A constant ratio between neighbours is the entire definition

What to assume they know

• A rule that turns a position number into a term — a sequence as a rule assigning a term to each position, and subscript notation
• Division of fractions, including fractions with a negative numerator
• Arithmetic with decimals that have several leading zeros after the point
• Arithmetic progression from the previous class — the constant difference, which this section is set against
• That division by zero has no value

What they should be able to do

• Test a given sequence for constant ratio by dividing consecutive terms, and state the verdict with the working shown
• Compute a common ratio that is negative, fractional or a small decimal
• State the chapter's definition of a geometric progression, including the requirement that no term is zero
• Explain what fails if a term of a candidate sequence is zero, and hence why neither the first term nor the ratio may be zero
• Write a G.P. in the standard form built from its first term and its common ratio
• Find a missing entry that makes three given numbers a G.P., and account for both sign choices
• Recognise that a constant ratio must hold at every step, not merely at the step you checked

Where it usually goes wrong

• "A G.P. is a sequence that grows fast." One of the chapter's own three lists shrinks towards zero and another alternates in sign while shrinking. Growth is not the criterion; constancy of the ratio is.
• "The common ratio has to be a whole number bigger than one." The section's own answers are 2, −1/3 and 0.01. Later items add √2 and √3.
• "Consecutive terms differ by the same amount." That is the previous class's arithmetic progression, and the section names it on the same page in order to be contrasted with it. Students blend the two constantly.
• "A zero term is harmless if the rest of the sequence behaves." It is not. The definition is built on dividing by each term, and the first zero makes that division meaningless from then on. Show the step where it breaks.
• "A ratio of zero is allowed — you just get a, 0, 0, 0, …" The non-zero clause rules this out, and so does a first term of zero. Both are worth naming explicitly, because neither is spelt out on the page and both follow from one short condition.
• "Checking one pair of neighbours settles it." The definition demands the same value at every step. Three terms in ratio prove nothing about the fourth.
• "Divide the earlier term by the later one." Half of all wrong ratios in this chapter are the reciprocal of the right one. Fix the direction with a drawn arrow that never changes.
• "The textbook cannot be wrong, so my minus sign must be." Section 9 exists for this. The three ratios printed for list two lost their signs; the student's own arithmetic is the authority.

Questions to check understanding

• Decide whether a listed sequence is a G.P., showing the division at more than one step
• Compute a common ratio that is negative, fractional, decimal or irrational
• Insert a missing term so that three given numbers form a G.P., reporting both admissible values
• State the definition and say what the non-zero condition excludes
• Prove that a sequence built from two G.P.s, term by term, is itself a G.P., and give its ratio
• Explain in words why a sequence containing zero cannot be a G.P.

Examples worth working on the board

Values marked verified are worked out here on the chapter's stated inputs.

• The three opening lists (§8.4, p. 139). List one: 2, 4, 8, 16, and onward. List two: 1/9, then −1/27, then 1/81, then −1/243, and onward. List three: 0.01, then 0.0001, then 0.000001, and onward. The page works the divisions out for the first two lists and hands the third to the reader. Verified: list one gives 2 at every step; list two gives −1/3 at every step, since −1/27 divided by 1/9 is −9/27; list three gives 0.01 at every step, since 0.0001 divided by 0.01 is 0.01. These three are the whole argument of sections 1 to 4 — an explanation that swaps in tidier numbers destroys the point, because tidier numbers are recognisable and these are chosen not to be.
• The definition (§8.4, p. 139). Stated as a condition on the sequence: the terms are all non-zero, and dividing the term at position k + 1 by the term at position k returns the same constant for every k from 1 upward.
• The standard form (§8.4, p. 139). Setting the opening term to a turns the sequence into a, ar, ar², ar³, and onward. Note: two numbers, a and r, pin down the entire progression — that is what makes every formula in the rest of the chapter possible.
• The notation list (§8.4, p. 140). The chapter fixes a for the first term, r for the common ratio, l for the last term, n for how many terms there are, and Sₙ for the total of the first n. Show this once and leave it available; the next three topics all draw on it.
• A misprint worth showing the student (§8.4, p. 139). Where the page works out the three consecutive ratios for list two, each is printed as 1/3 with no minus sign — yet two lines below, and again in the closing line of the section, the common ratio of that list is given as −1/3. Verified: −1/3 is correct at every step. This is a gift to a teacher: a student who does the division honestly will disagree with one line of the textbook and be right.
• A missing middle term (Exercise 8.2 item 6, p. 145). Inputs: −2/7, then an unknown, then −7/2, required to be a G.P. Verified: the unknown squared must equal the product of the outer two, which is 1, so the unknown is 1 or −1, and both genuinely work — one gives a ratio of −7/2, the other of 7/2. Zero is excluded twice over, by the equation and by the definition.
• Sequences to be tested (Exercise 8.2 item 5, p. 145). Inputs: 2, 2√2, 4, …; √3, 3, 3√3, …; and 1/3, 1/9, 1/27, …. Verified: the ratios are √2, √3 and 1/3 respectively. The first two are the useful ones here — an irrational ratio is still a constant ratio, which is exactly what the test is indifferent to.
• Two progressions multiplied term by term (Exercise 8.2 item 20, p. 146). Inputs: a, ar, ar², … arⁿ⁻¹ alongside A, AR, AR², … ARⁿ⁻¹, with the corresponding terms multiplied. Verified: each new term divided by its predecessor gives rR, so the result passes the test. This is the best item in the exercise for the thesis, because it can only be settled by running the test — the resulting list is not recognisable by eye.

Figures to have open

• A ladder of consecutive terms with a division arrow drawn between each neighbouring pair, all arrows pointing the same way. This carries sections 2, 3, 4 and 6 and must be reused unchanged so the direction never comes into question. Standard schematic; the chapter draws nothing.
• The same ladder with one rung set to zero and the arrow leaving it marked as having no value. Standard schematic.
• A side-by-side of a constant-difference ladder and a constant-ratio ladder for section 5's contrast. Standard schematic.
• Nothing can be lifted from the book. All sixteen printed pages were opened as images: this chapter carries no numbered figure, and its only illustration is the portrait on p. 135.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 8 "Sequences and Series", §8.4 Geometric Progression (G. P.), pp. 139–140, including the definition and the notation list at the head of p. 140
• Exercise 8.2 items 5, 6 and 20, pp. 145–146
• The Summary, p. 149, restates the constant-ratio definition and names a and r, but does not repeat the non-zero condition
• §8.1, p. 135, names arithmetic progression as prior learning