Common difference, and why the nth term is a + (n − 1)d

What to assume they know

• Sequences, terms and subscript notation (Order is the point: a sequence carries position as well as value)
• Explicit rules, and testing membership by solving an equation (An explicit rule computes any term straight from n)
• Recursive rules and the role of the seed (A recursive rule builds each term from the ones before it)
• A constant difference as the mark of a linear pattern (A constant difference is the signature of a linear pattern)
• Solving a pair of linear equations in two unknowns, for recovering a and d
• Arithmetic with negative numbers, fractions and decimals

What they should be able to do

• Test a given sequence for a constant difference and decide whether it is an arithmetic progression
• Identify the first term and the common difference of a given AP, including when the common difference is negative, fractional or decimal
• Derive tₙ = a + (n − 1)d by counting the additions between the first term and the nth
• Explain why the coefficient is (n − 1) and not n
• Write the recursive rule of an AP alongside its explicit rule
• Find any term of an AP, and find which term equals a given value
• Recover a and d from two stated terms by solving a pair of equations
• Model a fixed-charge-plus-fixed-rate situation as an AP and say which quantity is the first term

Where it usually goes wrong

• **"The nth term is a + nd."** Test it on the first term: that would make t₁ = a + d, one step too far along. Have students point at the first term and say how many additions they have performed to be standing there.
• "An arithmetic progression increases." 11, 7, 3, −1, −5 is one of the chapter's own examples, with d = −4. Falling is what a negative common difference looks like, not a different kind of object.
• **"d is the difference between any two terms."** It is the difference between consecutive terms. Positions 3 and 50 differ by 47d, and the item-4 hint on p. 185 depends on that.
• **"In 200 + 40n the 200 is the first term."** The first term is 240. The 200 is the booking fee — the value at zero kilometres, a position this sequence does not have. Two correct-looking constants, only one of them a term.
• "Reaching ₹7,00,000 as the eleventh figure means eleven years." Ten raises produce eleven salary figures. Same fence-post as the formula, in the wording an examination paper actually uses.
• "A common difference is enough to identify an AP." It is enough to identify the family. Two of the p. 182 exercises share d = 2 and are different sequences; you need a as well.
• **"Any sequence with a formula in n is an AP."** 1, 4, 9, 16 has a formula and gaps of 3, 5, 7. Constant difference is the test, not the existence of a rule.
• "0 cannot be a term of a progression." In 21, 18, 15, … it is the eighth term. Nothing privileges zero.

Questions to check understanding

• Decide whether a given sequence is an AP and state a and d
• Find a stated distant term of an AP
• Find which term of an AP equals a given value, and whether a given value occurs at all — including the "is 0 a term?" form
• Write both the explicit and the recursive rule for a given AP
• Find a term of an AP from two other stated terms, via a pair of equations
• Model a fixed-fee-plus-rate or fixed-rise situation as an AP and answer a "after how many years/kilometres" question
• Count how many terms an AP has between two bounds — the form End-of-Chapter problems 3 and 4 use for three-digit multiples of 7 and multiples of 4 between 10 and 250

Examples worth working on the board

Values marked verified are worked out here on the chapter's stated data; this book prints no answer key.

• Fig. 8.3 (§8.4, p. 180). Four stages of a tile pattern on a single pink panel, labelled Stage 1 to Stage 4 beneath them. Read off the printed page: each stage has one red tile at the centre and green tiles arranged in a diagonal cross, and each new stage adds four green tiles, one at each arm's end. The counts are 1, 5, 9, 13. The chapter's own Think and Reflect asks for stages 5 and 6, 10, 11 and 12, stage 20, and then any stage — a deliberate escalation from walking to needing a rule.
• The build-up written out (§8.4, pp. 180–181). The page prints the counts first as 1, 1 + 4, 1 + 4 + 4, 1 + 4 + 4 + 4, …, then regroups them as 1, 1 + 1 × 4, 1 + 2 × 4, 1 + 3 × 4, …, and reads the multipliers off as tₙ = 1 + (n − 1) × 4, simplifying to tₙ = 4n − 3. This is the derivation the whole topic turns on: the multiplier in each row is one less than the position number, because that row has had that many additions.
• The first six terms of that pattern (p. 181): 1, 5, 9, 13, 17, 21, with the gap stated to be a constant 4.
• Two more APs named on p. 181. The sequence 1, 4, 7, 10, … is given as tₙ = 1 + (n − 1) × 3, first term 1 and common difference 3. The sequence 11, 7, 3, −1, −5, … is called an AP whose opening value is 11 and whose fixed step is −4. Verified: its explicit rule is 15 − 4n.
• The general form (p. 181): the terms run a, a + d, a + 2d, a + 3d, …, a + (n − 1)d.
• The recursive form (§8.4.1, p. 183): the same progression is t₁ = a with tₙ = tₙ₋₁ + d for n ≥ 2. The page then re-derives the specific case t₁ = 1, tₙ = tₙ₋₁ + 4 for the tile pattern.
• Two exercises on recognising an AP (§8.4.1, p. 182): confirm that 2, 5, 8, 11, … and −5, −1, 3, 7, … are APs and write their nth terms. Verified: 3n − 1 and 4n − 9.
• Two exercises with non-integer data (p. 182): find the nth term of ½, 5/2, 9/2, 13/2, … and of 1.5, 3.5, 5.5, 7.5, …. Verified: the first has a = ½ and d = 2, giving (4n − 3)/2; the second has a = 1.5 and d = 2, giving 2n − 0.5. Note that both have the same common difference as each other and different first terms, which is the point worth making.
• Example 5, the taxi (§8.4.1, p. 183). A fixed booking fee of ₹200, plus ₹40 for each kilometre. The page computes the fare after 1 km as ₹240, after 2 km as ₹280, after 3 km as ₹320, calls the resulting sequence 240, 280, 320, … an AP whose opening value is 240 and whose fixed step is 40, and gives the nth term as 240 + (n − 1) × 40, simplified to 200 + 40n. It asks for the fare over 10 km. Verified: ₹600. Two things when explaining it: ₹200 is the fare at zero kilometres and is not a term of this sequence, and the two printed forms of the rule are the same rule — 240 + (n − 1) × 40 shows the first term, 200 + 40n shows the fixed fee.
• **Recovering a and d from two terms** (Exercise Set 8.2 item 4, p. 185). An AP of 50 terms has 12 in position 3 and 106 as its last term; the printed hint sets up a + 2d = 12 and a + 49d = 106 and asks for the 29th term. Verified: 47d = 94, so d = 2, a = 8, and the 29th term is 64. The hint is doing the work of section 10: subtracting the two equations counts the 47 steps between positions 3 and 50.
• Exercise Set 8.2 items 1–3 (p. 185). The 10th and 26th terms of 3, 8, 13, 18, …; which term of 21, 18, 15, … equals −81, and whether 0 occurs in it; and the nth term plus recursive rule of 11, 8, 5, 2, …. Verified: 48 and 128; −81 sits in position 35 and 0 does occur, in position 8; the third is 14 − 3n with the step rule t₁ = 11, tₙ = tₙ₋₁ − 3. The 0 question is worth attention — students expect the answer to be no because 0 looks special.
• The salary problem (Exercise Set 8.2 item 6, p. 186). A starting annual salary of ₹5,00,000 rising by ₹20,000 each year; after how many years does it reach ₹7,00,000? Verified: the gap is ₹2,00,000, which is ten increments, so ₹7,00,000 is the eleventh term of the sequence and is reached after 10 years. This is the (n − 1) lesson wearing everyday clothes, and it is the single best assessment item in the section.
• Two end-of-chapter items in the same mould (p. 194). Problem 1: the 11th term is 38 and the 16th is 73, find the 31st. Verified: five steps carry 35, so d = 7, a = −32, and the 31st term is 178. Problem 2: the third term is 16 and the 7th exceeds the 5th by 12. Verified: two steps carry 12, so d = 6, a = 4, and the AP runs 4, 10, 16, 22, ….

Figures to have open

• Fig. 8.3 redrawn as a schematic: four stages, the central tile in one colour and the four tiles added at each stage highlighted so the constant increase is visible rather than asserted. Must keep the counts 1, 5, 9, 13. The printed art is a decorative painted panel and does not need reproducing.
• A step-counting strip: the terms of a general AP in a row with the additions numbered 0, 1, 2, 3, … beneath the gaps. This is the figure the chapter does not draw and section 4 cannot do without. Standard schematic.
• A fare diagram for Example 5 with the ₹200 fee drawn at position zero, outside the sequence, and the fares 240, 280, 320 as the terms. Standard schematic.
• A two-equation panel for the item-4 hint, showing the subtraction that isolates 47d. Standard schematic.

Where this sits in the book

• Chapter 8, §8.4 Arithmetic Progressions, pp. 180–181, and the first half of §8.4.1 Visualising an AP, pp. 181, 183 — the graphing half of §8.4.1 belongs to An AP plots as points on a straight line.
• Fig. 8.3 and its caption, p. 180. The stage-and-count table is at the foot of p. 181. Example 5 is on p. 183.
• Two Think and Reflect boxes: predicting the tile counts at distant stages (p. 180) and sorting the chapter's earlier sequences into APs and non-APs (p. 181). The second is the section's best diagnostic question.
• Three inline Exercise prompts on pp. 182–183: verifying two sequences are APs, finding nth terms with fractional and decimal data, and writing recursive rules for them.
• Exercise Set 8.2, pp. 185–186, items 1–4 and 6. Items 5 and 7 need a sum and belong to Pairing from both ends: a closed form for 1 + 2 + ⋯ + n.
• End-of-Chapter Exercises, p. 194, problems 1 and 2, and p. 195 problems 3, 4 and 8; the last three carry a printed asterisk.
• The summary's statement of the AP definition and of a + (n − 1)d is on p. 196.