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Sequences and nth term rules - Mathematics Year 10

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Introduction

The term value and its position are different quantities. Find a rule that uses the position n, then check it against the stated pattern and a distant term.

Derive five linear nth-term rules and explore two clearly labelled Higher-tier quadratic rules. Use first and second differences, compare structural models and explain why a finite list alone cannot uniquely determine every future term.

Derive five linear rules and distant terms

An arithmetic sequence adds the same difference d each time. For first terma₁ and positionsn=1, 2, 3, …, the nth-term rule is $a_{n} = a_{1} + (n - 1) d .$ There are n −1 transitions between the first position and positionn. Using nd would add one extra step.

For the source 4, 7, 10, 13, the common difference is 3. Start with 3 n, which gives 3, 6, 9, 12 at positions 1 through 4. Each actual value is 1 higher, so the rule is 3 n+1. The first-term formula also gives 4+3(n −1)=3 n+1, providing a second derivation.

Five complete linear rules. In each task, assume the displayed constant-difference pattern continues as an arithmetic sequence and n starts at 1. Cover the last two columns, derive the rule and then find the twentieth term.

Task	First four terms	Common difference	Nth-term rule	Term 20
1	4, 7, 10, 13	3	3 n+1	61
2	9, 14, 19, 24	5	5 n+4	104
3	20, 16, 12, 8	−4	24−4 n	−56
4	1.5, 4, 6.5, 9	2.5	2.5 n −1	49
5	−6, −1, 4, 9	5	5 n −11	89

For row 2, use 9+5(n −1)=5 n+4, so term 20 is 104. In row 3, the difference is −4. Formula 20−4(n −1)=24−4 n gives −56 at n 20. Negative differences and negative terms are legitimate for an abstract sequence; no physical quantity has been specified that must remain nonnegative.

For row 4, the difference 2.5 is not a whole number, but the same derivation works: 1.5+2.5(n −1)=2.5 n −1. At 20, the answer 49 follows. For row 5, the first negative value does not make the difference negative: −6 to −1 to 4 to 9 repeatedly adds 5. Formula −6+5(n −1)=5 n −11 gives 89 at 20.

Check every proposed rule at more than one position. The source 3 n+1 produces 4 at 1, 10 at 3, and 61 at 20. A learner’s 3 n+4 produces 7 at 1 and therefore misses the defined starting term. The first invalid decision was treating the observed first term 4 as the constant added to 3 n without accounting for 3 n already being 3 at the first position.

The nth term is not the previous term. Rule 3 n+1 computes a value directly from a position; “add 3” needs a starting value and describes how to move from one term to the next. A direct formula is efficient for a distant term, while repeated addition can help inspect the local difference pattern.

Higher-tier extension: derive two quadratic rules

A blank hundred-square grid made from equal cells. — Count equal cells rather than measuring pixels. The original rectangle-count schematic below uses3,8 and15 dots;the Commons grid itself is not pre-shaded to represent those answers.

Higher-tier extension: quadratic nth terms. Constant first differences suggest a linear model; constant nonzero second differences suggest a quadratic model when that pattern is assumed to continue. For a rulean²+bn+c at consecutive integer positions, the second difference is 2 a, so half the second difference gives the quadratic coefficienta.

Two complete quadratic tasks. Positions begin at 1. Derive rather than merely match the final expression, and check the twentieth term independently.

Task	First four terms	First differences	Second difference	Nth-term rule	Term 20
1	3, 8, 15, 24	5, 7, 9	2	n²+2 n	440
2	4, 11, 22, 37	7, 11, 15	4	2 n²+n+1	821

For the source 3, 8, 15, 24, the second difference 2 givesa=1. Subtractn²=1, 4, 9, 16 from the terms to get 2, 4, 6, 8. This is 2 n, so the rule is n²+2 n. At 20, it gives 400+40=440. The alternative formn(n+2) agrees and provides a counting model.

The following original schematic has n rows with n+2 dots in each row. It gives 3, 8, 15 at the first three positions. Count the dots within each separate group; their physical screen size is irrelevant.

n=1    n=2     n=3
●●●    ●●●●    ●●●●●
       ●●●●    ●●●●●
               ●●●●●
3      8       15

Text alternative: one row of 3, two rows of 4, three rows of 5. The structural countn(n+2) expands to n²+2 n for any positive integern, providing more justification than a few matching values alone. Equivalently, (n+1)²−1 describes a larger square with one cell removed.

For row 2, the second difference 4 givesa=2. Subtract 2 n² from 4, 11, 22, 37 to get 2, 3, 4, 5. The remainder is n+1, so the whole rule is 2 n²+n+1. At 20, the value is 800+20+1=821. Substitutingn 1, 2, 3, 4 reproduces every supplied term.

An alternative coefficient method usesan²+bn+c. For row 2, a=2 leaves b+c=2 from the first term and 2 b+c=3 from the second. Subtraction gives b=1, then c=1. This agrees with the remainder-sequence route. The method is tied to an assumed quadratic family; without that model assumption, a finite list can admit many other extensions.

Test membership, models and independent transfer

For the source arithmetic rule 3 n+1, is 100 a term? Solve 3 n+1=100 to getn=33, a positive integer, so yes. For 101, the equation givesn=100/3, which is not an integer position, so 101 is not a term in this one-based sequence. A numerical solution for n is not sufficient unless it lies in the permitted position domain.

For a distant quadratic example, n²+2 n at n 10 gives 120. The rectangle model 10×12=120 provides an independent check. For 2 n²+n+1 at n 10, the result 211 agrees with subtracting the quadratic part 200 to leave the linear remainder 11. Distant-term checks test the rule directly rather than relying on a long list of repeatedly added differences.

The actual original Corbettmaths preview uses the different quadratic sequence 6, 9, 14, 21, 30 and obtainsn²+5, with tenth term 105. Compare its constant second difference 2 with our source sequence 3, 8, 15, 24. Equal second differences give the same quadratic coefficient, not the same complete rule. Original: corbettmaths, Quadratic Sequences Version 1 - Corbettmaths. The written explanations and native tasks work without sound or the external player.

Do not infer uniqueness from a few matched terms. Rules 3 n+1 and n²+3 both give 4 and 7 at n 1 and 2, but at n 3 they give 10 and 12. Even matching four terms would not prove an unrestricted infinite rule: adding a multiple of(n −1)(n −2)(n −3)(n −4) changes no first-four values but can change later ones. State a continuing arithmetic/quadratic assumption or justify a structural counting rule.

For comparison, 3, 6, 12, 24 has constant ratio 2 and changing first differences 3, 6, 12. It is geometric rather than arithmetic, and has rule 3×2^(n −1). The first term is still at n 1. Neither the linear first-difference formula nor the quadratic second-difference method should be used merely because a list increases.

The context may restrict a useful mathematical rule. If 20, 16, 12, 8 described a hypothetical remaining stock losing 4 units per step, 24−4 n would cease to describe a nonnegative stock after position 6. Its abstract twentieth term −56 is still algebraically correct, but the physical model needs a domain cutoff. The task table itself is explicitly abstract.

Retain five derived linear rules, two Higher-tier quadratic rules, all supplied-term checks, a distant term and a membership decision with integer-position reasoning. Show a first-error repair and distinguish a model assumption from a proof. Record later independent retrieval only after an actual new attempt.

Interactive Tasks

Quiz: Test Your Knowledge

Source 4, 7, 10, 13. Nth term with n 1 first? (3 n+1) (!3 n+4) (!4 n+3) (!n+3)

20, 16, 12, 8. Nth term? (24−4 n) (!20−4 n) (!4 n+16) (!20+4 n)

1.5, 4, 6.5, 9. Common difference? (2.5) (!1.5) (!4) (!3)

−6, −1, 4, 9. Nth term? (5 n −11) (!5 n −6) (!-5 n −1) (!n −7)

Twentieth source arithmetic term? (61) (!60) (!64) (!57)

Higher tier: second difference 2 gives quadratic coefficient? (1) (!2) (!4) (!0)

Higher tier: 3, 8, 15, 24 has rule? (n²+2 n) (!n²+3) (!2 n²+1) (!3 n)

Higher tier: 4, 11, 22, 37 has rule? (2 n²+n+1) (!n²+3 n) (!2 n²+2) (!4 n+1)

Is 100 a term of 3 n+1 with positive integer positions? (Yes; position 33) (!Yes; position 100) (!No; position is not an integer) (!No; the rule only has four terms)

Geometric 3, 6, 12, 24 has constant? (Ratio 2) (!Difference 3) (!Difference 6) (!Second difference 2)

Memory Game

Position	Input index locating a sequence term
Arithmetic	Sequence with constant first difference
Quadratic	Rule with highest position power two
Second difference	Difference between consecutive first differences
Geometric	Sequence with constant ratio between successive terms
Membership	Whether a value occurs at a permitted position

Drag and Drop

3 n+1	First arithmetic source rule
24−4 n	Decreasing arithmetic rule
2.5 n −1	Non-integer-difference arithmetic rule
n²+2 n	First Higher-tier quadratic rule
2 n²+n+1	Second Higher-tier quadratic rule

Crossword Puzzle

SEQUENCE	Ordered list governed by stated rules
POSITION	Index of a term
ARITHMETIC	Having a constant first difference
QUADRATIC	Having highest position power two
GEOMETRIC	Having a constant term ratio
DIFFERENCE	Subtraction between successive values

LearningApps

Optional extra practice. Choose an appropriate activity with your teacher. If the external collection is unavailable, use the native activities and independent tasks in this course.

Cloze Text

Open Tasks

Easy

Separate position from term: Use the first source value at n 1.
Find a difference: Explain positive versus negative change.
Check a proposed rule: Reject 3 n+4 using the first term.
Read the dot model: Count 1×3, 2×4 and 3×5.

Standard

Derive five linear rules: Use both first-term and coefficient-offset methods.
Calculate all twentieth terms: Retain direct substitution and a check.
Decide membership: Compare 100 and 101 for the source arithmetic rule.
Interpret a physical cutoff: Separate abstract negative terms from remaining stock.

Advanced

Higher tier: derive two quadratic rules: Use second differences and remainder sequences.
Higher tier: compare derivations: Use coefficient equations for the second quadratic.
Challenge uniqueness: Give two rules agreeing at two positions but differing at the third.
Contrast a geometric sequence: Explain why constant ratio is a different structure.

Learning Check

Five linear rules: State difference, first term and twentieth term.
Two Higher-tier quadratic rules: Use second differences and remaining linear parts.
Distant terms and structure: Check by a separate representation.
Membership and domain: Solve for a permitted position.
Model assumptions and contrast: Explain why matching a finite list is limited.
First error and evidence: Retain derivations rather than copied rules.

Worked checks and assessment guidance

3 n+1→61; 5 n+4→104; 24−4 n→−56; 2.5 n −1→49; 5 n −11→89. Each rule reproduces all four supplied terms with n 1 first. Abstract negative values are allowed unless a context restricts the domain.
Source n²+2 n gives 3, 8, 15, 24 and twentieth term 440. Second 2 n²+n+1 gives 4, 11, 22, 37 and twentieth term 821. Second differences 2 and 4 give leading coefficients 1 and 2; remaining parts are 2 n and n+1.
Atn 10, first quadratic gives 120=10×12; second gives 211=200+11. Source dot modeln rows ofn+2 proves countn(n+2)=n²+2 n for positive integern.
3 n+1=100 givesn 33, so 100 occurs. For 101, n 100/3 is not an integer, so it does not occur. A hypothetical nonnegative stock model 24−4 n is valid only through position 6.
3 n+1 and n²+3 agree at 1 and 2 but differ at 3. Arithmetic/quadratic continuation assumptions must be stated.3, 6, 12, 24 is geometric with rule 3×2^(n −1), not constant first difference.
3 n+4 misses the source starting term because it treats the first value as an offset. Keep five linear and two quadratic derivations, all supplied-term checks, distant terms and actual independent later work. Matching examples alone is not an unrestricted uniqueness proof.

Learning evidence

Keep five linear rules, two Higher-tier quadratic rules, all initial-term checks, twentieth and tenth terms, membership decisions, model assumptions and a first-error repair.

OER and Wikipedia

Department for Education:GCSE Mathematics subject content and assessment objectives. Source area:GCSE Mathematics subject content > Scope of study > Algebra - Sequences items 23 to 25. Year10 allocation is editorial within the GCSE pathway;Higher-only extensions are explicitly labelled. Original examples,teaching and feedback;Commons images retain their actual attribution,and videos identify original publishers. Reviewed12November2026.

Cuisenaire-Rods.jpg, Celcom at en.wikipedia, CC BY 3.0. Unchanged original; caption and learning activities are original.

100 grid.PNG, The original uploader was Leightwing at English Wikiversity., Public domain. Unchanged original; caption and learning activities are original.