The number that sits between two others multiplicatively

What to assume they know

• Reaching any term without walking through the earlier ones — the general term arⁿ⁻¹, and counting positions in a G.P.
• A constant ratio between neighbours is the entire definition — the constant-ratio definition and the non-zero condition
• Square roots, cube roots and higher roots, and which of them admit a negative answer
• Fractional exponents as roots, since the general case is written that way
• That a product of two positive numbers is positive

What they should be able to do

• State the condition that defines the geometric mean of two positive numbers, and derive the square-root formula from it
• Compute the geometric mean of a given pair and verify that the three numbers form a G.P.
• Explain why the section restricts the definition to positive numbers
• Insert a stated number of terms between two given numbers so the result is a G.P., and justify the ratio used
• Count positions correctly to see that the closing number occupies position n + 2 when n numbers are inserted
• Decide how many admissible ratios a given insertion problem has, from the order of the root involved
• Recognise the geometric-mean condition when it appears disguised as a relation between terms at equally spaced positions

Where it usually goes wrong

• "The geometric mean is the average." For 2 and 8 the average is 5 and the geometric mean is 4. Different questions, different answers; only one of them produces a constant ratio.
• "The geometric mean sits halfway between the two numbers." It sits halfway in the multiplicative sense — equally many multiplications from each end. On a plain number line it is closer to the smaller number.
• "Inserting numbers gives you a free choice." It does not. Once you say how many go in, the ratio is pinned to one root, and every insert follows.
• "With n numbers inserted, take the nth root." Take the (n + 1)th. The closing number lands at position n + 2, so it has been multiplied n + 1 times. This is the same off-by-one that governs the general term, appearing again.
• "Every insertion problem has two answers, because roots come in pairs." Only even-order roots do. Example 12 needs a 4th root and has two; the exercise item needs a cube root and has one.
• "Any two numbers have a geometric mean." The section defines it for two positive numbers. If the product is negative, no real number squares to it.
• "The negative branch of Example 12 is a mistake." It is a genuine G.P. and the page prints it. What it is not is a set of geometric means as §8.4.3 defined them, since two of its terms are negative.
• "q² = ps is a separate formula to learn." It is the geometric-mean condition in different clothes, arising because the three positions are equally spaced.

Questions to check understanding

• Compute the geometric mean of a given pair, and verify the resulting triple is a G.P.
• Insert a stated number of terms between two given numbers, with the ratio justified
• Say how many admissible ratios a given insertion problem has, and why
• Find an exponent making a given expression the geometric mean of two numbers
• Prove that terms at equally spaced positions of a G.P. satisfy a middle-squared relation
• Find a missing middle term making three given numbers a G.P., reporting all real values

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 definition and its instance (§8.4.3, p. 143). The geometric mean of two positive numbers is given as the square root of their product, and the section immediately takes 2 and 8, whose geometric mean is 4, observing that 2, 4, 8 then sit as neighbours inside a G.P. Verified: 2 × 8 = 16 and its root is 4; the ratios 4 ÷ 2 and 8 ÷ 4 are both 2. Present the condition before the formula, because the formula is the answer to the condition and looks arbitrary without it.
• A contrast worth showing. Verified: the half-sum of 2 and 8 is 5, and 2, 5, 8 has a constant difference rather than a constant ratio. Two different jobs, two different middle numbers, same pair of outer numbers. This contrast is added here — the chapter does not draw it here — and it sets up the comparison that Why one of the two means can never overtake the other is entirely about.
• The general insertion set-up (§8.4.3, p. 143). Numbers G₁ through Gₙ are to be placed between positive a and b so that a, G₁, G₂, …, Gₙ, b is a G.P. The page notes that b is then the term at position n + 2, so b equals a times r raised to n + 1, giving r as the (n + 1)th root of b ÷ a. Each inserted number is then a times a matching power of that root: the first carries exponent 1/(n + 1), the second 2/(n + 1), the third 3/(n + 1), and the last n/(n + 1).
• The position count is the trap. Verified by counting: a occupies position 1, the n inserted numbers occupy positions 2 through n + 1, so b occupies position n + 2 and has been multiplied by the ratio n + 1 times. Students who count the inserts and stop reach for the nth root and get everything wrong afterwards.
• Example 12 (p. 144). Inputs: three numbers to be inserted between 1 and 256 so the whole becomes a G.P. Verified: here n = 3, so the closing number is the 5th term and the ratio satisfies r⁴ = 256, giving r = 4 or r = −4 if real values only are taken. The positive ratio produces 4, 16, 64; the negative ratio produces −4, 16, −64. Check the second: 1, −4, 16, −64, 256 has ratio −4 at every one of its four steps.
• The sentence the page ends on (p. 144). Having listed both branches, the section closes by naming only the positive triple, while its wording still refers to more than one resulting progression. Both branches are genuine G.P.s; only the positive one consists of geometric means in the sense §8.4.3 defined, since that definition was stated for positive numbers. Say this explicitly rather than smoothing it over — the mismatch is on the page.
• A useful invariant. Verified: in both branches of Example 12 the middle insert is 16, which is the geometric mean of 1 and 256. That is not a coincidence: the middle slot is equally many steps from each end, so its value is fixed whichever sign the ratio takes. This observation is added here.
• Exercise 8.2 item 26 (p. 146). Inputs: two numbers to be inserted between 3 and 81. Verified: here the closing number is the 4th term, so r³ = 27 and r = 3, giving 9 and 27. Set this beside Example 12 deliberately — an odd-order root admits one real value and an even-order root admits two, which is why one problem has a single answer and the other has a pair.
• Exercise 8.2 item 27 (p. 146). Inputs: the expression formed by dividing a raised to n + 1 plus b raised to n + 1 by a raised to n plus b raised to n, required to equal the geometric mean of a and b, with n to be found. Verified: n = −1/2 works, since with that exponent the denominator is the numerator divided by the square root of ab. Worth showing because it is the only item solving for an index or an exponent whose answer is not a whole number, and students assume it must be. Do not widen that to the whole exercise: item 1's entry at position 20, item 12's ratio and item 14's first term are all fractions too.
• The geometric mean in disguise (Exercise 8.2 items 3, 6 and 17, pp. 145–146). Inputs: a G.P.'s entries at positions 5, 8 and 11, named p, q and s, with q² = ps to be shown; the numbers −2/7, an unknown, and −7/2 required to be a G.P.; a G.P.'s entries at positions 4, 10 and 16, named x, y and z, to be shown geometric themselves. Verified: in each case the middle quantity squared equals the product of the outer two, because the middle sits at the same position gap from each. The second is the interesting one — its outer numbers are negative, their product is 1, and the middle value is 1 or −1. The condition on the middle term survives there, but the section's square-root formula does not apply, because it was stated for positive numbers.

Figures to have open

• A slot diagram: two fixed end numbers with empty boxes between them, and multiplication arrows of equal weight spanning every gap. Sections 1, 5, 6 and 7 all run on it, and the count of arrows is what section 6 is about. Standard schematic.
• A number line carrying 2, 4, 5 and 8, showing the geometric mean and the half-sum in different places, for section 3 and the first misconception. Standard schematic.
• Two parallel chains for section 9, one with ratio 4 and one with ratio −4, both starting at 1 and ending at 256. Standard schematic.
• Nothing can be borrowed 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", §8.4.3 Geometric Mean (G.M.), pp. 143–144
• Example 12, p. 144
• Exercise 8.2 items 3, 6, 17, 26 and 27, pp. 145–146
• The Summary, p. 149, restates the geometric mean of two positive numbers and the three-term progression it completes
• §8.5, p. 144, immediately afterwards, sets the geometric mean against the arithmetic mean; that comparison is Why one of the two means can never overtake the other