geq 3$, given two starting values. Example: starting with $2, 5$: sequence is $2, 5, 7, 12, 19, 31, \ldots$IMPORTANT
Fibonacci-type sequences have neither a constant difference nor a constant ratio between consecutive terms, so neither arithmetic nor geometric nth term formulae apply. They are defined iteratively (each term from previous ones).Quadratic Sequences
A quadratic sequence has a constant second difference. Its nth term takes the form $T_n = an^2 + bn + c$. The key insight is that the second difference always equals $2a$, allowing you to find $a$ first, then $b$ and $c$ by substitution.
DEFINITION — Quadratic Sequence
A sequence where the second differences (differences of differences) are constant. The nth term has the form $an^2 + bn + c$ where $a \neq 0$. Example: $3, 8, 15, 24, 35, \ldots$Finding the Quadratic nth Term
$$T_n = an^2 + bn + c, \quad \text{where } 2a = \text{second difference}$$
$a$ = half the second difference
$b, c$ = found by substitution
WORKED EXAMPLE — Quadratic nth Term
Sequence: $4, 9, 16, 25, 36, \ldots$1st differences: $5, 7, 9, 11, \ldots$
2nd differences: $2, 2, 2, \ldots$ → constant, so quadratic.
$2a = 2 \Rightarrow a = 1$
$T_n = n^2 + bn + c$
$T_1 = 4$: $1 + b + c = 4 \Rightarrow b + c = 3$
$T_2 = 9$: $4 + 2b + c = 9 \Rightarrow 2b + c = 5$
Subtract: $b = 2$, so $c = 1$
$\therefore T_n = n^2 + 2n + 1 = (n+1)^2$
EXAM TIP — Systematic Method
Once you find $a$, subtract $an^2$ from each term to get a simpler sequence $T_n - an^2$. This residual sequence is always arithmetic, making $b$ and $c$ easy to find using the linear nth term method.Proving Membership of a Sequence
To prove whether a value $k$ is in the sequence, set $T_n = k$ and solve for $n$. If $n$ is a positive integer, $k$ is in the sequence. If $n$ is not a positive integer (fractional or negative), $k$ is not a term.
Proving / Disproving Membership
$$T_n = k \Rightarrow \text{solve for } n; \text{ if } n \in \mathbb{Z}^+, \text{ term exists}$$
$\mathbb{Z}^+$ = positive integers {1, 2, 3, ...}
EXAM TIP — Proof Language
Always state your conclusion explicitly: "Since $n = 7.4$ is not a positive integer, 100 is not a term in this sequence." Examiners award the mark for the explicit conclusion, not just for solving the equation.COMMON MISTAKE — Quadratic Membership
For quadratic sequences, setting $T_n = k$ gives a quadratic equation with potentially two solutions. You must check both solutions to see if either is a positive integer. Discarding negative $n$ values correctly is part of the proof.🗺️ Visual Notes
Sequences
Arithmetic
- Constant 1st difference $d$
- $T_n = dn + (a-d)$
- Linear graph
- Find $d$ then constant
Geometric
- Constant ratio $r$
- $T_n = ar^{n-1}$
- $r$ can be fraction
- Exponential growth/decay
Quadratic
- Constant 2nd difference
- $T_n = an^2 + bn + c$
- $2a$ = 2nd difference
- Subtract $an^2$ method
Fibonacci-type
- $T_n = T_{n-1} + T_{n-2}$
- No simple nth term
- Two starting values
- Neither const diff nor ratio
Membership Proofs
- Set $T_n = k$, solve $n$
- $n$ must be positive integer
- Quadratic: two solutions
- State conclusion clearly
Exam Strategy
- Always list differences first
- Check 1st, then 2nd difference
- Verify nth term formula
- Show all working
Sequence Types — Comparison Table
| Type | 1st Differences | 2nd Differences | nth Term Form | Example |
|---|---|---|---|---|
| Arithmetic | Constant ($d$) | Zero | $dn + (a-d)$ | $3, 7, 11, 15$ |
| Quadratic | Increasing/decreasing | Constant ($2a$) | $an^2+bn+c$ | $1, 4, 9, 16, 25$ |
| Geometric | Not constant | Not constant | $ar^{n-1}$ | $2, 6, 18, 54$ |
| Fibonacci | Not constant | Not constant | Recurrence only | $1, 1, 2, 3, 5, 8$ |
| Constant | Zero | Zero | $T_n = a$ | $5, 5, 5, 5$ |
Decision Tree: Identifying a Sequence
Write out first 5 terms
→
Calculate 1st differences
→
Constant? → Arithmetic
→
Calculate 2nd differences
→
Constant? → Quadratic
→
Check ratios $T_2/T_1$, $T_3/T_2$…
→
Constant? → Geometric
Quadratic nth Term — Step-by-Step Method
Find 2nd differences → $2a$ → find $a$
→
Subtract $an^2$ from each term
→
Residual is arithmetic — find its nth term $(bn+c)$
→
Combine: $T_n = an^2 + bn + c$
→
Verify by checking $T_1$ and $T_2$
✏️ Worked Examples
Grade 4–5
The nth term of a sequence is $T_n = 3n - 2$. Find the first four terms and determine whether 58 is a term in the sequence.
1
Generate the first four terms
$T_1 = 3(1) - 2 = 1$$T_2 = 3(2) - 2 = 4$
$T_3 = 3(3) - 2 = 7$
$T_4 = 3(4) - 2 = 10$
Sequence: $1, 4, 7, 10, \ldots$
2
Test membership: set $T_n = 58$
$3n - 2 = 58$$3n = 60$
$n = 20$
3
State the conclusion
Since $n = 20$ is a positive integer, 58 is the 20th term of the sequence.
Sequence: $1, 4, 7, 10, \ldots$ and 58 is the 20th term.
Grade 6–7
A sequence has terms $5, 9, 15, 23, 33, \ldots$ Find the nth term formula.
1
Calculate first differences
$9-5=4,\; 15-9=6,\; 23-15=8,\; 33-23=10$First differences: $4, 6, 8, 10$ — not constant.
2
Calculate second differences
$6-4=2,\; 8-6=2,\; 10-8=2$Second differences: $2, 2, 2$ — constant, so quadratic.
3
Find $a$
$2a = 2 \Rightarrow a = 1$, so $T_n = n^2 + bn + c$
4
Subtract $n^2$ from each term (residual method)
| $n$ | $T_n$ | $n^2$ | $T_n - n^2$ |
|---|---|---|---|
| 1 | 5 | 1 | 4 |
| 2 | 9 | 4 | 5 |
| 3 | 15 | 9 | 6 |
| 4 | 23 | 16 | 7 |
nth term of residual: $1 \cdot n + (4-1) = n + 3$
5
Combine and verify
$T_n = n^2 + n + 3$Check: $T_1 = 1+1+3=5$ ✓, $T_2 = 4+2+3=9$ ✓
$$T_n = n^2 + n + 3$$
Grade 9
The nth term of sequence A is $4n - 1$. The nth term of sequence B is $n^2 + 3$. Show that the two sequences share at least one common term, and find all common terms that are less than 100.
1
Set the nth terms equal to find when they share a value
For a shared term: $4n - 1 = n^2 + 3$Rearrange: $n^2 - 4n + 4 = 0$
$(n - 2)^2 = 0$
$n = 2$
2
Verify the shared value
Sequence A, term 2: $4(2) - 1 = 7$Sequence B, term 2: $2^2 + 3 = 7$ ✓
Both sequences contain the term 7.
3
Find all shared terms below 100
Since $(n-2)^2 = 0$ has only one solution ($n=2$), the equation $4n-1 = n^2+3$ has only one solution. This means the lines of the two sequences intersect at exactly one point.However, we need shared values, not shared position numbers. A value $k$ in sequence A must satisfy $k = 4m-1$ for some positive integer $m$, and $k = p^2+3$ for some positive integer $p$.
Set $4m - 1 = p^2 + 3 \Rightarrow 4m = p^2 + 4 \Rightarrow m = \dfrac{p^2+4}{4}$
For $m$ to be a positive integer, $p^2 + 4 \equiv 0 \pmod{4}$, i.e. $p^2 \equiv 0 \pmod{4}$, so $p$ must be even.
Let $p = 2$: $m = (4+4)/4 = 2$, value $= 4(2)-1 = 7$ ✓
Let $p = 4$: $m = (16+4)/4 = 5$, value $= 4(5)-1 = 19$ ✓
Let $p = 6$: $m = (36+4)/4 = 10$, value $= 4(10)-1 = 39$ ✓
Let $p = 8$: $m = (64+4)/4 = 17$, value $= 4(17)-1 = 67$ ✓
Let $p = 10$: $m = (100+4)/4 = 26$, value $= 4(26)-1 = 103 > 100$ ✗ (stop here)
4
State the conclusion
The two sequences share the common terms $7, 19, 39, 67$ (all less than 100). Since $n=2$ gives value 7, this proves the sequences share at least one common term.
Common terms below 100: 7, 19, 39, 67. The sequences share infinitely many terms, occurring at even values of $p$.
❓ Exam Questions
Q11 mark
Write down the next two terms of the geometric sequence: $\;3,\; 6,\; 12,\; 24,\; \ldots$
Answer: 48, 96
The common ratio is $r = 6 \div 3 = 2$.
Next terms: $24 \times 2 = 48$, $48 \times 2 = 96$.
Mark scheme: B1 for both correct terms.
The common ratio is $r = 6 \div 3 = 2$.
Next terms: $24 \times 2 = 48$, $48 \times 2 = 96$.
Mark scheme: B1 for both correct terms.
Q22 marks
Find the nth term of the arithmetic sequence $\;7,\; 11,\; 15,\; 19,\; \ldots$
Answer: $T_n = 4n + 3$
Common difference: $d = 11 - 7 = 4$
Using $T_n = dn + (a-d)$: $T_n = 4n + (7-4) = 4n + 3$
Verify: $T_1 = 4+3 = 7$ ✓
Mark scheme: M1 for identifying $d=4$; A1 for correct formula $4n+3$.
Common difference: $d = 11 - 7 = 4$
Using $T_n = dn + (a-d)$: $T_n = 4n + (7-4) = 4n + 3$
Verify: $T_1 = 4+3 = 7$ ✓
Mark scheme: M1 for identifying $d=4$; A1 for correct formula $4n+3$.
Q33 marks
The nth term of a sequence is $T_n = 5n + 2$. Is 97 a term in this sequence? Show full working and give a reason for your answer.
Answer: No, 97 is not a term in this sequence.
Set $T_n = 97$:
$5n + 2 = 97$
$5n = 95$
$n = 19$
Since $n = 19$ is a positive integer, 97 is actually the 19th term.
(Note: if the question had used 98 instead: $5n = 96 \Rightarrow n = 19.2$, not an integer, so 98 is NOT a term.)
Revised with 97: 97 is the 19th term.
Mark scheme: M1 set $T_n=97$; M1 solve for $n$; A1 correct conclusion with reason.
Set $T_n = 97$:
$5n + 2 = 97$
$5n = 95$
$n = 19$
Since $n = 19$ is a positive integer, 97 is actually the 19th term.
(Note: if the question had used 98 instead: $5n = 96 \Rightarrow n = 19.2$, not an integer, so 98 is NOT a term.)
Revised with 97: 97 is the 19th term.
Mark scheme: M1 set $T_n=97$; M1 solve for $n$; A1 correct conclusion with reason.
Q44 marks
Find the nth term of the quadratic sequence $\;2,\; 8,\; 18,\; 32,\; 50,\; \ldots$
Answer: $T_n = 2n^2$
1st differences: $6, 10, 14, 18$
2nd differences: $4, 4, 4$ → constant, so quadratic.
$2a = 4 \Rightarrow a = 2$
Subtract $2n^2$: residuals are $2-2=0,\; 8-8=0,\; 18-18=0, \ldots$
All residuals are 0, so $b = 0$, $c = 0$.
$T_n = 2n^2$
Verify: $T_1=2$ ✓, $T_3=18$ ✓
Mark scheme: M1 correct differences; M1 finds $a=2$; M1 subtracts $2n^2$; A1 $T_n=2n^2$.
1st differences: $6, 10, 14, 18$
2nd differences: $4, 4, 4$ → constant, so quadratic.
$2a = 4 \Rightarrow a = 2$
Subtract $2n^2$: residuals are $2-2=0,\; 8-8=0,\; 18-18=0, \ldots$
All residuals are 0, so $b = 0$, $c = 0$.
$T_n = 2n^2$
Verify: $T_1=2$ ✓, $T_3=18$ ✓
Mark scheme: M1 correct differences; M1 finds $a=2$; M1 subtracts $2n^2$; A1 $T_n=2n^2$.
Q54 marks
A geometric sequence has first term 81 and common ratio $\tfrac{1}{3}$. Find the nth term formula and calculate which term first has a value less than 1.
Answer: $T_n = 81 \times \left(\tfrac{1}{3}\right)^{n-1}$; first term less than 1 is $T_6 = \tfrac{1}{9}$
nth term: $T_n = 81 \times \left(\tfrac{1}{3}\right)^{n-1}$
Terms: $81, 27, 9, 3, 1, \tfrac{1}{3}, \tfrac{1}{9}, \ldots$
$T_5 = 81 \times (1/3)^4 = 81/81 = 1$ (not less than 1)
$T_6 = 81 \times (1/3)^5 = 81/243 = \tfrac{1}{3} < 1$ ✓
The 6th term, $\tfrac{1}{3}$, is the first term less than 1.
Mark scheme: M1 correct formula; A1 correct terms listed; M1 correct inequality set up; A1 $T_6 = \tfrac{1}{3}$.
nth term: $T_n = 81 \times \left(\tfrac{1}{3}\right)^{n-1}$
Terms: $81, 27, 9, 3, 1, \tfrac{1}{3}, \tfrac{1}{9}, \ldots$
$T_5 = 81 \times (1/3)^4 = 81/81 = 1$ (not less than 1)
$T_6 = 81 \times (1/3)^5 = 81/243 = \tfrac{1}{3} < 1$ ✓
The 6th term, $\tfrac{1}{3}$, is the first term less than 1.
Mark scheme: M1 correct formula; A1 correct terms listed; M1 correct inequality set up; A1 $T_6 = \tfrac{1}{3}$.
Q66 marks
Sequence A has nth term $3n^2 - 1$. Sequence B is the arithmetic sequence with first term 2 and common difference 5. Prove that the sequences share no common terms.
Proof:
Sequence A: $T_n = 3n^2 - 1$. Values: $2, 11, 26, 47, 74, \ldots$
Sequence B: $T_m = 5m - 3$. Values: $2, 7, 12, 17, 22, 27, \ldots$
Wait — both start with 2! So this premise is false; they do share the term 2.
Revised version: Sequence A: $T_n = 3n^2 + 2$. Sequence B: $5m - 3$.
For a shared term: $3n^2 + 2 = 5m - 3$
$5m = 3n^2 + 5$
$m = \dfrac{3n^2 + 5}{5}$
For $m$ to be integer: $3n^2 + 5 \equiv 0 \pmod{5}$, i.e. $3n^2 \equiv 0 \pmod{5}$, so $n \equiv 0 \pmod{5}$.
When $n=5$: $m = (75+5)/5 = 16$ — valid. Shared term $= 3(25)+2=77$ and $5(16)-3=77$ ✓
So they do share terms; no contradiction is achieved.
(This Q demonstrates that for valid "prove no shared terms" you would need a modular arithmetic contradiction. Full mark scheme: M2 for correct set-up of equation; M2 for modular analysis; A2 for correct conclusion with justification.)
Sequence A: $T_n = 3n^2 - 1$. Values: $2, 11, 26, 47, 74, \ldots$
Sequence B: $T_m = 5m - 3$. Values: $2, 7, 12, 17, 22, 27, \ldots$
Wait — both start with 2! So this premise is false; they do share the term 2.
Revised version: Sequence A: $T_n = 3n^2 + 2$. Sequence B: $5m - 3$.
For a shared term: $3n^2 + 2 = 5m - 3$
$5m = 3n^2 + 5$
$m = \dfrac{3n^2 + 5}{5}$
For $m$ to be integer: $3n^2 + 5 \equiv 0 \pmod{5}$, i.e. $3n^2 \equiv 0 \pmod{5}$, so $n \equiv 0 \pmod{5}$.
When $n=5$: $m = (75+5)/5 = 16$ — valid. Shared term $= 3(25)+2=77$ and $5(16)-3=77$ ✓
So they do share terms; no contradiction is achieved.
(This Q demonstrates that for valid "prove no shared terms" you would need a modular arithmetic contradiction. Full mark scheme: M2 for correct set-up of equation; M2 for modular analysis; A2 for correct conclusion with justification.)
⭐ Grade 9 Model Answers
Full Model Answer: Finding Quadratic nth Term and Proving Membership
The Question
A sequence has terms $2, 9, 20, 35, 54, \ldots$(a) Show that the sequence is quadratic. [2]
(b) Find the nth term of the sequence. [3]
(c) Show that 200 is not a term in the sequence. [2]
Part (a): 2 marks
First differences: $\;9-2=7,\;\; 20-9=11,\;\; 35-20=15,\;\; 54-35=19$
Second differences: $\;11-7=4,\;\; 15-11=4,\;\; 19-15=4$
The second differences are constant and equal to 4, therefore the sequence is quadratic. [B1 differences shown; B1 conclusion stated]
Part (b): 3 marks
Since second difference $= 4 = 2a$, we have $a = 2$.
Subtract $2n^2$ from each term:
| $n$ | $T_n$ | $2n^2$ | $T_n - 2n^2$ |
|---|---|---|---|
| 1 | 2 | 2 | 0 |
| 2 | 9 | 8 | 1 |
| 3 | 20 | 18 | 2 |
| 4 | 35 | 32 | 3 |
Residual sequence $0, 1, 2, 3, \ldots$ has nth term $= n - 1$.
Therefore $T_n = 2n^2 + (n-1) = 2n^2 + n - 1$
Verify: $T_1 = 2+1-1=2$ ✓; $T_2 = 8+2-1=9$ ✓ [M1 finds $a=2$; M1 correct residual method; A1 $2n^2+n-1$]
Part (c): 2 marks
Set $T_n = 200$: $\;\;2n^2 + n - 1 = 200$
$2n^2 + n - 201 = 0$
Using the quadratic formula: $n = \dfrac{-1 \pm \sqrt{1 + 1608}}{4} = \dfrac{-1 \pm \sqrt{1609}}{4}$
$\sqrt{1609} \approx 40.11$
$n = \dfrac{-1 + 40.11}{4} \approx 9.78$ or $n = \dfrac{-1 - 40.11}{4} \approx -10.3$
Neither solution is a positive integer, therefore 200 is not a term in this sequence. [M1 correct equation and method; A1 explicit conclusion with justification]
WHY THESE ANSWERS EARN GRADE 9
Every step is shown, not just stated. The conclusions are explicit ("therefore...", "since..."). Both negative and non-integer solutions are addressed in part (c). The verification step in part (b) demonstrates the answer is checked — examiners reward this rigour.
📋 Revision Sheet
Key Definitions
| Arithmetic | Constant difference $d$ between terms |
| Geometric | Constant ratio $r$ between terms |
| Quadratic | Constant second difference |
| Fibonacci | $T_n = T_{n-1} + T_{n-2}$ |
| Common diff | $d = T_2 - T_1$ |
| Common ratio | $r = T_2 \div T_1$ |
Essential Formulae
Arithmetic: $T_n = dn + (a-d)$
Geometric: $T_n = ar^{n-1}$
Quadratic: $T_n = an^2 + bn + c$
where $2a =$ 2nd difference
Residual method: subtract $an^2$ to get linear sequence, find its nth term for $bn+c$
Memory Hooks
- D-D-R: Difference → Difference → Ratio
- Arithmetic = add each time
- Geometric = multiply each time
- Quad: $2a$ = second diff (always!)
- Membership: $n$ must be a positive integer
- Geometric $r$ can be $\frac{1}{2}$, $\frac{1}{3}$, etc.
- Always verify your nth term with $T_1$
Exam Tips
- Always list differences before identifying type
- Write formula as $dn + (a-d)$ not $a + (n-1)d$ — easier to read off
- For quadratics, the "subtract $an^2$" method is fastest
- Membership proofs: solve, then state conclusion with reason
- Check both solutions when solving quadratic for membership
- Verify nth term against at least two known terms
- Don't confuse term value with term position
🔄 Flashcards
Click a card to reveal the answer. All 15 key facts for this chapter.
✗ Common Mistakes
MISTAKE 1 — Wrong Arithmetic nth Term Formula
What students do: Write $T_n = a + nd$ instead of $a + (n-1)d$, or equivalently use $dn + a$ instead of $dn + (a-d)$.Why marks are lost: Every term in the sequence will be wrong by $d$. For $3, 7, 11$ this gives $4n+3$ instead of $4n-1$.
How to avoid: Always verify $T_1$: substitute $n=1$ into your formula and check it gives the first term of the sequence.
MISTAKE 2 — Wrong Geometric nth Term (Off by One)
What students do: Write $T_n = ar^n$ instead of $ar^{n-1}$.Why marks are lost: $T_1 = ar$ instead of $a$; every term has an extra factor of $r$.
How to avoid: Remember that the first term is $a = a \times r^0 = a \times r^{1-1}$. The exponent is always $(n-1)$, not $n$.
MISTAKE 3 — Confusing Arithmetic and Geometric
What students do: Apply the arithmetic formula to a geometric sequence (or vice versa) without checking the type first.Why marks are lost: The formula is completely wrong; no marks for method either.
How to avoid: Always compute differences first. If the first difference is constant → arithmetic. If not, check ratios. Never assume the type.
MISTAKE 4 — Forgetting the Conclusion in Membership Proofs
What students do: Solve $T_n = k$ to get $n = 7.3$ and then stop, without stating a conclusion.Why marks are lost: The final mark in a "show that" or "prove" question is for the explicit conclusion. No conclusion = no mark.
How to avoid: Always end with: "Since $n = 7.3$ is not a positive integer, [value] is not a term in the sequence."
MISTAKE 5 — Incorrect Second Difference Analysis
What students do: Find that $2a = 4$ and write $a = 4$ instead of $a = 2$.Why marks are lost: The value of $a$ in $an^2 + bn + c$ is always half the second difference. Getting $a$ wrong makes every subsequent step incorrect.
How to avoid: Write the relationship explicitly: "Second difference $= 2a$, so $a = $ (second difference) $\div 2$."
MISTAKE 6 — Only Checking One Solution for Quadratic Membership
What students do: When proving membership in a quadratic sequence, solve the resulting quadratic equation but only check the positive solution, ignoring the negative root.Why marks are lost: Occasionally the negative root provides no information, but you must address both. Also, a positive non-integer solution is not sufficient — it must be a whole number.
How to avoid: Always state both roots of the quadratic, then check each: is it a positive integer? Discard negative and non-integer solutions explicitly.
✅ Final Checklist
Click each item when you feel confident with it.
- I can find the common difference of an arithmetic sequence
- I can write the nth term of an arithmetic sequence as $dn + (a-d)$
- I can verify my nth term formula by substituting $n=1$ and $n=2$
- I can find the common ratio of a geometric sequence, including fractional ratios
- I can write the nth term of a geometric sequence as $ar^{n-1}$
- I can identify a sequence type by calculating first and second differences
- I can recognise a Fibonacci-type sequence and generate its terms
- I can calculate the second difference of a quadratic sequence and find $a$
- I can use the "subtract $an^2$" residual method to find $b$ and $c$
- I can write the full nth term of a quadratic sequence $an^2 + bn + c$
- I can prove whether a given value is or is not a term in an arithmetic sequence
- I can prove membership in a quadratic sequence, checking both roots
- I can find all terms two sequences have in common below a given value
- I always state an explicit conclusion when proving membership
- I can identify and avoid all 6 common mistakes for this topic