A constant difference is the signature of a linear pattern

What to assume they know

• What makes a polynomial linear, and the equation you get by fixing its value — linear polynomials, and that equal input steps give equal output steps
• Univariate polynomials and what degree names — degree, and that degree 1 means linear
• A polynomial as an input–output machine — substituting a value into a rule to get an output
• Counting squares in a drawn arrangement
• Multiplying out a bracket, as in 15 × (n – 2)
• Subtracting repeatedly, and reading a negative step as a decrease

What they should be able to do

• Extend a drawn growing pattern by two or three further stages and tabulate the counts
• Compute the differences between consecutive terms of a sequence and decide whether they are all equal
• Derive a rule for the nth term of a growing pattern from the size of its repeated step and its first value
• Justify the rule geometrically by pointing at what gets added at each stage
• Use a rule in both directions: find the term at a given stage, and find the stage that yields a given term
• Recognise a decreasing linear pattern and write its rule with a negative step
• Identify the stretch of a table over which a linear rule is valid, and say what happens outside it
• Check a claimed nth-term rule by substituting a value the table already shows

Where it usually goes wrong

• "Find the next term, and you have found the pattern." The next term is one number. The rule has to work at stage 26 as well, and it has to be checkable against stages you have already drawn. Substitute back into the table every time.
• "The rule is 'add two', so the rule is 2n." 2n gives 2, 4, 6 — the right gaps and the wrong values. The step fixes the coefficient; the starting value still has to be paid for, and here it costs a – 1.
• "A decreasing pattern is not linear." Bela's ₹5 a day is as linear as the tiles. What matters is that the gap never changes size, not which way it points.
• "If the gaps are equal somewhere, the whole table is linear." The fare table is the counterexample the chapter itself supplies. Its first gap is zero. Draw the gaps as labelled arcs between cells so the odd one out is visible.
• "n ≥ 2 is just the textbook being careful." Substitute n = 1 and the formula returns ₹10 against a printed ₹25. The restriction is load-bearing.
• "The fare rises by ₹15 per km from the start." It rises by ₹15 for each kilometre after the second. Ten kilometres buys eight increments, not ten — which is exactly why the chapter's calculation reads 15 × 8.
• "Stage 0 must be part of the pattern." The tile pattern is numbered from stage 1 and 2n – 1 would give –1 at n = 0, which is not a count of tiles. Bela's table does start at day 0 — so whether the count starts at 0 or 1 is a fact about each situation, not a rule.
• "Any target number has a stage." 47 tiles happens at stage 24; 48 tiles happens at no stage at all, because 2n – 1 is always odd. Worth one slide.
• "The area of a rectangle is not linear — it has two dimensions." With the length pinned at 13 cm, the area depends on one quantity and is linear in it. Item 3 fixes the length precisely so that this is true.

Questions to check understanding

• Continue a drawn or tabulated pattern for two or three further stages
• Find a rule for the nth term of a linear pattern and verify it against a given stage
• Use a rule in reverse to find which stage produces a stated value, and justify when no stage does — End-of-Chapter item 12 on pp. 38–39 asks exactly this about 200 matchsticks in a hexagon pattern
• Express a described situation as a linear pattern, including a decreasing one
• Given a table, decide whether the pattern is linear and say over which stretch
• Compute the common difference and relate it to the coefficient in the rule

Examples worth working on the board

Inputs below. Values marked verified are worked out here on the chapter's data; the chapter prints answers only where noted, and this book has no appended answer key.

• Fig. 2.4, the tile pattern (p. 21, captioned as a growing pattern of square tiles). The tiles were counted on a close-up of this figure. Stage 1 is a single square. Stage 2 is a row of two squares with one further square resting on the right-hand one — three tiles. Stage 3 has two rows of two with the same single square capping the right column — five tiles. Stage 4 has three rows of two plus the cap — seven tiles. So the shape is a two-wide block that gains one row per stage, with one lone square always sitting on top of the right column.
• The chapter's table (p. 22). Stages 1 to 7 across the top; the tile counts beneath them are 1, 3, 5, 7, 9, 11 and 13.
• The chapter's own generalisation (p. 22). It states that each stage adds two tiles to the previous one; that each stage's count falls one short of double the stage number, giving 2 × 2 – 1 = 3 at stage 2 and 2 × 5 – 1 = 9 at stage 5; that the rule for stage n is 2n – 1; that 2n – 1 has degree 1 and so is linear; that the gaps in 1, 3, 5, 7, 9 … are all 2; and that the pairing of stage number with tile count is a linear relationship.
• The geometric derivation, which the chapter does not give. Verified, and an added argument — read off the figure, stage n is a two-wide block of n – 1 rows, which is 2(n – 1) tiles, plus the one capping square. So the count is 2(n – 1) + 1, and multiplying out gives 2n – 1. Check: at n = 4, 2 × 3 + 1 = 7, which matches the table. This form is worth showing because each of its three pieces points at something in the drawing — the 2 is the row width, the n – 1 is the number of rows added since stage 1, and the + 1 is the cap. The chapter reaches 2n – 1 by inspecting the numbers instead.
• Think and Reflect, p. 21. Predict the counts for the next three stages and write the sequence out to stage 7. The inputs are the four drawn stages and the target stage 7.
• Think and Reflect, p. 22. Using 2n – 1, find the tile counts at stage 15 and stage 26, and find which stages hold 21 tiles and 47 tiles. Hand over 15, 26, 21 and 47. Verified — 29 and 51 tiles; stages 11 and 24. The last two are the rule run backwards, and both targets are odd, which is not an accident: 2n – 1 can never be even, so an even target would have no stage at all. The chapter does not point this out.
• Example 7, Bela's pocket money (p. 22). Bela starts with ₹100 and spends ₹5 a day. The question asks after how many days she is left with ₹40. The printed table gives day 0 with ₹100, then day 1 as 100 – 1 × 5 = 95, day 2 as 100 – 2 × 5 = 90, day 3 as 100 – 3 × 5 = 85, and day 4 as 100 – 4 × 5 = 80. The chapter states the rule for the nth day as ₹(100 – 5n) and states that day 12 leaves ₹40, computed as 100 – 12 × 5.
• Why Bela's table is the same idea inverted. Verified — the gaps are all –5. Nothing about the reasoning changes; only the sign of the step does. Note that this table does show day 0, unlike the chess club table two topics back, so here the constant term 100 is visible in the data.
• Think and Reflect, p. 23. What is left on the 15th day, and how many days until the money is gone? Verified — ₹25 on day 15, and 100 – 5n = 0 at n = 20.
• Example 8, the auto-rickshaw fare (p. 23). The fare opens at ₹25 and stays there for the first 2 km; after that each further kilometre adds ₹15. The chapter asks for the fare over 10 km and computes 25 + 15 × 8 = ₹145. The printed table runs: 1 km → ₹25; 2 km → ₹25; 3 km → 25 + 1 × 15 = 40; 4 km → 25 + 2 × 15 = 55; 5 km → 25 + 3 × 15 = 70; 6 km → 25 + 4 × 15 = 85. The chapter then states the fare for n km as 25 + 15 × (n – 2) = 15n – 5, valid when n ≥ 2, and calls the fare a function of the distance.
• The two forms agree. Verified — expanding 25 + 15(n – 2) gives 25 + 15n – 30, which is 15n – 5. Check at n = 10: 150 – 5 = 145, the chapter's own figure.
• Why the restriction is real, not decorative. Verified — at n = 1 the formula returns 15 – 5 = 10, while the table plainly shows ₹25. So the rule is not merely unproven below 2 km; it is wrong there. And the sequence of fares 25, 25, 40, 55, 70, 85 has a first gap of 0 and every later gap 15, so it is not a constant-difference run over its whole length. This is the chapter's own demonstration that the linear description applies to a stretch of a table.
• Think and Reflect, p. 23. For what distance is the fare ₹130? Hand over the 130. Verified — 15n – 5 = 130 gives n = 9, comfortably inside the valid range.
• The chapter's definition (p. 23). It states that each of these rules is an expression of degree 1 in n, and defines a linear pattern by the equality of the gaps between successive terms. It then points forward to a later chapter on sequences and progressions.
• Exercise Set 2.3 (pp. 23–24), five items, all inputs to hand over:
  • A student holds ₹500 in a savings account and receives ₹150 of pocket money each month. Give the amount at the end of each month from the second onward, and a linear expression for the nth month.
  • A rally begins with 120 members and loses 9 of them every hour. Give the numbers remaining after 1, 2, 3 … hours, and a linear expression for the nth hour.
  • A rectangle of length 13 cm: find the area when the breadth is 12 cm, 10 cm and 8 cm, and give the linear pattern for the area.
  • A rectangular box of length 7 cm and breadth 11 cm: find the volume when the height is 5 cm, 9 cm and 13 cm, and give the linear pattern for the volume.
  • Sarita reads 20 pages a day of a 500-page book: how many pages remain after 15 days, expressed as a linear pattern.
• What items 3 and 4 are quietly doing. Verified — the breadths 12, 10, 8 step by –2 and the heights 5, 9, 13 step by 4, so neither list steps by one. Areas come out 156, 130, 104 and volumes 385, 693, 1001, so the area gaps are –26 and the volume gaps 308 — each the coefficient times the input step (13 × 2 and 77 × 4). These two items are the best evidence in the chapter that the step in the output tracks the step you chose in the input.

Figures to have open

• The four tile stages, redrawn as a schematic rather than reproduced from Fig. 2.4 (p. 21), but keeping the arrangement exactly: a two-wide block that gains a row per stage, with one square capping the right column. Sections 3 and 4 depend on the arrangement, not just the counts, so it must be right.
• A gap-annotation layer that draws labelled arcs between consecutive cells of a table. Needed three times — tiles, Bela, fares — and it is what makes the fare table's odd first gap visible. Standard schematic.
• The fare table with a shadeable valid range, so n ≥ 2 can be seen rather than only stated. Standard schematic.
• No photograph or artwork from the book is required.

Where this sits in the book

• NCERT Ganita Manjari, Class 9, printed Chapter 2, "Introduction to Linear Polynomials", §2.3 "Exploring linear patterns", pp. 21–24. Fig. 2.4 is at the foot of p. 21; the table and the derivation of 2n – 1 are on p. 22; Example 7 begins on p. 22 and concludes on p. 23; Example 8 and the definition of a linear pattern are on p. 23; Exercise Set 2.3 runs from the foot of p. 23 to the top of p. 24.
• Four Think and Reflect boxes bear on this topic: p. 21 (extend the pattern), p. 22 (stages 15 and 26, targets 21 and 47), and two separate boxes on p. 23 — Bela's day 15 and the days to zero, after Example 7, then the ₹130 fare, after Example 8. The Worked examples section above lists those last two individually.
• Backward link inside the chapter: the constant-step property is first stated on p. 19 for the square and the chess club, covered in What makes a polynomial linear, and the equation you get by fixing its value.
• Forward links inside the chapter: the tile pattern returns on p. 26 written as y = 2x – 1, and on p. 31 the chapter states that the slope of that line is the constant gap in the sequence. That is the payoff this topic sets up, and it lands in What a and b do to the line: slope, y-intercept, and parallel families.
• Forward pointer out of the chapter: p. 23 names a later chapter on sequences and progressions.
• End-of-Chapter Exercises, item 5 on p. 37 and starred item 12 on pp. 38–39, are the assessment forms.