Working backwards from a term to its position, or to a or d

What to assume they know

• Building the rule for the nth term out of repeated addition — the nth-term rule and why the step is taken n − 1 times
• Testing a given list, and what "finite" versus "infinite" changes — confirming that a given list really is an AP before using any AP machinery on it
• Solving a linear equation in one unknown, including one whose coefficient is negative
• Solving a pair of simultaneous linear equations by elimination
• The simple-interest relationship from earlier classes

What they should be able to do

• Rearrange the nth-term rule to make any one of the four quantities the subject
• Find the position of a stated entry in a given AP
• Decide whether a given number belongs to a given AP, justifying it by the two demands on the position — a whole number, and not before the first entry — rather than by how big the number looks
• Count the entries of a finite AP whose first entry, step and last entry are known
• Recover the first entry and the step from any two stated entries by solving a pair of equations
• Locate an entry counted from the end of a finite AP, by two independent methods
• Turn a described situation — interest, plantings, wages — into an AP and answer a positional question about it
• Complete a partly filled table of a, d, n and aₙ by choosing which quantity each row leaves unknown

Where it usually goes wrong

• "301 is too big to be on the list, so the answer is no." The list is infinite and passes 301 without stopping; every entry beyond 299 is larger. The reason 301 is absent is that 296 is not a multiple of 6. Size reasoning gets the right verdict here by accident and will get the wrong one next time.
• "n came out as a fraction, so I made an arithmetic mistake." Sometimes a fractional n is the answer, and what it tells you is that the number is not an entry. The chapter treats the fraction as information, not as an error.
• "The eleventh from the end is the fourteenth from the start." With 25 entries it is the fifteenth, because the last entry is the first one counted backwards. The chapter raises this trap deliberately.
• "A decreasing AP cannot reach zero or go negative." Example 4 does both, and Example 8 ends at −62. Negative entries are ordinary.
• "Reversing the AP is a different problem." It is the same list read the other way; the first and last entries swap and the step changes sign. Solving it twice and getting −32 both times is the demonstration.
• "Simple interest and compound interest both give APs." Only simple interest does, because the amount added each year is computed on the unchanging original sum. Do not credit Exercise 5.1 with this contrast: its four situations are a taxi fare, a pump taking a fixed fraction of what is left, a digging cost rising by a fixed amount per metre, and a compound-interest deposit — so what it separates is fixed addition from fixed multiplication. Simple interest does not appear in this chapter until Example 9, which is where the contrast is actually available.
• "Two stated entries are not enough to find the AP." They are two equations in two unknowns, and they always suffice unless they name the same position twice.

Questions to check understanding

• Asked which position of a stated AP holds a given value — the form set most often
• Asked whether some named number appears anywhere in a stated AP, justification required
• Count the terms of a finite AP given its first entry, step and last entry
• Find a and d from two stated entries, then answer a further question about the same AP
• Find a term counted from the last term, by either method
• Complete a table with one of a, d, n, aₙ missing per row
• Word problems on wages, savings and plantings that hide a positional question
• Compare two APs — equal terms, differing terms, shared common difference

Examples worth working on the board

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

• Example 4 (p. 58). The AP 21, 18, 15, … Two questions are asked of it: which position holds −81, and whether any position holds 0. Verified: a = 21 and d = −3; setting the nth entry to −81 gives 21 − 3(n − 1) = −81, so 3(n − 1) = 102 and n = 35. Setting it to 0 gives 3(n − 1) = 21, so n = 8. Both answers are positive whole numbers, so both are genuine entries — note that a decreasing AP passes through zero without anything special happening.
• Example 5 (p. 58). An AP whose third entry is 5 and whose seventh is 9. Verified: a + 2d = 5 and a + 6d = 9, so subtracting gives 4d = 4, hence d = 1 and a = 3, and the AP opens 3, 4, 5, 6, 7, … The structural point is that two stated entries are two equations, and two equations settle two unknowns.
• Example 6 (p. 59). Is 301 an entry of 5, 11, 17, 23, …? Verified: the differences are 6, 6, 6, so a = 5 and d = 6; solving 5 + 6(n − 1) = 301 gives 6n = 302, so n = 151/3. That is not a whole number, so 301 is not on the list. This is the topic's centre of gravity — 301 sits comfortably inside the range the list eventually covers, and is still not on it, because 301 − 5 = 296 is not a multiple of 6.
• Example 7 (p. 59). How many two-digit numbers does 3 divide? Verified: the list runs 12, 15, 18, …, 99, with a = 12, d = 3 and last entry 99; solving 12 + 3(n − 1) = 99 gives n − 1 = 29 and n = 30. Here the unknown is the count rather than a value, and it is the same equation.
• Example 8 (pp. 59–60). The AP 10, 7, 4, …, −62, and the entry lying eleventh from the end. Verified: a = 10, d = −3, last entry −62; solving 10 − 3(n − 1) = −62 gives n − 1 = 24 and n = 25 entries in all. The eleventh from the end is therefore the fifteenth from the start, because 25 − 11 + 1 = 15 — the last entry is itself the first one counted from the end. Its value is 10 − 3 × 14 = −32. Verified by the second method too: written backwards the AP starts at −62 with step +3, and −62 + 10 × 3 = −32. The chapter flags the fourteenth-versus-fifteenth trap explicitly.
• Example 9 (pp. 60–61). ₹1000 at 8% simple interest per year; the interest earned by the end of each year. Verified: the standard simple-interest computation gives 80 after one year, 160 after two and 240 after three, so the yearly totals form an AP with a = 80 and d = 80; the total by the end of thirty years is 80 + 29 × 80 = 2400. Worth pointing out that d here equals a, which is a special feature of simple interest and not a general rule.
• Example 10 (p. 61). A flower bed whose rows carry 23, then 21, then 19 rose plants, dropping by twos until a final row holding 5. The question is the number of rows. Verified: a = 23, d = −2, final entry 5, and solving 23 − 2(n − 1) = 5 gives n − 1 = 9 and n = 10 rows. The unknown is again a count.
• Exercise 5.2 question 1 (p. 61), a five-row table with one blank per row. Read off the printed page, the rows are: a = 7, d = 3, n = 8, aₙ blank; a = −18, d blank, n = 10, aₙ = 0; a blank, d = −3, n = 18, aₙ = −5; a = −18.9, d = 2.5, n blank, aₙ = 3.6; a = 3.5, d = 0, n = 105, aₙ blank. This table is the topic — five rows, one equation, and the blank moving between four of the five places. It does not move on every row: rows (i) and (v) both leave the entry itself blank, so the last row returns to where the first started, after d, a and n have each had a turn.
• Exercise 5.2 question 3 (p. 62), five short APs with boxed gaps: 2, ▢, 26; ▢, 13, ▢, 3; 5, ▢, ▢, 9½; −4, ▢, ▢, ▢, ▢, 6; ▢, 38, ▢, ▢, ▢, −22. Verified on the printed page as five items with 1, 2, 2, 4 and 4 boxes respectively. These are solved by treating the two known entries as two equations, exactly as in Example 5.
• Further Exercise 5.2 data worth handing over (pp. 62–63): which entry of 3, 8, 13, 18, … equals 78; how many entries in 7, 13, 19, …, 205 and in 18, 15½, 13, …, −47; whether −150 belongs to 11, 8, 5, 2, …; an AP with eleventh entry 38 and sixteenth entry 73, asked for its thirty-first; a fifty-entry AP with third entry 12 and last entry 106, asked for its twenty-ninth; an AP whose third and ninth entries are 4 and −8, asked which entry is zero; an AP whose seventeenth entry exceeds its tenth by 7; which entry of 3, 15, 27, 39, … exceeds the fifty-fourth by 132; two APs sharing a step whose hundredth entries differ by 100, asked about their thousandth entries; how many three-digit numbers 7 divides; how many multiples of 4 lie between 10 and 250; the position at which 63, 65, 67, … and 3, 10, 17, … agree; an AP with third entry 16 whose seventh exceeds its fifth by 12; the twentieth entry from the end of 3, 8, 13, …, 253; an AP whose fourth and eighth entries sum to 24 and whose sixth and tenth sum to 44; Subba Rao starting in 1995 at ₹5000 a year with ₹200 added annually, asked when he reaches ₹7000; Ramkali saving ₹5 in the first week and adding ₹1.75 each week, asked for the week in which she saves ₹20.75.

Figures to have open

• A four-slot diagram of the nth-term relation with a movable "unknown" marker. Standard schematic; it is the spine of the whole topic.
• A number-line strip for 5, 11, 17, 23, … with the entries marked and 301 placed between two of them, plus the remainder computation beside it. Standard schematic, and it is what makes section 5 stick.
• A double-numbered strip for a 25-entry AP: positions 1 to 25 above, positions counted from the end below, with the pair (15, 11) picked out. Standard schematic.
• The flower-bed rows as a simple stepped block diagram, 23 down to 5 in twos. The chapter prints no figure for Example 10 — so this must be drawn fresh.
• The Exercise 5.2 table itself, reproduced as data with its blanks intact.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, Chapter 5 "Arithmetic Progressions", §5.3, pp. 58–61 — Examples 4 to 10
• Exercise 5.2, pp. 61–63, questions 1 to 20
• The rule being rearranged throughout is derived on p. 57 and restated as point 3 of the summary, §5.5, p. 72
• Example 9 leans on the simple-interest relationship from earlier classes, which the chapter restates on p. 60 rather than deriving