Subtracting a scaled copy of the total to collapse it to two terms

What to assume they know

• Reaching any term without walking through the earlier ones — the general term arⁿ⁻¹, and the standard form
• What changes once the terms are added instead of listed — a series as the indicated addition, and Sₙ as the total of the first n terms
• Multiplying a whole sum through by a factor, term by term
• Subtracting one equation from another and collecting what remains
• Factorising a quadratic, or applying the quadratic formula, with fractional roots
• Recognising a number as a power of a small base

What they should be able to do

• Derive a G.P.'s total over its opening n terms by writing the scaled copy and then subtracting, without quoting the result
• Explain which terms cancel and why exactly two survive
• Say why a ratio of one has to be handled separately, and total that case by counting
• Show that the two printed forms of the formula are one formula, and choose between them to keep the signs convenient
• Use the formula forwards to total a stated number of terms
• Use the formula backwards to find how many terms produce a stated total
• Convert a sequence that is not geometric, such as one built from repeated digits, into an expression involving a G.P. and total it
• Solve for a geometric triple given its total and its product, using the symmetric substitution

Where it usually goes wrong

• "The formula has to be memorised." Two lines and a subtraction rebuild it in fifteen seconds. Make the class rebuild it rather than recall it, because the rebuild also tells them where the r = 1 exclusion came from.
• "Substitute r = 1 into the formula and take a limit." At r = 1 the derivation gives zero equals zero — true and useless. That is why the chapter splits the case out before dividing, and the honest total there is n copies of the first term.
• "There are two different formulas for the sum." There is one. Negating the numerator and the denominator together changes nothing. Show the two forms transforming into each other.
• "Use the first form when the ratio is small and the second when it is large." That is a habit for keeping both parts positive, not a mathematical rule. Either form returns the same number for any admissible ratio.
• "You need the last term to total a G.P." You need the first term, the ratio and how many terms there are. The chapter lists a symbol for the last term, but the derivation never calls on it.
• "7, 77, 777 is a geometric progression." The ratios are not constant. Have students check two of them before Example 10 begins, or the trick looks like sleight of hand.
• "The middle terms cancel because they are small." They cancel because they are identical. Colour the shared block in both lines and remove it in one motion.
• "The exclusions attached to two exercise items are pedantry." Each one names exactly the value that would make that G.P. have ratio 1 — the same case the derivation had to carve out. They are the derivation showing up in the exercise.

Questions to check understanding

• Total a stated number of terms of a given G.P.
• Find how many terms of a given G.P. produce a stated total
• Total a G.P. whose ratio is irrational, negative, or a letter with an exclusion attached
• Total a sequence of repeated digits by converting it into a G.P. first
• Find a geometric triple from its total and its product
• Word problems on doubling populations, compound interest, depreciation and chain letters, in each case identifying the first term, the ratio and the number of terms before computing
• State and justify the separate treatment of a ratio equal to one

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 (§8.4.2, p. 140). Line one sets Sₙ as the addition of a, ar, ar², up to arⁿ⁻¹. The ratio-one case is separated off first and totalled as n copies of a. For any other ratio, line two is line one multiplied through by r, running from ar up to arⁿ. Subtracting leaves (1 − r)Sₙ equal to a − arⁿ, which the page factors as a(1 − rⁿ). Count the terms: each line holds n of them, the middle n − 1 are common to both, so 2n − 2 of the 2n terms disappear.
• A formula that is missing from its own section, and this matters (§8.4.2, p. 140). After the subtraction the page prints the words that introduce a result, then leaves blank space where the result should be, then prints "or" followed by the form whose denominator is r − 1. The companion form, with 1 − r underneath, is not printed in §8.4.2 at all — the "or" has nothing to be an alternative to. The missing form then turns up unannounced inside Example 7 on p. 141, is used again in Example 8 on the same page, and finally both forms appear together in the Summary on p. 149. A student reading straight through meets a formula in a worked example that the derivation never produced.
• Example 7 (p. 141). Inputs: the geometric series beginning 1, then 2/3, then 4/9, with the total of n terms and of 5 terms wanted. Verified: a = 1 and r = 2/3, so the total of n terms is 3 times the quantity 1 − (2/3)ⁿ; at n = 5, (2/3)⁵ is 32/243, the bracket is 211/243, and the total is 211/81.
• Example 8 (p. 141). Inputs: the G.P. 3, 3/2, 3/4, and onward, with the target total 3069/512 and the number of terms wanted. Verified: a = 3 and r = 1/2, so the total of n terms is 6 times the quantity 1 − 2⁻ⁿ; setting that equal to the target gives 2⁻ⁿ = 1/1024, so 2ⁿ = 1024 and n = 10. The useful teaching beat is that the unknown ends up in an exponent, so the last step is again a matching of powers.
• Example 9 (p. 142). Inputs: a geometric triple of neighbouring entries whose total is 13/12 and whose product is −1. The page writes the three terms as a ÷ r, a, and ar, so that the product collapses to a³. Verified: a³ = −1 gives a = −1 taking real values only; substituting into the total produces 12r² + 25r + 12 = 0, whose roots are −3/4 and −4/3; the two triples are 4/3, −1, 3/4 and the same three reversed. Check both ways: 4/3 − 1 + 3/4 = 13/12 and the product of those three is −1. The symmetric substitution is the whole point — centring the unknowns on a is what makes the product lose r.
• Example 10 (pp. 142–143). Inputs: the sequence 7, 77, 777, 7777, and onward, totalled to n terms. It is not a G.P. — the ratios are not constant. The page converts each term into a string of nines over nine, then each nine-string into a power of ten less one, splitting the total into a genuine G.P. of powers of ten minus n copies of one. Verified at two terms: the resulting expression returns 84, and 7 + 77 = 84. Use that check; it is the cheapest possible reassurance that a long manipulation landed.
• Example 11 (p. 143). Inputs: 2 parents, then 4 grandparents, then 8 great grandparents, doubling onward, totalled over ten generations. Verified: a = 2, r = 2, n = 10, and the total is 2 × 1023 = 2046. This closes the question §8.2 opened on p. 135.
• Exercise 8.2, totals to a stated number of terms (p. 145). Inputs: 0.15, 0.015, 0.0015, … to 20 terms; √7, √21, 3√7, … to n terms; 1, −a, a², −a³, … to n terms with a not equal to −1; x³, x⁵, x⁷, … to n terms with x not equal to 1 or −1. Verified ratios: 0.1, √3, −a and x². The two stated exclusions are worth a beat — each one is precisely the value that would make the ratio equal 1 and send the denominator to zero.
• Exercise 8.2, totals run backwards and other inputs (pp. 145–146). Inputs: how many terms of 3, 3², 3³, … reach the total 120; three terms with total 39/10 and product 1; the first three terms totalling 16 while the next three total 128; a G.P. with first term 729 whose 7th term is 64, its seven-term total wanted; a G.P. whose first two terms total −4 and whose 5th term is four times its 3rd; the total to n terms of 8, 88, 888, 8888, …; the term-by-term products of 2, 4, 8, 16, 32 with 128, 32, 8, 2, 1/2, totalled. Verified: the first gives 3ⁿ = 81 and n = 4; the second gives a = 1 with ratio 5/2 or 2/5 and the triple 2/5, 1, 5/2; the third gives r³ = 8, so r = 2 and a = 16/7; the fifth gives r² = 4, so the two admissible progressions start −4/3 with ratio 2 and 4 with ratio −2; the last has products 256, 128, 64, 32, 16, themselves a G.P. of ratio 1/2, totalling 496.
• Growth and decay items (Exercise 8.2 items 30 and 31, pp. 146–147, and Miscellaneous Exercise items 15 and 17, p. 148). Inputs: 30 bacteria doubling every hour, counted at the 2nd hour, the 4th and the nth; Rs 500 at 10 per cent compounded annually over 10 years; a chain letter in which each person writes to four others, costing 50 paise a letter, followed to the 8th set; a machine costing Rs 15625 losing 20 per cent of its value each year, valued after 5 years. Verified: the bacteria give 120, 480 and 30 × 2ⁿ; the deposit grows to 500 × 1.1¹⁰, which is about Rs 1296.87; the chain mails 4 + 4² + ⋯ + 4⁸ = 87380 letters, costing Rs 43690; the machine falls to 15625 × 0.8⁵ = Rs 5120.

Figures to have open

• Two stacked rows of term cards, the second shifted right by one place, with the overlapping cards drawn identically. This single image carries sections 2, 3 and 4, and the shift by one place is the reason the cancellation happens. Standard schematic.
• A panel showing both printed forms with arrows indicating that numerator and denominator have each been negated. Standard schematic.
• A reproduction-free depiction of the p. 140 gap for section 7: the introducing words, an empty box, and the surviving alternative. Describe it rather than photographing the page.
• No artwork exists to borrow. 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", §8.4.2, p. 140, which derives the total of a G.P.'s opening run of terms
• Examples 7 to 11, pp. 141–143
• Exercise 8.2 items 7–16, 18, 19, 24, 30 and 31, pp. 145–147
• Miscellaneous Exercise on Chapter 8 items 1, 2, 11, 15 and 17, pp. 147–148
• The Summary, p. 149, which prints both forms of the total together — the only place in the chapter where they appear side by side