Reaching any term without walking through the earlier ones

What to assume they know

• A constant ratio between neighbours is the entire definition — the constant-ratio definition, the standard form a, ar, ar², ar³, …, and the requirement that no term is zero
• Laws of exponents: multiplying powers of the same base, and dividing them
• Recognising a number as a power of a small base — that 65536 is a power of 4, that 19683 is a power of 3
• Solving a simple equation in an exponent by matching both sides to a common base
• Square roots written as fractional powers, needed for two of the exercise items

What they should be able to do

• Derive the rule for the term at position n of a G.P. by counting multiplications, rather than quoting it
• Explain why the exponent is n − 1 and predict a named far-off term before the rule is formally stated
• Compute a distant term of a given G.P. by direct substitution
• Given a term's value, solve for its position by reducing both sides of the equation to a common base
• Recover both defining constants of a G.P. — its opening term and its ratio — from two terms at known but non-adjacent positions
• Show that the quotient of two terms of a G.P. depends only on the difference of their positions, and use that to prove short relations between named terms
• Write a finite and an infinite G.P. in standard form, and the geometric series each one names

Where it usually goes wrong

• **"The term at position n is a times r to the n."** This is the single commonest error on this topic, and it comes from counting terms where the formula counts steps. Every time the rule appears, put the step count beside it.
• **"Example 4 proves the general term is the ratio to the power n."** It does not — the tidy answer there is an accident of the first term and the ratio being equal. Say so at the moment it appears.
• "To find which term has a given value, list terms until you hit it." For 131072 that is eight rounds of arithmetic and it does not scale. Reducing both sides to a common base is the method.
• "Dividing two terms gives the common ratio." It gives the ratio raised to the gap between their positions. Example 6 divides terms three positions apart and gets a cube; a student who skips that reads r = 8.
• "Two given terms are not enough to fix a G.P." They are, provided their positions are known: one division fixes r, one substitution fixes a.
• **"You cannot divide the equations, because a might be zero."** In a G.P. it cannot be — the non-zero condition from the definition is what licenses every division in Example 6 and in the fourth exercise item.
• "A fractional ratio means the terms are not really a progression." The first exercise item has ratio one half and shrinks throughout.
• "With a negative ratio you cannot say the sign of a distant term." You can: the sign is decided by whether the step count is even or odd, which is another place the n − 1 matters.

Questions to check understanding

• Find a named term of a G.P. given the first term and the ratio
• Find the general term of a G.P. given its opening few terms
• Given a value, find which position it occupies, with the common-base step shown
• Given two terms at stated non-adjacent positions, find the first term, the ratio, and a third named term
• Prove a short relation among three terms at equally spaced positions
• Given a condition linking two terms, such as one being the square of another, solve for the ratio

Examples worth working on the board

Values marked verified are worked out here on the chapter's stated inputs. Exercise items are inputs; this chapter prints no answers to them.

• The derivation as the page runs it (§8.4.1, p. 140). The opening term is a, the next is a times r, the next is that times r again, and so on. The page then rewrites the first five terms so that each exponent is displayed as its position minus one, asks the reader to predict the sixteenth term, gives it as a times r to the fifteenth, and only then states the rule. Reproduce that order exactly — the prediction is placed before the rule on purpose, and moving the rule earlier destroys the section.
• Standard form and the series (§8.4.1, p. 140). A finite G.P. runs a, ar, ar², …, arⁿ⁻¹; an infinite one continues past that point. The corresponding additions are named as finite and infinite geometric series.
• Example 4 (p. 140). Inputs: the G.P. 5, 25, 125, …, with the tenth term and the general term wanted. Verified: a = 5 and r = 5, so the tenth term is 5 × 5⁹ = 5¹⁰, which is 9765625, and the general term collapses to 5ⁿ. Flag the collapse when explaining it: it happens only because the first term and the ratio are the same number here, and students routinely generalise it into the wrong rule.
• Example 5 (p. 141). Inputs: the G.P. 2, 8, 32, …, and the value 131072, whose position is wanted. Verified: a = 2 and r = 4; dividing gives 4ⁿ⁻¹ = 65536, and since 65536 = 4⁸ the position is 9. Note that 2 × 65536 does return 131072, so the check closes.
• Example 6 (p. 141). Inputs: 24 sits at position 3, 192 sits at position 6, and the value at position 10 is wanted. Verified: dividing the second by the first leaves r³ = 8, so r = 2; substituting back gives 4a = 24 and a = 6; the tenth term is 6 × 2⁹ = 3072. The page compresses the cube-root step into a single line — open it up, because the exponent 3 is exactly the gap between positions 6 and 3, and that is the point of the example.
• Exercise 8.2, direct substitution items (p. 145). Inputs: the G.P. 5/2, 5/4, 5/8, … with the 20th and general terms wanted; a progression of ratio 2 holding 192 at position 8, its 12th entry wanted; a progression whose 4th entry is the square of its 2nd, opening at −3, its 7th entry wanted. Verified: the first has a = 5/2 and r = 1/2, so the general term reduces to 5 ÷ 2ⁿ, putting 5 ÷ 1048576 at position 20. The second needs four more steps, giving 192 × 16 = 3072. In the third, the condition forces the ratio to equal the first term, so r = −3 and the 7th term is −3 × 729 = −2187.
• Exercise 8.2, find-the-position items (p. 145). Inputs: 2, 2√2, 4, … with target 128; √3, 3, 3√3, … with target 729; 1/3, 1/9, 1/27, … with target 1/19683. Verified: the ratios are √2, √3 and 1/3; matching powers gives the positions 13, 12 and 9 respectively, using 64 = 2⁶, 729 = 3⁶ and 19683 = 3⁹. The first two are the best items in the chapter for showing that an exponent equation is solved by choosing a common base, not by guessing.
• Exercise 8.2, position-gap items (pp. 145–146). Inputs: a progression's entries at positions 5, 8 and 11, named p, q and s, with q² = ps to be shown; its entries at positions 4, 10 and 16, named x, y and z, to be shown geometric themselves; entries at three unspecified positions p, q and r, named a, b and c, with a product of three powers to be shown equal to 1; the first and nth terms called a and b with P the product of all n terms, and P² = (ab)ⁿ to be shown. Verified for the first two: both sit on equal gaps — three positions in one case, six in the other — so each quotient is the same power of r, which is precisely what makes the three terms geometric.

Figures to have open

• A number line of positions 1 to 16 with a multiplication arrow between each neighbouring pair, and a running count of arrows displayed separately from the position. Sections 2, 3 and 5 all depend on this and it must persist unchanged. Standard schematic.
• The same line with two marked positions and the arrows between them bracketed, labelled with the gap, for section 9. Standard schematic.
• A pair of aligned exponent towers for section 8, showing both sides of the equation reduced to the same base. Standard schematic.
• Nothing here can be taken from the book. All sixteen printed pages were opened as images: the 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", §8.4.1 General term of a G.P., p. 140
• Examples 4, 5 and 6, pp. 140–141
• Exercise 8.2 items 1–5, 15, 17, 22 and 23, pp. 145–146
• The Summary, p. 149, states the rule for the general term in one line
• The notation list at the head of p. 140, established in §8.4, is used throughout