Two versions of the total, and choosing between them

What to assume they know

• Gauss's pairing trick, and why it generalises — the reversal-and-stacking derivation, and why the halving is there
• Working backwards from a term to its position, or to a or d — solving the nth-term relation for whichever quantity is unknown
• Solving a quadratic equation by factorising
• Rearranging a formula containing a bracket, and substituting an expression for a letter

What they should be able to do

• Show that the two printed totals are the same statement, by substituting the nth-term expression for the last entry
• Choose the appropriate version of the formula from the quantities a question supplies
• Compute a total by direct substitution when the first entry, step and count are known
• Solve for the step, the first entry or the count when the total is known instead
• Recognise that solving for the count produces a quadratic, and interpret both roots when both are positive whole numbers
• Explain why two different counts can share a total, in terms of the entries in between summing to zero
• Recover a single entry from consecutive totals, and justify why the difference of two totals is one entry
• Build the list first from a given general term, then total it

Where it usually goes wrong

• "There are two formulas, so there are two rules to memorise." There is one rule. The second version is what the first becomes once the last entry is known, and deriving it in two lines removes the second memorisation entirely.
• **"Use the l version when the numbers are nicer."** Use it when you were given a last entry and no step. The chapter says exactly what it is for, and the 1-to-1000 example is the case where the step is never needed.
• "Two answers means one of them is wrong." Both 4 and 13 are positive whole numbers and both really do give 78. Rejecting one is the error here — though the check itself still matters, since a negative or fractional root would have to go.
• "The entries between must be small if they cancel." They are not small; they run from 12 down to −12. Nor does a positive start with a falling step make them cancel — that only lets a run cross zero at all. The block cancels because the question's own condition forces it to: two different counts reaching the same total means everything between them adds to nothing. For an AP, adding to nothing and sitting symmetrically about zero are the same fact, so the symmetry is what you see, not why it happens.
• "Sₙ and aₙ are interchangeable." One is a total, the other a single entry. Their relationship is a subtraction of consecutive totals, and Exercise 5.3 question 11 is built to expose the confusion.
• "A list given by a formula is automatically an AP." It is not — the formula has to be linear in the position. Example 15 checks the differences before using any AP result, and that check is not a formality.
• "The total of a decreasing AP must be positive because it starts positive." Example 11 totals −979.
• "The step in a word problem is whatever number appears in the question." In the television problem the yearly rise is 25, and no number in the question says 25 — it has to be solved for.

Questions to check understanding

• Total a stated number of terms of a given AP
• Total an AP presented by its first and last terms
• Given a total, solve for the step, the first term or the number of terms
• Interpret a quadratic with two admissible roots and justify keeping both
• Given the total to n terms as an expression in n, extract particular terms
• Show a list defined by a formula is an AP, then total it
• Word problems on penalties, prizes, plantings, stacked objects and repeated journeys
• Optional-exercise work: a first negative term, a sum from a stated sum and product of two terms, and physical totals from a ladder or a terrace

Examples worth working on the board

Inputs only. Values marked verified are worked out here on the chapter's printed data.

• The two versions, and the bridge between them (§5.4, p. 64). The chapter gets from one to the other in two printed lines: it splits the bracket so that the first entry stands apart from the expression a + (n − 1)d, then names that expression as the entry in position n, and finally as the last entry l when the AP stops there. Verified as algebra: 2a + (n − 1)d is a + [a + (n − 1)d], and the second half is exactly the nth entry, so the two versions are the same statement. The chapter notes that the l version is what you want when the step was never given to you.
• Example 11 (p. 65). Total the first 22 entries of 8, 3, −2, … Verified: a = 8 and d = −5, so the total is 11 × [16 + 21 × (−5)] = 11 × (−89) = −979. A negative total is worth dwelling on: the entries turn negative early and overwhelm the positive start.
• Example 12 (pp. 65–66). An AP whose first 14 entries total 1050, with first entry 10; find the twentieth entry. Verified: 1050 = 7 × (20 + 13d), so 20 + 13d = 150, giving d = 10; then the twentieth entry is 10 + 19 × 10 = 200. This is the topic's two-stage shape — the total gives the step, the step gives the entry.
• Example 13 (p. 66). How many entries of 24, 21, 18, … total 78? Verified: a = 24, d = −3, so the total condition becomes n(51 − 3n)/2 = 78, that is 3n² − 51n + 156 = 0, which reduces to n² − 17n + 52 = 0 and factorises as (n − 4)(n − 13). Both 4 and 13 are positive whole numbers, so both are genuine answers. Verified directly: 24 + 21 + 18 + 15 = 78, and the first thirteen entries also total 78. The chapter attaches two remarks explaining that the entries from the fifth to the thirteenth cancel out — verified: those nine entries run 12 down to −12 in steps of −3, symmetric about 0, so they total zero. Show the cancellation; it is what makes two answers reasonable rather than suspicious.
• Example 14 (pp. 66–67). Total the first 1000 counting numbers, then the first n. Verified: with first entry 1 and last entry 1000, the total is 500 × 1001 = 500500; in general the same version of the formula gives n(n + 1)/2. Note that this uses the l version and never mentions the step, which is the cleanest possible advertisement for section 2's argument.
• Example 15 (p. 67). A list whose entry in position n is given by the expression 3 + 2n; total its first 24 entries. Verified: the entries begin 5, 7, 9, 11, so the step is 2 and the first entry is 5; the total is 12 × [10 + 23 × 2] = 12 × 56 = 672. The order matters — the list has to be shown to be an AP before any AP formula is allowed near it.
• Example 16 (pp. 67–68). A manufacturer makes 600 television sets in the third year and 700 in the seventh, output rising by a fixed number annually. Asked for the first year's output, the tenth year's, and the total across the first seven. Verified: a + 2d = 600 and a + 6d = 700 give d = 25 and a = 550; the tenth year is 550 + 9 × 25 = 775; the seven-year total is (7/2) × [1100 + 150] = 4375.
• The remark connecting totals to entries (p. 65). The entry in position n equals the total to n less the total to n − 1. Verified as an argument: the two totals contain the same entries except that the longer one also contains the nth, so subtracting leaves precisely that entry. Exercise 5.3 question 11 sets this up as a whole problem: the total of the first n entries is 4n − n², and the first entry, the first two entries' total, the second entry, and then the third, tenth and general entries are wanted. Verified as a consistency check: that total corresponds to a = 3 and d = −2, since 3 + 1 = 4 and the running totals 3, 4, 3, 0 match 4n − n² at n = 1, 2, 3, 4.
• Exercise 5.3 data worth handing over (pp. 68–69): totals of 2, 7, 12, … over 10 entries; −37, −33, −29, … over 12; 0.6, 1.7, 2.8, … over 100; 1/15, 1/12, 1/10, … over 11. Totals with a stated last entry: 7 + 10½ + 14 + … + 84; 34 + 32 + 30 + … + 10; −5 + (−8) + (−11) + … + (−230). A ten-part question supplying three of the quantities and asking for the rest, including the rows a = 5, d = 3, aₙ = 50; a = 7, a₁₃ = 35; a₁₂ = 37, d = 3; a₃ = 15, S₁₀ = 125; d = 5, S₉ = 75; a = 2, d = 8, Sₙ = 90; a = 8, aₙ = 62, Sₙ = 210; aₙ = 4, d = 2, Sₙ = −14; a = 3, n = 8, S = 192; and l = 28, S = 144 with 9 entries. Then: how many entries of 9, 17, 25, … reach 636; an AP with first entry 5, last 45 and total 400; an AP with first entry 17, last 350, step 9; 22 entries with step 7 and twenty-second entry 149; 51 entries whose second and third are 14 and 18; an AP whose first 7 entries total 49 and first 17 total 289; lists given by the expressions 3 + 4n and 9 − 5n, to be shown to be APs and totalled over 15 entries; the first 40 positive multiples of 6; the first 15 multiples of 8; the odd numbers strictly between 0 and 50; a delay penalty starting at ₹200 and rising ₹50 a day over 30 days; ₹700 split into seven prizes each ₹20 below the one before; three sections per class each planting as many trees as their class number, Classes I to XII; a spiral of thirteen semicircles with radii 0.5, 1.0, 1.5, 2.0 cm and so on, taking π as 22/7 (Fig. 5.4); 200 logs stacked 20 in the bottom row, then 19, then 18 (Fig. 5.5); and a potato race with the bucket 5 m from the first potato, the remaining potatoes 3 m apart, ten in all (Fig. 5.6).
• Exercise 5.4 (p. 71), printed with a footnote marking it as outside the examination's scope — verified on the printed page, where an asterisk on the heading is answered by a rule and a footnote at the foot of the page. Its five items are: the first negative entry of 121, 117, 113, …; an AP whose third and seventh entries sum to 6 and multiply to 8, asked for its first sixteen entries' total; a ladder whose rungs sit 25 cm apart and shorten evenly from 45 cm to 25 cm across a span of 2½ m, asked for the total length of rung wood (Fig. 5.7, which labels 45 cm, 25 cm, the 25 cm spacing and the 2½ m span, and carries no other numbers); houses numbered 1 to 49 with a house whose predecessors and successors total equally; and a terrace of 15 concrete steps each 50 m long, rising ¼ m and treading ½ m (Fig. 5.8).

Figures to have open

• A split-bracket movement for section 1: the bracket opening, the nth-term expression lifting out of it and being relabelled. Standard schematic; this is the whole argument of the topic and the chapter does it in printed text only.
• A paired bar chart for Example 13: four bars totalling 78 beside thirteen bars totalling 78, with bars five to thirteen shaded as a symmetric block about zero. Standard schematic and entirely not in the book — the chapter gives this as two sentences of remark, and the picture is what makes it obvious.
• An inputs-versus-formula matrix: rows for the quantities supplied, columns for the two versions, ticks where a version can run. Standard schematic.
• Fig. 5.4, the spiral of semicircles with centres alternating between two points (p. 70), if the spiral question is used. On the printed page the diagram labels four arcs and marks the two centres on a horizontal line; the radii are given in the question text, not on the drawing.
• Fig. 5.5, the stacked logs (p. 70), if the log question is used. Verified on the printed page: the drawing shows the stack with no numbers printed on it, so the row counts must be added if the figure is reused.
• Fig. 5.6, the potato race (p. 70), if that question is used. Verified on the printed page: the bucket and the line of potatoes are drawn with 5 m and two 3 m gaps labelled, and the remaining gaps left unlabelled.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 5 "Arithmetic Progressions", §5.4, pp. 64–68 — the second version of the formula on p. 64, the remark on p. 65, and Examples 11 to 16
• Exercise 5.3, pp. 68–70, questions 1 to 20, with Figs. 5.4, 5.5 and 5.6 on p. 70
• Exercise 5.4, p. 71, questions 1 to 5, with Figs. 5.7 and 5.8; the heading carries an asterisk and a footnote placing it outside examination scope
• Both versions of the formula are restated as points 4 and 5 of the summary, §5.5, p. 72