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The nth term is a rule that gives a sequence’s term value from its position. For the arithmetic sequence 5, 8, 11, 14, …, the rule is 3n + 2 when the first position is n = 1. Substituting n = 4 gives 14. The letter n represents the position number, not the previous term and not a missing answer chosen at random.
This guide helps Sengkang families move from continuing a visible pattern to describing a rule that works at any allowed position. It focuses on arithmetic sequences, then compares them with a few other patterns so the linear method is used with care. The activities are original teaching examples, not actual school assignments, local construction plans or claims about a particular child’s progress. Match the depth to the learner’s readiness and school instructions.
Start with the core examples if the topic is new. Later contrasts are optional extensions; this guide is not a statement that every section belongs to one particular school year or subject level.
Algebraic expressions for families · Solving equations for families · Secondary 1 algebra and nth-term classroom
Choose a chapter
Separate position and value · Chapters 1–4
Derive a general rule · Chapters 5–9
Check harder patterns · Chapters 10–15
Practise independently · Chapters 16–20
Choose a continuation · Chapters 21–24
CHAPTER 1 OF 24 · Separate position and value
1. Read a sequence as an ordered list
A sequence is an ordered collection of terms. In 5, 8, 11, 14, the first term is 5 and the fourth term is 14. The order matters: rearranging the same numbers gives a different sequence. Position tells us where a term sits, while value tells us the number at that position.
Draw two rows on paper. Label the top row “position” and write 1, 2, 3, 4. Label the bottom row “term value” and write 5, 8, 11, 14. Read vertically: position 3 has value 11. Read horizontally along the lower row to inspect how consecutive values change.
Ask “What is the third term?” and “At which position does 11 occur?” These questions reverse the relationship. A child may answer both correctly with a short list while still confusing the two meanings in algebra. Keep the labels visible when introducing n so the symbol attaches to an already understood role.
The dots after a finite display suggest continuation, but the rule still needs to be specified or inferred within a stated pattern type. A handful of numbers does not force one unique infinite sequence. At this stage, say “Continue using the stated add-three rule” rather than implying that every list has an inevitable next value discoverable by intuition alone.
CHAPTER 2 OF 24 · Separate position and value
2. Find the difference in the right direction
For 5, 8, 11, 14, subtract each earlier term from the next: 8 − 5 = 3, 11 − 8 = 3 and 14 − 11 = 3. The repeated difference is 3. An arithmetic sequence changes by the same amount between consecutive terms, so this is a candidate arithmetic pattern.
The subtraction direction matters when the sequence decreases. In 17, 13, 9, 5, the next-minus-previous differences are all −4. Saying only “the gap is four” omits the direction. The term-to-term rule is subtract 4, or equivalently add −4. That sign must carry into any later nth-term formula.
Have the child draw an arrow between adjacent terms and write the change above it. Check every displayed interval, not only the first. The list 2, 5, 8, 12 begins with two differences of 3 but then changes by 4. The evidence does not support treating all its displayed terms as one arithmetic sequence.
Equal differences across the shown terms support an arithmetic model when that is the intended family of rules. They do not prove what every future term must be without an assumption about continuation. That qualification is useful rather than fussy: mathematics distinguishes a stated rule, a tested conjecture and a conclusion that follows from the rule.
The term-to-term rule tells how to move from one term to the next. For the sequence beginning 5 and increasing by 3, start at 5, then repeatedly add 3. To reach the second term requires one addition. To reach the third requires two additions. To reach the tenth requires nine additions.
Use ten position boxes and count the arrows between them. There are nine arrows because the first box is already occupied by the starting value. This observation explains the n − 1 in the arithmetic-sequence formula later. It is not a trick inserted to make the symbols look more complicated.
A term-to-term rule needs a starting term to identify the sequence. “Add 3” can describe 5, 8, 11, … or 1, 4, 7, … or −8, −5, −2, …. The change is shared, but the sequences begin at different values. Ask the child for two examples with the same difference to make that distinction explicit.
Term-to-term reasoning is excellent for nearby terms and for checking a generated list. It becomes inefficient for a distant position if every intermediate term must be written. That is the job of a position-to-term rule. Teach both as useful descriptions of the same structure, with different practical strengths, instead of discarding the familiar method as soon as algebra appears.
An nth-term rule maps a position to its term value. In the rule 3n + 2, replace n by the required positive integer position, multiply by 3 and add 2. At n = 1 the value is 5; at n = 2 it is 8; at n = 20 it is 62.
The expression does not mean “add three to the previous value and then add two”. There is no previous term in this explicit calculation. The input is the position number. Put a small label under n saying “position” until the child can state that meaning without prompting.
Use a simple input-output table. Positions 1, 2, 3 and 4 enter the rule and values 5, 8, 11 and 14 emerge. The table is not a new sequence of operations; each column is a fresh substitution into the same rule. Read one complete column before moving to the next.
The phrase “nth term” can feel abstract because n is unspecified. Explain that it lets us discuss any allowed position at once. Replacing n by 100 asks for the hundredth term; leaving n as a letter gives the general instruction. The rule is a compact statement about all positive integer positions in this particular infinite arithmetic model, not an equation that has to be solved for one unknown immediately.
Let the first term be a and the common difference be d, with position numbering beginning at 1. The nth term is a + (n − 1)d. It begins with the first value and adds the common difference once for each step after position 1. There are n − 1 such steps.
For a = 5 and d = 3, the rule is 5 + (n − 1) × 3. Expanding gives 5 + 3n − 3 = 3n + 2. The two forms are equivalent. The first emphasises the start and number of jumps; the second makes the linear position relationship concise.
Check n = 1 directly. The jump count becomes zero, so the result is a, as required. Check n = 2: there is one jump and the result is a + d. These boundary checks explain the formula more effectively than simply memorising the letters and hoping the first term appears correctly.
If a child writes a + nd, ask what it gives at n = 1. It adds an extra jump before reaching the first term. Use the boxes-and-arrows drawing to locate the mismatch. A formula error often comes from an indexing error rather than weak multiplication, so the correction should reconnect position counting to the sequence instead of repeating arithmetic drills.
For 5, 8, 11, 14, the common difference 3 suggests the base sequence 3n, which gives 3, 6, 9, 12. Each target term is 2 greater than its matching base term. Therefore the target rule is 3n + 2. The constant adjustment is checked column by column.
For 6, 10, 14, 18, compare with 4n, which gives 4, 8, 12, 16. The adjustment is +2 again, so the rule is 4n + 2. Matching adjustments do not mean the sequences are identical: their rates of change differ. Keep the difference and adjustment as separate decisions.
A child may guess 4n + 6 because the first term is 6. Substitute n = 1 and the result is 10, immediately revealing the error. The constant in the simplified form dn + c is generally not the first term. The first term is d + c when numbering starts at 1.
Use this method only after checking constant differences. A similar-looking list such as 3, 6, 11, 18 has changes 3, 5 and 7, so comparing it with 3n does not yield one fixed adjustment. The method’s condition matters. Naming that condition helps a learner choose when the technique applies, rather than treating every sequence question as a search for a multiple plus a number.
| Position n | 3n | Target value | Adjustment |
|---|---|---|---|
| 1 | 3 | 5 | +2 |
| 2 | 6 | 8 | +2 |
| 3 | 9 | 11 | +2 |
| 4 | 12 | 14 | +2 |
Consider 18, 14, 10, 6, 2, …. The common difference is −4. Starting from the first term gives 18 + (n − 1)(−4). Expanding carefully gives 18 − 4n + 4 = 22 − 4n. The positive constant 22 is an adjustment in the simplified expression, not a displayed first term.
At position 1, 22 − 4 = 18. At position 5, 22 − 20 = 2. At position 6, the value is −2. The mathematical arithmetic model can continue into negative values even if the first displayed terms are positive. A real context might impose an earlier stopping condition.
Ask the child to predict whether the twentieth term is larger or smaller than the tenth before calculating. The values are −58 and −18 respectively, so the twentieth is smaller. If the child expects a positive answer merely because position numbers are positive, return to the difference between input and output.
A decreasing sequence does not automatically represent an impossible situation. It may describe signed numbers, temperature changes in a fictional model or a purely numerical list. If it represents a stock of physical objects, however, negative counts would signal that the rule cannot describe the situation indefinitely. Separate the algebraic continuation from the context’s domain and explain both in the final answer.
For the rule 4n + 2, the fiftieth term is 4 × 50 + 2 = 202. There is no need to write the preceding forty-nine terms. The position-to-term rule provides direct access because the relationship has already been established for the arithmetic sequence.
Keep substitution visible, especially with negative coefficients. For 22 − 4n at n = 30, write 22 − 4 × 30 = 22 − 120 = −98. Do not read 22 − 4n as (22 − 4)n. The grouping and normal order of operations determine which multiplication happens first.
Use an adjacent-position check. If the fiftieth term of 4n + 2 is 202, the fifty-first should be 206 because the difference remains 4. Substitution gives 4 × 51 + 2 = 206. This does not prove every part of the model, but it can expose a copying or arithmetic error in a particular calculation.
Ask for an answer sentence that includes the position: “The fiftieth term is 202.” A bare 202 may be mathematically acceptable in a short exercise, but the sentence helps a learner keep the relationship clear during practice. When the rule describes objects, include the relevant unit or label as well. Good notation should reduce ambiguity rather than add unnecessary writing to every simple task.
For the arithmetic sequence with rule 3n + 2, ask whether 80 occurs. Set 3n + 2 = 80, subtract 2 and divide by 3 to obtain n = 26. Because 26 is a positive integer, 80 is the twenty-sixth term of the stated sequence.
Now test 81. The equation gives n = 79/3, which is not an integer. It is not an allowed position in a sequence numbered 1, 2, 3, …. Therefore 81 does not occur, even though it lies between some of the sequence’s term values. The rule defines discrete positions, not every point on a continuous line.
The positivity check matters too. In 3n + 2, a value of −1 gives n = −1, which is outside the stated domain. An algebraic solution can satisfy the expression’s equation while failing the sequence’s position restriction. State both parts of the conclusion: what n solves the equation and whether that n is allowed.
Ask the child to check 80 by substitution and then inspect its neighbours. Positions 25 and 27 give 77 and 83. This confirms the ordering around the value and reconnects equation solving to the original list. The aim is not merely to solve for n, but to interpret what the solution says about membership and position.
Suppose an arithmetic sequence has fourth term 17 and ninth term 37. The positions differ by five, so there are five equal jumps between those values. The total change is 37 − 17 = 20, giving common difference 20 ÷ 5 = 4. There are five intervals, not six.
Move back three jumps from position 4 to position 1. The first term is 17 − 3 × 4 = 5. Therefore the rule is 5 + (n − 1)4 = 4n + 1. Check the supplied terms: at n = 4 it gives 17; at n = 9 it gives 37.
An alternative is to use the simplified form 4n + c. Substitute n = 4 and value 17: 16 + c = 17, so c = 1. Both methods use the same relationship. Choose the one the child can explain, then show the other as a check or extension when useful.
The assumption “arithmetic” is essential. Two separated term values alone do not determine an arbitrary sequence. Many non-arithmetic rules could pass through those positions. State the model before deriving the common difference, and resist turning a method for one family of patterns into a claim about every possible list. Knowing when information is sufficient is part of the lesson.
CHAPTER 11 OF 24 · Check harder patterns
11. Fill missing terms without losing intervals
In the arithmetic display 7, __, __, 22, the change from the first to fourth term is 15 across three jumps. Divide 15 by 3 to get a common difference of 5. The missing terms are 12 and 17. Check all transitions: 7 to 12, 12 to 17 and 17 to 22 each add 5.
A common mistake divides by the number of blanks, giving 7.5. The blanks count missing values, while the divisor must count equal intervals. Draw four boxes and three arrows. This representation makes the relevant count visible without asking the child to memorise a separate rule for every arrangement of blanks.
For a decreasing version, 20, __, __, 2 has change −18 over three jumps, giving difference −6 and missing terms 14 and 8. Keep the negative sign throughout. If a learner subtracts six successfully but writes positive six as the common difference, ask them to distinguish the amount removed from the signed next-minus-previous change.
Missing terms can also sit at the start or end. With __, 11, 15, __, the difference is 4, so the values are 7 and 19. Ask the learner to explain which direction they travelled. Moving backwards through an add-four rule means subtracting four; the inverse movement depends on direction along the sequence.
Draw a row of connected unit squares using line segments. One square uses four segments. Adding the next square in the same row shares one side with the previous square, so only three new segments are needed. For one, two, three and four squares, the counts are 4, 7, 10 and 13.
With n squares, the segment count is 3n + 1. One explanation begins with four segments and adds three for each of the n − 1 later squares: 4 + 3(n − 1). Another counts n top segments, n bottom segments and n + 1 vertical segments, giving n + n + n + 1.
These explanations show why the rule works for the entire stated construction, not merely the first four pictures. Ask the child to point to the parts represented by each term in the expression. A diagram-based justification is stronger than noticing a numerical difference and assuming the same pattern must continue.
The construction rule must stay fixed: a single row of equal squares sharing complete sides. Separate squares, a two-row arrangement or overlapping shapes would produce a different count. Label the pictures as invented practice diagrams. They do not describe the dimensions or design of a particular Sengkang structure. The mathematical value lies in connecting a visible building rule to a general expression.
CHAPTER 13 OF 24 · Check harder patterns
13. Compare two patterns at the same position
Let pattern A have rule 3n + 1 and pattern B have rule 2n + 6, both for positive integer positions. At position 1, A has 4 and B has 8. A begins lower but increases by 3 each step, while B increases by 2. Their difference changes by one per position.
To find when the counts are equal at the same position, solve 3n + 1 = 2n + 6. This gives n = 5, with both values equal to 16. Check each rule independently at position 5. At position 6, A gives 19 and B gives 18, so A has moved ahead.
Distinguish this question from asking whether the two sequences ever contain the same value at different positions. The value 10 occurs in A at position 3 and in B at position 2. That shared value is real, but it is not an equal-at-the-same-position solution. The wording decides which relationship to solve.
Use a small table with separate position and value columns to expose the distinction. Ask the child to read one row at a time when comparing equal positions. Then invite a search across rows for shared values as a different activity. Precision about the comparison prevents correct algebra from being used to answer a question that was never asked.
CHAPTER 14 OF 24 · Check harder patterns
14. Handle fractions and decimals as differences
An arithmetic sequence need not use whole numbers. Consider 1.5, 2.25, 3, 3.75, …. The common difference is 0.75, so the rule is 1.5 + 0.75(n − 1) = 0.75n + 0.75. At position 10, it gives 8.25.
Write the same sequence in fractions if that makes the structure clearer: 3/2, 9/4, 3, 15/4. The difference is 3/4, and the rule is 3n/4 + 3/4. These forms describe the same values. Converting representation should preserve the rule, not create a new pattern.
For a decreasing example, start at 5/2 and subtract 1/2 each time. The nth term is 5/2 − (n − 1)/2 = 3 − n/2. The sixth term is zero, and the seventh is −1/2. Positive integer positions can correspond to fractional, zero or negative term values.
If decimal subtraction is distracting, use quarter-sized paper units or write all values with a common denominator. The sequence concept should remain visible while arithmetic is supported. Once the relationship is secure, return to the notation required by the task. A child who understands the positional structure but slips in fraction subtraction needs focused number work rather than a complete restart of the topic.
CHAPTER 15 OF 24 · Check harder patterns
15. Recognise constant and alternating patterns
The sequence 6, 6, 6, 6, … is arithmetic with common difference zero. Its nth term is 6. In the general formula, a = 6 and d = 0, so a + (n − 1)d remains 6 for every allowed position. A sequence can be arithmetic without increasing or decreasing.
The pattern 2, 5, 2, 5, … alternates. Its consecutive differences are 3, −3, 3, −3, so it is not arithmetic. A direct description is “2 at odd positions and 5 at even positions”. That is a valid position-based rule even though it is not one linear expression of the form dn + c.
Ask for the twentieth and twenty-first terms of the alternating pattern. Their values are 5 and 2 because the positions are even and odd. This can be solved without listing all earlier values. It shows that an nth-term idea is broader than the specific arithmetic formula taught in the main part of this guide.
Use contrast cases to strengthen method selection. Place an increasing arithmetic list, a constant list and an alternating list side by side. Ask which share constant differences and why. Do not label every unfamiliar pattern “wrong” or force it into a linear method. The useful question is which structural family the stated rule belongs to and which reasoning that family supports.
CHAPTER 16 OF 24 · Practise independently
16. Distinguish repeated addition from multiplication
In 3, 6, 12, 24, …, each term is twice the previous one. The differences are 3, 6 and 12, so the sequence is not arithmetic. Multiplying by 2 is a term-to-term rule, but it does not create a constant additive difference.
With first position 1, a position rule is 3 × 2 to the power n − 1, written 3 × 2^(n − 1). At n = 1 the exponent is zero and the value is 3. At n = 4 the rule gives 3 × 8 = 24. This is an optional extension for learners already comfortable with powers.
A child who writes 2n + 1 from the first two terms obtains 3 and 5, so the second displayed term immediately fails. Testing more than the first term is essential. Even a rule matching several terms should be justified against the intended construction or stated sequence family before distant predictions are treated as secure.
For a younger learner, keep the comparison verbal: “add the same amount” versus “multiply by the same factor”. There is no need to rush into exponent notation. The purpose of the contrast is to protect the arithmetic method’s meaning, not to turn a family introduction into a survey of every sequence type. Follow readiness when deciding whether the optional formula helps.
The square numbers 1, 4, 9, 16, 25, … have first differences 3, 5, 7 and 9. Those differences increase by 2. The sequence is not arithmetic, although its second differences are constant. Its familiar position rule is n², meaning n multiplied by itself.
Draw square arrays with side lengths 1, 2, 3 and 4. The total number of unit cells is 1, 4, 9 and 16. Adding the next outer L-shaped layer contributes 3, then 5, then 7 cells. The drawing gives a structural explanation for the changing first differences.
Avoid claiming that any short list with a constant second difference must continue quadratically forever. In a question that specifies a quadratic sequence or a square-array construction, that model is justified. Without a model, the finite evidence supports a candidate rather than a unique infinite continuation. The same caution applies to arithmetic patterns.
This extension helps a child reject an unsuitable linear rule. If first differences vary, stop before using dn + c. Ask whether a diagram, multiplication pattern or another supplied instruction explains the sequence. Being able to say “the arithmetic method does not apply here” is a valuable piece of understanding, even before the learner can derive every alternative formula independently.
Most school-style examples here number the first term as position 1. Some contexts begin at position 0, such as a model recording an initial amount before any changes occur. The expression must match that convention. A list beginning 5, 8, 11 at positions 0, 1, 2 has rule 3n + 5.
The same values numbered at positions 1, 2, 3 have rule 3n + 2. Neither formula is inherently wrong. They attach different position labels to the same ordered values. Before substituting, read the table headings or define what n counts in the context.
Consider an invented collection with five cards initially and three added after each completed session. After n sessions, the total is 5 + 3n, with n = 0 allowed. If instead n names the first displayed stage and that stage already shows five cards, the stage-number formula is 3n + 2. State which interpretation is being used.
A parent can diagnose this issue by asking for the value at the starting position. If the child gives the right first displayed value under the wrong position label, the problem is indexing. Draw a timeline with the initial point and completed steps. This makes the difference between “stage number” and “number of changes completed” visible without relying on a verbal warning alone.
One more check concerns a constant sequence. For the rule 9, every positive integer position has value 9. Asking “At which position does 9 occur?” therefore has infinitely many answers in the infinite model. Asking where 10 occurs has none. The usual membership equation can become 9 = 9 or 9 = 10 rather than an equation with one numerical solution for n. This is a useful boundary case once the learner understands zero common difference.
Do not tell a child that every term value has exactly one position without stating a condition. A nonconstant arithmetic sequence has distinct values at distinct positions, but a constant or repeating sequence does not. The alternating pattern 2, 5, 2, 5 has repeated values as well. Ask whether the question requests one position, all positions or the first occurrence. The intended answer changes with that wording. A careful response checks the rule’s structure before assuming that solving for one n is always the right final step.
CHAPTER 19 OF 24 · Practise independently
19. Diagnose a plausible but incorrect rule
Suppose a learner gives 5n for the sequence 5, 8, 11, 14. The rule matches the first term but produces 10 at position 2. Ask the child to generate its first four values: 5, 10, 15, 20. Comparing the generated list with the target shows the mismatch directly.
Another learner gives n + 4. That also matches the first term but produces consecutive values 5, 6, 7, 8. The coefficient of n controls the change per one-position step in a linear rule. Here the required difference is 3, so a coefficient of 1 cannot fit the arithmetic model.
A third learner gives 3n + 5, which has the correct difference but wrong start. Its values are 8, 11, 14, 17. This suggests the child has confused the first term with the constant adjustment or used a zero-based rule with one-based positions. Ask which position makes the rule give 5.
These errors need different prompts. Multiplication practice alone will not fix a misunderstood position input. Keep the child’s original expression and test it rather than replacing it immediately with the adult’s formula. The test gives the learner a reusable checking tool: generate a few terms, compare the differences, and verify the starting convention before using a distant position.
A further diagnostic asks a learner to create, rather than recover, an arithmetic sequence. Specify a common difference of 6 and a fourth term of 25. Moving back three intervals gives a first term of 7, so the rule is 7 + 6(n − 1) = 6n + 1. The first four values are 7, 13, 19 and 25. Designing the sequence requires the same structural decisions as solving a standard question, in a less familiar direction.
Next specify that a sequence has common difference −2 and tenth term zero. There are nine intervals from the first to tenth term, so the first value is 18. The rule is 18 − 2(n − 1) = 20 − 2n. At position 10 it gives zero; at position 11 it gives −2. The child should explain both the backwards recovery and the forward check, with the signed difference preserved throughout.
Finally, ask for two distinct arithmetic sequences that share the same tenth term. The rules n + 5 and 2n − 5 both give 15 at position 10, but their first terms are 6 and −3 and their differences are 1 and 2. One shared term does not identify a unique arithmetic rule. A second condition, such as another term or the common difference, is needed. This task helps a child distinguish knowing one output from knowing the whole relationship and gives a parent useful evidence of whether the formula has become more than a memorised procedure.
Task A: for 7, 11, 15, 19, …, find the common difference, the nth term and the twentieth term. Task B: for 20, 17, 14, 11, …, find the nth term and decide whether −10 occurs. State its position if it does.
Task C: an arithmetic sequence has third term 12 and eighth term 32. Find its first term and nth term. Task D: fill the blanks in 4, __, __, 25, assuming an arithmetic sequence. Task E: a sequence has rule 5n − 2 for positive integer positions. Decide whether 73 and 74 are terms.
For every rule, test the first term and one later displayed term. For membership questions, solve the equation and check that the position is an allowed positive integer. For separated terms, count intervals rather than displayed boxes. These checks target different sources of error and should be selected deliberately.
Work without looking at the answers first. If one part is difficult, keep the useful representation instead of erasing everything. A table with correct positions can support a new attempt at the formula. A correct difference can support a repaired starting value. The aim is to make the reasoning inspectable, so the child can identify what remains secure and what needs adjustment.
CHAPTER 21 OF 24 · Choose a continuation
21. Explain the arithmetic practice answers
For A, the difference is 4. The rule is 7 + 4(n − 1) = 4n + 3. At n = 20, it gives 83. Substituting n = 1 returns 7 and n = 4 returns 19. These checks connect the derived rule to both the start and a later displayed value.
For B, the difference is −3 and the rule is 20 − 3(n − 1) = 23 − 3n. Set 23 − 3n = −10. Then −3n = −33 and n = 11, so −10 is the eleventh term. The positive position does not require a positive term value.
For C, the total change is 32 − 12 = 20 over five intervals, so d = 4. The first term is 12 − 2 × 4 = 4, giving rule 4n. For D, the change is 21 over three intervals, so the common difference is 7 and the blanks are 11 and 18.
For E, 5n − 2 = 73 gives n = 15, so 73 occurs. The equation 5n − 2 = 74 gives n = 76/5 = 15.2, which is not an allowed position. Therefore 74 does not occur. Explain the integer restriction explicitly; rounding 15.2 to a nearby integer would change the equation and would not establish membership.
Task F: connected squares in a single row use 3n + 1 line segments for n squares. How many segments are needed for 12 squares, and can exactly 50 segments form a complete pattern under this rule? Task G: patterns A and B have rules 4n + 2 and 2n + 12. Find their equal value at the same position.
Task H: explain why 2, 4, 8, 16 is not arithmetic. Task I: a model begins with 8 tokens at session count zero and gains 5 after each completed session. Write a rule for the total after n sessions and calculate the total after six sessions. Task J: an alternating sequence is 9, 4, 9, 4, …. Find its thirtieth and thirty-first terms.
For F, a numerical solution must also represent a whole number of squares. For G, use the same n in both expressions because the question compares the same position. For I, the starting value belongs to n = 0. For J, use odd and even position numbers rather than forcing a constant-difference formula.
These tasks test whether the child selects a model, interprets its domain and notices when the main method should be set aside. A correct answer supported by the wrong rule may fail as soon as the position changes. Ask for a short explanation tied to the construction, input or sequence family, rather than judging success only by the final number.
CHAPTER 23 OF 24 · Choose a continuation
23. Explain the pattern practice and teach calmly
For F, 12 squares require 3 × 12 + 1 = 37 segments. Solving 3n + 1 = 50 gives n = 49/3, so exactly 50 segments do not make a complete row under the rule. For G, 4n + 2 = 2n + 12 gives n = 5 and common value 22.
For H, the differences are 2, 4 and 8, not one constant amount. The displayed terms follow doubling. For I, the rule is 8 + 5n, giving 38 tokens after six sessions. For J, even positions contain 4 and odd positions contain 9, so the thirtieth term is 4 and the thirty-first is 9.
For a family session, choose one arithmetic pattern and one contrast. Build a position table, explain the common difference, derive the rule and test it at a distant position. On another day, ask a membership question. Spacing these decisions across short sessions can make the child’s independent reasoning easier to see than completing a long page of nearly identical substitutions.
If difficulty persists, bring the original task and first attempt to the teacher or tutor. A child who confuses n with a term value needs a position-input activity; a child who derives the rule correctly but mishandles negative multiplication needs a different intervention. Use the actual working to locate the first uncertain step and keep the next practice focused on that step.
Is the constant in dn + c always the first term? No. With first position 1, the first term is d + c. The constant c is the value the linear expression would have at n = 0, which may or may not be an allowed position in the sequence’s stated context.
Must the nth term use one particular form? No. The forms a + (n − 1)d and dn + (a − d) are equivalent. Use the form requested by the task, and check equivalence by expansion or substitution. Can a sequence have negative values? Yes. Position numbers and term values have different roles.
Does a finite list prove its next term? Not without a rule or an assumption about the pattern family. Many rules can match a finite beginning. In a task stating an arithmetic sequence, constant difference supplies the intended model. In a picture problem, the construction rule can justify continuation structurally.
Use the linked OpenStax arithmetic-sequence section for a formal reference on common differences and explicit rules. The existing algebraic-expression and equation family guides support substitution and membership checks. Return to a two-row position table whenever the symbols become detached from meaning. A secure sequence explanation tells the reader what n counts, how the values change, why the rule fits the model and which positions are permitted.
References and further reading
The teaching cases and practice answers above are original examples. These references provide definitions, standard usage and further study; the connected eduKate guides offer easier foundations or school-focused continuation.
OpenStax: arithmetic sequences · Algebraic expressions for families · Solving equations for families · Secondary 1 algebra and nth-term classroom
Continue with a connected family guide
Probability and Sample Spaces for Sengkang Families: List Outcomes Before Calculating Chance
