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The sequence 7, 11, 15, 19, … increases by four each time, but its nth term is 4n + 3, not 4n. The expression 4n has the right step size and the wrong starting value. The extra three fixes the first term while preserving the common difference.
For an arithmetic sequence starting at a and changing by d at every step, the nth term is a + (n − 1)d when positions begin at one. There are n − 1 steps from the first term to the nth term. Rewriting gives dn + (a − d), where a − d is the offset.
This guide explains that starting adjustment through tables, repeated addition, graphs and answered practice. It is written for G2 readers learning linear sequence rules. The sequences used as arithmetic examples are explicitly defined to continue with a constant difference; a finite list by itself does not force one unique rule for every later term.
Contents
Sections 1–5
Sections 6–10
Sections 11–15
Sections 16–20
Sections 21–25
Sections 26–28
Find the common difference d, then write dn as a starting comparison. Substitute n = 1 and compare d with the actual first term a. Add the constant a − d to correct the starting value. For first term seven and difference four, that constant is three.
The equivalent form a + (n − 1)d makes the step count visible. At n = 1, there have been no changes yet, so it returns a. At n = 2, there has been one change, so it returns a + d.
Check the rule at the first term and at another position. A correct first-term value alone does not verify the difference, while a correct difference alone does not verify the starting value. Both pieces determine the arithmetic sequence once its indexing and constant-step assumption are established.
The position n says where a term occurs. The value Tₙ says what number is at that position. In 7, 11, 15, 19, the third position has value fifteen. The number three and the value fifteen are not interchangeable.
An nth-term rule takes a permitted position and returns its value. With Tₙ = 4n + 3, substituting n = 3 gives fifteen. The expression is a formula for the values, not a statement that every value is a multiple of four.
The ordinary sequence indexing in this guide starts with n = 1. If a problem starts with n = 0, the constant in a dn + c description changes because the first listed value now occurs at a different input.
Write the position row before building a rule. This makes the indexing visible and prevents a learner from using the first term as though it were the coefficient of n merely because it appears first in the list.
The underlying relationship connects two quantities: step number and term value. Naming both is the first protection against a formula that has plausible arithmetic but assigns values to the wrong positions.
An arithmetic sequence has a fixed difference between consecutive terms. For 7, 11, 15, 19, the differences are all four. Under the stated arithmetic rule, every move one position forward adds four to the value.
The common difference can be positive, negative or zero. A negative difference produces a decreasing arithmetic sequence; zero produces a constant one. It is the consistency of the step, not its sign, that defines the structure.
The difference tells you how values change but not where they begin. The sequences 4, 8, 12, 16 and 7, 11, 15, 19 both have difference four. They cannot have the same nth-term values because their first terms differ.
This is why identifying d is necessary but incomplete. The formula dn reproduces one particular sequence with that difference, namely the one whose first term is d when indexing begins at one.
To identify another sequence with the same step, the starting position must be adjusted. That adjustment is the offset, a constant added to every term of the comparison sequence. It changes the level without changing the difference between neighbouring values.
For d = 4, the expression 4n gives four when n = 1. It then gives eight, twelve and sixteen. It does not begin at zero under one-based indexing; the first positive position already contributes one copy of four.
If the intended first term is seven, the comparison formula is three too small at every position. Adding three produces seven, eleven, fifteen and nineteen. Because the same constant is added everywhere, each difference remains four.
This explains why adding the first term to dn is usually wrong. The formula 4n + 7 would give eleven at n = 1, not seven. It counts the starting value and an additional first step before the sequence has moved anywhere.
The correct first-term form is 7 + 4(n − 1), which subtracts the unnecessary step. Expanding gives 7 + 4n − 4 = 4n + 3.
The two descriptions highlight the same correction from different directions: either adjust the comparison sequence by three, or begin at seven and count only the steps actually taken after the first term.
From term one to term two there is one gap. From term one to term three there are two gaps. From term one to term ten there are nine gaps. The number of changes is one less than the destination’s position when counting from the first term.
That is why the formula is a + (n − 1)d. The first term contributes a once, and each of the n − 1 gaps contributes d. At n = 1, the multiplier on d is zero.
For first term seven and difference four, the tenth term is 7 + 9 × 4 = 43. Adding ten changes instead would give forty-seven, which is the eleventh term.
This off-by-one error often survives otherwise accurate multiplication. The learner may know the difference and multiply correctly but count the wrong number of transitions. A small position sketch can repair that decision before any larger calculation is attempted.
Use the word gaps or changes if it helps. The formula’s n − 1 is not a mysterious algebraic adjustment; it records how many consecutive moves separate the first position from the target position.
Assume the sequence continues arithmetically with common difference four and positions starting at one. The first-term formula is Tₙ = 7 + (n − 1) × 4. Expand and collect to obtain Tₙ = 4n + 3.
Check n = 1: 4 + 3 = 7. Check n = 4: 16 + 3 = 19. The common difference also follows algebraically: Tₙ₊₁ − Tₙ = [4(n + 1) + 3] − [4n + 3] = 4.
The rule therefore has both the required starting value and the required constant step. Those conditions identify the stated arithmetic sequence. The first few numerical checks illustrate them; the algebraic difference calculation verifies the step for all successive allowed positions.
If a proposed answer is 4n, it has the correct difference but begins at four. If it is 4n + 7, it begins at eleven. If it is 7n + 4, it begins at eleven and has difference seven.
A complete explanation distinguishes these errors rather than calling every incorrect rule a failure to spot the pattern. The coefficient and offset play different roles, so each can be checked separately.
A table makes the constant adjustment visible. Compare the intended term values with the sequence produced by 4n at the same positions.
| Position n | 4n | Intended Tₙ | Tₙ − 4n |
|---|---|---|---|
| 1 | 4 | 7 | 3 |
| 2 | 8 | 11 | 3 |
| 3 | 12 | 15 | 3 |
| 4 | 16 | 19 | 3 |
The final column is constant. That is why one offset of three works across the sequence. If the differences in that column were not constant, the proposed linear comparison would need to be reconsidered.
The table should use matching positions. Comparing the first target term with the second multiple of four would introduce a different indexing shift and hide the intended correspondence.
This method can be convenient when the difference is easy to identify. It is equivalent to the first-term formula, not a competing rule. Choose the representation that makes the reasoning clearest, and be able to explain why both lead to the same expression.
Consider the arithmetic sequence 2, 7, 12, 17, … with difference five. The comparison sequence 5n begins five, ten, fifteen, twenty. Each target value is three less, so Tₙ = 5n − 3.
The offset is negative because the intended first term lies below d. Using a − d gives 2 − 5 = −3 directly. There is no requirement that the constant in an nth-term rule be positive.
Check at n = 1: five minus three is two. At n = 4: twenty minus three is seventeen. The difference remains five because subtracting the same three from every comparison value does not alter the gaps.
A common error is to use 5n + 3 because the learner recognises the magnitude three but loses its direction. Ask whether 5n is too high or too low at the first position. If it is too high, the correction must subtract.
The sign of the offset describes the sequence’s level relative to dn. It does not by itself tell whether the sequence increases or decreases; that direction comes from the coefficient d.
For 18, 15, 12, 9, … the common difference is negative three. The first-term form is Tₙ = 18 − 3(n − 1), which expands to 21 − 3n.
The coefficient of n is negative because each forward step lowers the value by three. The offset twenty-one is positive, showing again that coefficient and offset describe different features.
Check n = 1: 21 − 3 = 18. Check n = 4: 21 − 12 = 9. At n = 7, the value is zero; at later positions it becomes negative if the arithmetic sequence is defined to continue indefinitely.
Whether negative term values make sense in a real context depends on the model. A pure number sequence can continue through zero. A count of physical objects may require a stopping condition before negative values occur.
Retain the domain from the question. The algebraic rule can produce values for many inputs, but the sequence or context determines which positions are allowed. A correct formula should be interpreted within that domain rather than treated as permission to create impossible physical counts.
For 6, 6, 6, 6, … the common difference is zero. The general formula gives Tₙ = 6 + (n − 1) × 0 = 6. In dn + c form, d = 0 and c = 6.
The rule does not need to contain n visibly to describe a sequence. It assigns the same value to every permitted position. Position still matters as an input label even though the output does not change.
This example is useful because some learners assume an nth-term answer must include a nonzero multiple of n. That assumption would incorrectly exclude constant arithmetic sequences.
It also shows why reverse questions may behave differently when d = 0. Asking which term has value six has many answers if the sequence continues indefinitely. Asking which term has value seven has none. Dividing by d to solve for n is unavailable because d is zero.
The nonzero condition belongs to that reverse-solving step, not to the definition of an arithmetic sequence. Zero difference is allowed; it simply requires interpreting membership and position questions according to the constant rule.
Suppose the values seven, eleven, fifteen are assigned to positions n = 0, 1, 2. The rule is now Tₙ = 4n + 7. At zero it returns the first listed value seven.
Under the one-based indexing used earlier, the same ordered values were described by 4n + 3. The sequences of values are the same, but the input labels have shifted by one.
Neither formula is intrinsically wrong. It is wrong only when paired with the wrong indexing convention. This is why a position row or an explicit statement about the first index matters.
If a problem gives a diagram labelled Stage 0, do not automatically apply a formula derived for Stage 1 without adjustment. Count how many transitions separate the labelled starting stage from the requested stage.
The common difference remains four because adjacent index labels still differ by one. The offset changes to ensure the correct value at the chosen starting input. This is the same separation of change and starting level that governs every linear arithmetic rule in the guide.
For Tₙ = 4n + 3, the fiftieth term is 4 × 50 + 3 = 203. The formula reaches the requested position directly, without writing the forty-nine preceding terms.
The first-term route gives 7 + 49 × 4 = 203. Its forty-nine changes make the indexing explicit, while the simplified form is quicker to evaluate. Both should agree because they are algebraically equivalent.
If a learner obtains 207, inspect whether they added fifty changes to the first term. If they obtain 200, inspect whether they omitted the offset. The wrong numerical result can reveal which part of the structure was misunderstood.
For a decreasing example Tₙ = 21 − 3n, the twentieth term is negative thirty-nine. In a pure arithmetic sequence that value is valid. In a context that stops at nonnegative counts, the twentieth position might lie outside the model’s allowed range.
Always finish with the requested quantity. The fiftieth term is a value, not a position equation to solve. Keeping the roles distinct prevents a forward-substitution task from being confused with finding where a particular value occurs.
In the arithmetic sequence Tₙ = 4n + 3, determine which term equals ninety-five. Set 4n + 3 = 95, subtract three and divide by four, giving n = 23. The value ninety-five occurs at the twenty-third position.
Check by substitution: 4 × 23 + 3 = 95. Also check that twenty-three is a permitted positive integer position. Solving the equation is necessary, but the sequence domain completes the answer.
If the target is ninety-six instead, the same algebra gives n = 93/4 = 23.25. That is not a permitted position in this one-based discrete sequence, so ninety-six is not a term.
The formula may define a straight-line value at a fractional input when viewed as a real function, but that input is not automatically a sequence index. A sequence consists of its specified discrete positions.
Do not round 23.25 to twenty-three or twenty-four and claim membership. Those positions give ninety-five and ninety-nine. The exact integer condition is what distinguishes a value near the sequence from a value actually in it.
Suppose an arithmetic sequence has T₃ = 11 and T₈ = 31. There are five gaps from position three to position eight, and the value increases by twenty. Therefore d = 20/5 = 4.
Now work back from the third term to the first. Two gaps separate them, so T₁ = 11 − 2 × 4 = 3. The rule is Tₙ = 3 + 4(n − 1) = 4n − 1.
Check the supplied positions: at three, the formula gives eleven; at eight, it gives thirty-one. The offset negative one ensures those values line up with the correct indices.
A common error divides the value change by eight or by three rather than by the difference in positions. The rate of change compares changes in value with changes in position, so the denominator is 8 − 3.
This connection resembles finding the gradient of a straight line. The arithmetic sequence is still discrete, but its points lie on a line with slope d. That graph perspective can support the calculation when the matching positions and values are kept clear.
Plotting the points (n, Tₙ) for Tₙ = 4n + 3 places the sequence on the line y = 4x + 3. The gradient four represents the constant change per unit increase in position, while the intercept three describes the line’s value at input zero.
The sequence itself may begin at n = 1, so the intercept is not necessarily a listed term. It is the zero-position extrapolation of the linear rule. The first listed value is seven because one step of four has already been added to that intercept.
This is why the offset need not equal the first term. A learner familiar with y = mx + c can use that distinction to understand the role of c in dn + c.
Do not connect all fractional-input points and then claim the sequence contains them. The line is a useful continuous representation of the rule; the sequence uses its stated integer positions.
The Mathematics hub connects sequence rules with linear graphs and gradients. The shared structure is constant change plus a starting adjustment, while the permitted input set depends on whether the task concerns a sequence, a graph or a contextual model.
Find nth-term rules for these explicitly arithmetic sequences, indexed from one: 6, 10, 14, …; 3, 8, 13, …; and 20, 16, 12, …. State both the difference and the offset.
For the first, d = 4 and a − d = 6 − 4 = 2, so Tₙ = 4n + 2. For the second, d = 5 and the offset is −2, giving Tₙ = 5n − 2.
For the third, d = −4 and a − d = 20 − (−4) = 24, giving Tₙ = 24 − 4n. The subtraction of a negative difference is a common place for a sign error.
Check the first term in each rule, then the next difference. The first rule gives six and ten; the second gives three and eight; the third gives twenty and sixteen.
Explaining the two checks is more useful than reporting three expressions without support. Each rule must have the correct starting alignment and the correct change. A formula can match one feature while failing the other.
For Tₙ = 5n − 2, find T₂₀. Substitute n = 20 to get ninety-eight. This is a forward calculation from a position to a value.
Now find the position of the term whose value is sixty-three. Solve 5n − 2 = 63, giving n = 13. This reverses the relationship, so the answer is a position rather than a term value.
Decide whether sixty-four is a term. Solving gives n = 66/5 = 13.2, which is not an allowed positive integer index. Therefore sixty-four is not a term of this sequence.
The same expression appears in all three questions, but the tasks differ. A careful reader identifies whether n is supplied, Tₙ is supplied, or membership is being tested.
When checking, use the answer in the original relationship. Twenty gives ninety-eight, thirteen gives sixty-three, and neither thirteen nor fourteen gives sixty-four. Rounding an index would replace the membership question with a nearby-value question the exercise did not ask.
An arithmetic sequence has T₂ = 9 and T₆ = 29. The value change is twenty across four gaps, so d = 5. The first term is 9 − 5 = 4.
The nth-term rule is 4 + 5(n − 1) = 5n − 1. Check at n = 2 and n = 6 to recover nine and twenty-nine.
For a second example, T₄ = 12 and T₉ = −3. The value change is negative fifteen across five gaps, so d = −3. Working back three gaps from T₄ gives first term twenty-one.
Thus Tₙ = 21 − 3(n − 1) = 24 − 3n. At the supplied positions it gives twelve and negative three. The signs are consistent with a decreasing sequence.
These reverse examples show that the offset is determined by alignment with position, not by a visual guess from a short list. Once the constant difference is known and one term is located, the first term and offset can be recovered systematically.
Suppose a stated pattern has three fixed markers and four additional markers for each numbered stage. At stage n, the count is 4n + 3. The four describes growth per stage, while the three describes a contribution that stays unchanged.
If stage one therefore has seven markers, the rule matches the earlier sequence. The diagram supplies a structural reason for the formula rather than only a list of observed counts.
Another pattern might begin with one segment and add four for each new stage after stage one. Its count is 1 + 4(n − 1) = 4n − 3. The difference is still four, but the fixed starting structure is different.
When deriving a formula from a diagram, explain which objects are counted by each term. If objects are shared between adjacent units, ensure the formula does not count them twice.
This is stronger than guessing a rule from several totals because the construction can justify every stage under its stated process. The formula’s coefficient and offset should correspond to actual repeated and fixed parts, making the resulting rule easier to verify and remember.
The values seven, eleven, fifteen and nineteen suggest an arithmetic sequence, but a finite prefix alone does not logically determine every later value. A different rule could match those four positions and then depart.
For example, 4n + 3 + (n − 1)(n − 2)(n − 3)(n − 4) matches 4n + 3 at n = 1, 2, 3 and 4 because the added product is zero at each of those inputs. At n = 5 it adds twenty-four.
The alternative rule is not being proposed as the intended school answer. It demonstrates why the constant-difference assumption or the pattern’s construction matters when making a general claim.
In an exercise explicitly asking for an arithmetic sequence, use that stated structure. In a diagram problem, use the construction to justify the pattern. In an exploratory list, describe an arithmetic continuation as a conjecture unless additional information establishes it.
This distinction keeps pattern recognition honest. Recognising a likely rule is useful, but proof that the rule applies to every later stage requires a condition or argument covering those stages, not only successful substitutions into the visible prefix.
To check Tₙ = dn + c, evaluate the first allowed position and calculate the change between consecutive positions. The first check tests alignment; the second tests the coefficient’s role as common difference.
For 4n + 3, the first value is seven and the consecutive difference is four. For 4n + 7, the difference is still four but the first value is eleven. For 7n, the first value is seven but the difference is seven.
These contrasts explain why one successful check can be insufficient. The wrong rule 7n matches the first term of the target sequence, while the wrong rule 4n + 7 matches its growth rate. Each misses a different requirement.
An algebraic difference calculation verifies the constant step across all consecutive integer inputs. Once the starting value and step are established, the arithmetic sequence is determined under the stated indexing.
Use numerical checks as convenient confirmation, especially after expansion or sign arithmetic. Use the structure of the formula to explain general validity. Checking is most informative when each calculation is chosen to test a specific possible error.
For Tₙ = 4n + 3, the ratio Tₙ/n is 4 + 3/n. It changes with n, even though the difference between consecutive terms is always four. The sequence is linear in n but not directly proportional to n.
At n = 1, the ratio is seven. At n = 2, it is 11/2. At n = 3, it is five. These changing ratios do not contradict the constant additive difference; they describe a different comparison.
Only when the offset is zero does Tₙ = dn express direct proportion between term value and position over the relevant nonzero inputs. This connects the sequence offset with the fixed-starting-amount distinction in other mathematical models.
Do not use a ratio shortcut merely because the values lie on a straight line. A line with a nonzero intercept needs its offset included when calculating new values.
The Mathematics explainer directory includes related lessons on fixed starting amounts and proportion. The shared lesson is that constant change and constant ratio are different relationships, and the starting term determines which one applies.
If every term of an arithmetic sequence is increased by six, the common difference stays the same. For 7, 11, 15, … the new sequence is 13, 17, 21, … with rule 4n + 9.
Subtracting consecutive terms shows why: (Tₙ₊₁ + 6) − (Tₙ + 6) = Tₙ₊₁ − Tₙ. The added constants cancel, so the difference remains four.
Multiplying every term by two changes both the difference and the offset. The rule becomes 8n + 6, giving 14, 22, 30, …. Its difference is eight, not four.
These transformations help distinguish a vertical adjustment from a change in growth rate. They also provide a way to generate related practice sequences without treating each as an unrelated pattern.
When modifying a rule, apply the stated operation to the whole term expression. Twice the term 4n + 3 is 2(4n + 3) = 8n + 6, not 8n + 3. Brackets preserve the meaning of the transformation and prevent an offset from being accidentally left unchanged.
Suppose a learner identifies difference four and writes Tₙ = 4n for the target sequence seven, eleven, fifteen, …. A precise correction is that the growth rate is correct but the starting alignment is three too low.
Ask for the value at n = 1, then compare it with the actual first term. Adding three repairs the alignment while leaving the difference unchanged. Follow by deriving 7 + 4(n − 1) so the learner sees why the same correction appears in the step-count method.
Next use a sequence with a negative offset, such as two, seven, twelve. This tests whether the learner can decide the direction of the adjustment rather than always adding a positive number.
Then use a decreasing sequence and a constant sequence. The central procedure should survive those changes: identify the stated constant difference, count gaps correctly and align the starting term.
After a delay, ask the learner to explain what d and c each do in dn + c. An answer that distinguishes change from starting level is stronger evidence of understanding than a correctly guessed expression with no account of its parts.
This guide focuses on arithmetic sequences, linear nth-term expressions and position-value interpretation. More involved reverse problems and alternative-continuation arguments are supporting extensions to choose according to the learner’s course and readiness.
For official context, consult the 2027 G2 Mathematics syllabus. Use school instructions to establish the expected scope and methods; this explanation does not promise that every extension is assessed in the same way.
The existing Arithmetic Sequences: Constant Difference as Structure guide is the broader topic owner. The Sequences and the nth Term lesson connects pattern recognition with general expressions.
Return to the Mathematics hub for related work on linear graphs, substitution and algebraic brackets. If a learner knows the rule but substitutes incorrectly, practise evaluation. If the difference is correct but the first term fails, practise offsets. The evidence in the working should determine the next step.
An arithmetic sequence begins nine, fifteen, twenty-one, …. Its difference is six, so its offset is 9 − 6 = 3. The rule is Tₙ = 6n + 3. The thirtieth term is 183.
To decide whether 123 is a term, solve 6n + 3 = 123. This gives n = 20, a permitted positive integer. For 124, the equation gives n = 121/6, which is not an integer, so that value is not a term.
Now consider positions beginning at zero with the same first listed value nine and difference six. The rule becomes 6n + 9. The coefficient remains six while the offset changes to match the indexing.
Finally, explain why six times the position is not enough in the one-based version. The formula 6n begins at six, while the target begins at nine. The constant three supplies the required alignment. A complete answer identifies both the growth rate and the starting adjustment instead of treating the common difference as the entire rule.
The first term and common difference also explain uniqueness within the arithmetic model. Once the first value is fixed, the second must be that value plus d, the third must add another d, and each later position is determined in turn. A different arithmetic formula with the same first value and difference cannot diverge later without breaking the stated constant-step rule.
For an explicitly arithmetic sequence, identify its first term, common difference and starting index. With positions beginning at one, use Tₙ = a + (n − 1)d or the equivalent Tₙ = dn + (a − d).
The coefficient d determines the constant change. The offset a − d aligns the values with their positions. It may be positive, negative or zero, and it is not generally the same as the first term.
Check the first value and the consecutive difference. For reverse questions, check that a calculated position belongs to the sequence’s allowed integer domain. For contextual patterns, retain any stopping conditions.
The useful question is, “How many changes have occurred before this position?” That question explains the n − 1 and prevents a correct step size from being attached to the wrong starting point. The offset is the information that turns a growth pattern into the intended sequence.
These answers assume an arithmetic sequence indexed from one unless another starting index is stated. Indexing is part of the rule, so check it before comparing apparently different formulas.
Why is the nth term not just the common difference times n?
That expression has the right step size but begins at d. It works without an offset only when the actual first term equals d. Otherwise add a − d to align the starting value.
Why does the first-term formula use n − 1?
There are n − 1 gaps between the first position and position n. The first term is already present before any changes occur. Adding n changes would move one position too far.
Can the offset be negative?
Yes. For two, seven, twelve, … the rule is 5n − 3. The comparison sequence 5n is three too high at every position, so subtracting three gives the intended values without changing the difference.
Does a fractional solution for n identify a sequence term?
Not in a sequence restricted to positive integer positions. If solving Tₙ = k gives a noninteger n, the target value k is not a term under that indexing. Do not round the position to manufacture membership.