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Lesson 3 of 4

Sequences and Series · Lesson 3 of 4

Mixed Applications

“Choose representations that reveal hidden geometric structure in equations, products and changing quantities.”

Learning Objectives

• Translate term, sum and product conditions into equations and retain all valid real branches. • Explain geometric structure in equally spaced terms, shifted sums, products and reciprocals. • Use sum-of-squares reasoning to prove that a list forms a geometric progression. • Transform repeated-digit and repeated-decimal sums without misclassifying the original sequence. • Connect means, ratios and quadratic roots through sums and products. • Distinguish constant additions from constant percentage changes in practical models.

Start with the information, not with a formula

A mixed problem may describe some terms, a sum, a product or an update rule. Each description supplies a different equation. Before calculating, decide what the unknowns represent and what the given information tells you about them. A useful representation can make several of those equations simpler at once.

For a G.P., the standard unknowns are its first term a and its nonzero common ratio r. A term at position k gives arᵏ⁻¹; a finite total gives Sₙ. For three terms whose product is known, m/r, m, mr is often convenient because the product becomes m³. If you change representations, state whether your symbol names the first or middle term.

Information suppliedUseful translationWhat to watch
Two specified termsaₚ = arᵖ⁻¹ and a_q = ar^(q−1)Subtract positions when dividing the equations.
A total over n termsSₙ = a(1 − rⁿ)/(1 − r), or na for r = 1n must be a positive integer.
Three consecutive terms and their productm/r, m, mr; product m³m denotes the middle term.
Two endpoint terms with inserted meansb = arⁿ⁺¹ for n insertionsCount the n + 1 gaps.
Two positive meansa + b = 2A; ab = G²Check A ≥ G > 0.
Example — Sum conditions for consecutive blocks

Problem
The first three terms of a G.P. sum to 16 and the next three sum to 128. Find a, r and Sₙ.

  1. 1.The first block is a + ar + ar² = a(1 + r + r²) = 16.
  2. 2.The next block is ar³ + ar⁴ + ar⁵ = r³a(1 + r + r²) = 128.
  3. 3.Divide the second block sum by the first: r³ = 128/16 = 8, giving the real ratio r = 2.
  4. 4.Now a(1 + 2 + 4) = 16, so a = 16/7.
  5. 5.Since r ≠ 1, Sₙ = (16/7)(2ⁿ − 1).
  6. 6.The next block is eight times the first because each of its terms is three positions later.
Example — Even-power ambiguity in a term condition

Problem
A G.P. has first-two-term sum −4, and its fifth term is four times its third. Find the possible progressions.

  1. 1.The first-two-term condition is a(1 + r) = −4.
  2. 2.The fifth and third terms give ar⁴ = 4ar². Since a and r are nonzero, divide to obtain r² = 4.
  3. 3.Thus r = 2 or r = −2.
  4. 4.For r = 2, 3a = −4, so a = −4/3. The list begins −4/3, −8/3, −16/3, … .
  5. 5.For r = −2, −a = −4, so a = 4. The list begins 4, −8, 16, −32, 64, … .
  6. 6.Both satisfy the conditions. Keeping both branches is necessary because the term equation fixes an even power of the ratio.

A real solution of an algebraic equation is only a candidate. It must also satisfy the original conditions. Check nonzero denominators, the sign restrictions on means, and whether a calculated term count is a positive integer. These checks prevent a transformed equation from producing an inadmissible answer.

Equally spaced positions retain geometric structure

Choosing terms at equally spaced positions means making the same number of ratio steps each time. The new list therefore has its own constant ratio. This idea is useful for proofs involving distant terms, not only for numerical calculations.

If we select positions p, p + d and p + 2d, their values are arᵖ⁻¹, arᵖ⁺ᵈ⁻¹ and arᵖ⁺²ᵈ⁻¹. Each selected term is rᵈ times the preceding selected term. Consequently the square of the middle selected value equals the product of the outer selected values.

Three equally spaced termsLaTeX
The position gap is d. This holds for nonzero real G.P. terms, including signed terms, because the powers on both sides agree.
Example — A proof using distant positions

Problem
The fifth, eighth and eleventh terms of a G.P. are p, q and s. Show q² = ps.

  1. 1.Write p = ar⁴, q = ar⁷ and s = ar¹⁰.
  2. 2.Then q² = a²r¹⁴.
  3. 3.Also ps = (ar⁴)(ar¹⁰) = a²r¹⁴.
  4. 4.The expressions are equal, so q² = ps. Equivalently, the position gaps are both 3 and the selected list has ratio r³.

The same argument works for positions 4, 10 and 16, where the gap is 6. Their values form a G.P. with ratio r⁶. A useful related identity comes from eliminating both the first term and the ratio from three arbitrary specified positions.

Eliminating a common first term and ratioLaTeX
p, q and s are positive integer positions; u, v and w are nonzero terms. Integer exponents allow negative term values, and negative exponents require nonzero bases.

To justify the identity, substitute u = arᵖ⁻¹, v = arᑫ⁻¹ and w = arˢ⁻¹. The total exponent of a is (q − s) + (s − p) + (p − q) = 0. The total exponent of r is (p − 1)(q − s) + (q − 1)(s − p) + (s − 1)(p − q), whose terms cancel in pairs. The expression is therefore a⁰r⁰ = 1. The reasoning is cancellation of exponents, not a new memorised formula.

Products, reciprocals and sums of paired terms

Geometric sequences have a predictable form at every position. Multiplying corresponding terms adds their exponents, and taking reciprocals reverses them. These operations can reveal a new G.P. and make a sum easier to evaluate.

Let one G.P. have terms arᵏ⁻¹ and another have terms ARᵏ⁻¹. Their corresponding products are aA(rR)ᵏ⁻¹, so the product list has first term aA and ratio rR. If rR = 1, all products are equal; use the constant-term sum rather than the fraction formula.

Corresponding products form a G.P.LaTeX
The operation is term-by-term multiplication of two lists, not multiplication of their sums.
Example — Adding corresponding products

Problem
Find the sum of the corresponding products of 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2.

  1. 1.Multiply the first pair: 2 × 128 = 256. The product ratio is 2 × (1/4) = 1/2.
  2. 2.The products are therefore 256, 128, 64, 32, 16.
  3. 3.Their sum is 256[1 − (1/2)⁵]/(1 − 1/2).
  4. 4.This is 512 × 31/32 = 496.
  5. 5.The sum of corresponding products is different from multiplying the two separate list sums.

Taking reciprocals of nonzero G.P. terms gives 1/a, 1/(ar), 1/(ar²), … . The new ratio is 1/r. Similarly, for four consecutive terms a, ar, ar², ar³, for a positive integer exponent N, the quantities aᴺ + (ar)ᴺ, (ar)ᴺ + (ar²)ᴺ and (ar²)ᴺ + (ar³)ᴺ have the common factor aᴺ(1 + rᴺ) and successive factors 1, rᴺ, r²ᴺ. When the common factor is nonzero, they form a G.P. with ratio rᴺ. If it vanishes, all three are zero; the algebraic equality remains valid, but the nonzero-term definition does not call the zero list a G.P.

Pairing endpoints explains products of all terms

The first and last terms of a finite G.P. have the same product as the second and second-last terms. The reason is that each pair uses the same total number of ratio factors. This pairing can relate a total product to endpoints without calculating every individual value.

If the first term is a and the nth term is b = arⁿ⁻¹, then the kth term times the term at position n + 1 − k is a²rⁿ⁻¹ = ab. Write the product P once in the forward order and once in the reverse order; multiplying these lists pairs each term with its opposite term.

Product and endpointsLaTeX
P is the product of all n nonzero terms, while a and b are the first and last values. The squared relation avoids choosing an unjustified sign for P.

There is a similar relation for a sum of reciprocals. Let S be the sum of the terms and R be the sum of their reciprocals. Dividing ab by the kth term gives its opposite term, so multiplying R by ab gives the original sum in reverse order. Thus abR = S. Raising this relation to the integer power n and using P² = (ab)ⁿ gives P²Rⁿ = Sⁿ. No division by S or R is needed, so zero sums cause no difficulty in this proof.

Sum, product and reciprocal sumLaTeX
R denotes a reciprocal sum here, not a common ratio. The distinct meanings of the symbols should be stated whenever a problem introduces them.
Example — Testing the paired relationships

Problem
Use 2, 6, 18 to check the endpoint-product and reciprocal-sum relationships.

  1. 1.The endpoint product is 2 × 18 = 36. The total product is P = 2 × 6 × 18 = 216.
  2. 2.P² = 46656 and 36³ = 46656, so the endpoint relation holds.
  3. 3.The term sum is S = 26. The reciprocal sum is R = 1/2 + 1/6 + 1/18 = 13/18.
  4. 4.Multiplying abR gives 36 × 13/18 = 26 = S.
  5. 5.The numerical check illustrates the relationships; the paired-term argument establishes them generally.

Whole blocks and alternating positions

A shifted block of terms is obtained by multiplying every earlier term by the same power of r. Likewise, choosing every second term forms a new G.P. with ratio r². Recognising the entire block transformation is often quicker than expanding several sum formulas separately.

Let S be the first-n-term sum. The sum T of positions n + 1 through 2n is rⁿS because each corresponding term moves n positions forward. Therefore S/T = 1/rⁿ when the sums are nonzero. For a signed G.P., a block sum can be zero; in that case the equation T = rⁿS still holds, but the quotient 0/0 is not defined.

Equal-length blocksLaTeX
The ratio statement is conditional on the denominator being nonzero. Under the nonzero ratio assumption, S ≠ 0 also guarantees T ≠ 0.
Example — Odd and even positions

Problem
A finite G.P. has 2k terms. Its whole sum is five times the sum of terms in odd positions. What ratio follows when the odd-position sum is nonzero?

  1. 1.The odd-position sum is O = a + ar² + … + ar²ᵏ⁻².
  2. 2.Each even-position term is r times its corresponding odd-position term, so the even-position sum is E = rO.
  3. 3.The total is O + E = (1 + r)O. The condition gives (1 + r)O = 5O.
  4. 4.Since O is nonzero, divide to get 1 + r = 5, so r = 4.
  5. 5.If the problem permits O = 0, the condition alone would need separate analysis. For real r, the odd-position sum a(1 + r² + … + r²ᵏ⁻²) cannot be zero when a ≠ 0, because the parenthesis is positive. Thus the division is justified in this real G.P. setting.

Proving geometric structure with nonnegative squares

Some problems hide equal ratios inside a quadratic expression. A useful method is to rewrite the expression as a sum of real squares. If that sum is at most zero, every square must be zero. The resulting equalities can force consecutive terms to have a common ratio.

Consider the expression (a² + b² + c²)p² − 2(ab + bc + cd)p + (b² + c² + d²). Group its terms so that a²p² − 2abp + b² becomes (ap − b)², and do the same with the next two pairs. This exposes three quantities that cannot be negative individually.

Example — The sum-of-squares argument

Problem
Suppose a, b, c and d are distinct real numbers and (a² + b² + c²)p² − 2(ab + bc + cd)p + (b² + c² + d²) ≤ 0. Explain why a, b, c, d form a G.P.

  1. 1.Rewrite the left side as (ap − b)² + (bp − c)² + (cp − d)².
  2. 2.Every term is a nonnegative real square. A sum of such squares can be ≤ 0 only if the sum is 0.
  3. 3.Therefore ap = b, bp = c and cp = d.
  4. 4.If a = 0, those equations force b = c = d = 0, contradicting distinctness. If p = 0, they force b = c = d = 0, also contradicting distinctness.
  5. 5.Thus a and p are nonzero, and b = ap, c = ap², d = ap³ are nonzero.
  6. 6.The ratios b/a, c/b and d/c all equal p, so the list is a G.P.
  7. 7.The nonzero check is what makes the quotient-based conclusion legitimate.

The same pattern explains an identity involving four consecutive G.P. terms. Write b = ar, c = ar² and d = ar³. Then b² + c² + d² = r²(a² + b² + c²), and ab + bc + cd = r(a² + b² + c²). Squaring the second expression gives exactly the product of the first two sums.

A four-term identityLaTeX
Here a, b, c and d are consecutive G.P. terms. The proof works by factoring the common ratio from the shifted expressions.

Generating a sequence from a function rule

A function condition can turn into a recurrence when one input is fixed. If the domain is the positive integers, replacing one input by 1 relates neighbouring function values. This connects function notation to the sequence viewpoint developed earlier.

Example — An additive input creates a multiplicative update

Problem
A function on the positive integers satisfies f(x + y) = f(x)f(y), with f(1) = 3. If f(1) + … + f(n) = 120, find n.

  1. 1.Put y = 1: f(x + 1) = f(x)f(1) = 3f(x).
  2. 2.Starting from f(1) = 3, the values are 3, 9, 27, …, so f(k) = 3ᵏ for positive integers k.
  3. 3.The total is 3(3ⁿ − 1)/(3 − 1) = 120.
  4. 4.Multiply by 2 and divide by 3: 3ⁿ − 1 = 80, giving 3ⁿ = 81.
  5. 5.Hence n = 4. Check 3 + 9 + 27 + 81 = 120.
  6. 6.Only positive-integer inputs were used; no value of f(0) or assumption about all real inputs is needed.

Changing a G.P. into an A.P.

A geometric progression changes by multiplication, while an arithmetic progression changes by equal additions. A problem may describe how separate adjustments to three geometric terms make the adjusted terms arithmetic. The adjustment condition becomes a second equation alongside the geometric sum or product.

For three numbers u, v, w in an arithmetic progression, v − u = w − v, or 2v = u + w. This is a short recall of a previously studied relationship. It is useful here because it expresses the condition without introducing a separate common-difference unknown.

Example — Two progressions in one problem

Problem
Three G.P. terms sum to 56. Subtracting 1, 7 and 21 respectively makes an A.P. Find the possible ordered triples.

  1. 1.Let the G.P. terms be u, v, w, so u + v + w = 56 and v² = uw.
  2. 2.The adjusted A.P. condition is 2(v − 7) = (u − 1) + (w − 21).
  3. 3.Simplify to 2v = u + w − 8. Since u + w = 56 − v, this gives 3v = 48, so v = 16.
  4. 4.Then u + w = 40 and uw = 16² = 256. The endpoint values solve t² − 40t + 256 = 0.
  5. 5.Factor as (t − 8)(t − 32) = 0. The endpoints are 8 and 32.
  6. 6.Both 8, 16, 32 and 32, 16, 8 are G.P.s. Their adjusted lists are 7, 9, 11 and 31, 9, −13 respectively, both A.P.s.
  7. 7.The conditions permit both orders; their common ratios are 2 and 1/2.
Example — Four terms from two differences

Problem
Four consecutive G.P. terms have the third greater than the first by 9, and the second greater than the fourth by 18. Find them.

  1. 1.Represent the terms as a, ar, ar², ar³. The conditions become a(r² − 1) = 9 and ar(1 − r²) = 18.
  2. 2.The second left side is −r times the first, so −9r = 18, giving r = −2.
  3. 3.Now a(4 − 1) = 9, so a = 3.
  4. 4.The terms are 3, −6, 12, −24. Check 12 − 3 = 9 and (−6) − (−24) = 18.
  5. 5.Differences between selected terms can determine a negative ratio even though the described differences are positive.

Cross-multiplication is useful when a problem presents equal rational expressions rather than explicit ratios of consecutive terms. If (a + bx)/(a − bx) = (b + cx)/(b − cx), with x ≠ 0 and the displayed denominators nonzero, expanding both cross-products yields 2x(b² − ac) = 0, hence b² = ac. The next equality similarly gives c² = bd. For nonzero a, b, c, d these imply b/a = c/b = d/c, so the list is a G.P. The nonzero requirements must be checked before the quotients are taken.

Repeated decimals are related to a geometric sum

The decimal 0.6, followed by 0.66, then 0.666, forms a sequence whose values are not obtained by multiplying by a fixed ratio. Instead, each additional digit contributes a new power of one-tenth. A finite geometric sum describes the value of each term and then the total of those terms.

The kth entry is 6/10 + 6/10² + … + 6/10ᵏ. Its sum is (2/3)(1 − 10⁻ᵏ). This formula describes a finite decimal, not the infinitely repeating decimal 0.666… . Summing these finite entries adds a constant contribution n times and subtracts a geometric correction.

Example — A sum of growing finite decimals

Problem
Find 0.6 + 0.66 + 0.666 + … through the entry with n sixes.

  1. 1.The kth term is (2/3)(1 − 10⁻ᵏ).
  2. 2.The required total is (2/3)[n − (10⁻¹ + 10⁻² + … + 10⁻ⁿ)].
  3. 3.The correction is a G.P. with a = 1/10 and r = 1/10, so its sum is (1 − 10⁻ⁿ)/9.
  4. 4.Thus Sₙ = (2/3)[n − (1 − 10⁻ⁿ)/9].
  5. 5.For n = 2, the formula gives (2/3)[2 − 0.99/9] = (2/3)(1.89) = 1.26, agreeing with 0.6 + 0.66.
Example — A term formed by a product

Problem
Find the twentieth term of the series 2 × 4 + 4 × 6 + 6 × 8 + … .

  1. 1.The first factor at position k is 2k and the second is 2(k + 1).
  2. 2.Therefore the kth term is aₖ = 4k(k + 1).
  3. 3.At k = 20, a₂₀ = 4 × 20 × 21 = 1680.
  4. 4.This question asks for one term, not the twentieth partial sum. Also, this product sequence is not a G.P.; the factors change additively rather than multiplicatively.

Means, ratios and quadratic roots

Two means encode two different pieces of information: the arithmetic mean gives a sum and the geometric mean gives a product. The sum and product also determine a quadratic equation with the original numbers as roots. This gives a bridge between sequences and familiar algebra.

Example — Building a quadratic from means

Problem
Positive roots have arithmetic mean 8 and geometric mean 5. Find a monic quadratic equation with those roots.

  1. 1.If the roots are u and v, then u + v = 16 and uv = 25.
  2. 2.Expand (x − u)(x − v) = x² − (u + v)x + uv.
  3. 3.The required equation is x² − 16x + 25 = 0.
  4. 4.Its roots are 8 ± √39. Both are positive because √39 < 8.
  5. 5.Their sum is 16 and their product is 64 − 39 = 25, so the stated means are recovered.
Example — Two quadratic pairs in a single G.P.

Problem
The roots a, b of x² − 3x + p = 0 and the roots c, d of x² − 12x + q = 0 occur in that order in a nonzero real G.P. Show (q + p):(q − p) = 17:15.

  1. 1.The root sums give a + b = 3 and c + d = 12; the root products give ab = p and cd = q.
  2. 2.In the G.P., c = ar² and d = ar³, so c + d = r²(a + b).
  3. 3.Thus 12 = 3r² and r² = 4.
  4. 4.Also cd = r⁴ab, so q = 16p. Since all G.P. terms are nonzero, p ≠ 0.
  5. 5.Hence (q + p):(q − p) = 17p:15p = 17:15.
  6. 6.There is no need to choose between r = 2 and r = −2: the squared ratio information is enough for the requested relationship.

If the ratio of the arithmetic and geometric means is m:n with m ≥ n > 0, choose a positive scale t so that A = mt and G = nt. The endpoint formula from the earlier lesson gives t[m + √(m² − n²)] and t[m − √(m² − n²)]. The common scale disappears when the ratio of the numbers is taken.

Ratio of numbers from a ratio of meansLaTeX
This lists one order; reversing the two numbers reverses the ratio. The mean ratio requires m ≥ n > 0.
Example — A sum tied to a geometric mean

Problem
Two positive numbers have sum six times their geometric mean. Find their ratio.

  1. 1.Their arithmetic mean is half their sum, so A = 3G. Thus A:G = 3:1.
  2. 2.Using m = 3 and n = 1, the ratio is (3 + √8):(3 − √8).
  3. 3.Therefore u:v = (3 + 2√2):(3 − 2√2), or its reverse.
  4. 4.Both quantities are positive and their product is 9 − 8 = 1. Their sum is 6, so this representative pair satisfies the condition.
Example — When a quotient is a geometric mean

Problem
For positive unequal a and b, find the real value n for which (aⁿ⁺¹ + bⁿ⁺¹)/(aⁿ + bⁿ) = √ab.

  1. 1.Set G = √ab and multiply by the positive denominator: aⁿ(a − G) + bⁿ(b − G) = 0.
  2. 2.Write a = u² and b = v² with u, v > 0 and u ≠ v. Then G = uv.
  3. 3.The equation becomes u²ⁿu(u − v) − v²ⁿv(u − v) = 0.
  4. 4.Since u − v ≠ 0, cancel it: u²ⁿ⁺¹ = v²ⁿ⁺¹.
  5. 5.For unequal positive bases this requires 2n + 1 = 0, so n = −1/2.
  6. 6.Check directly: (√a + √b)/(1/√a + 1/√b) = √ab. If a = b, the original quotient already equals a for every real n.
  7. 7.If n had been restricted to positive integers, there would be no solution for unequal endpoints; always check the stated domain.

Equal additions and equal percentage changes

A fixed increase each period creates an arithmetic pattern. A fixed percentage of the current value creates a geometric pattern. The verbal distinction matters: adding ₹500 repeatedly is not the same operation as multiplying repeatedly by 1.05. Realistic stories can contain both kinds of change, so specify which quantity is updated.

For a starting value V and a decrease of 20% each year, the retained fraction is 0.80. The value after t completed updates is V(0.8)ᵗ. For simple interest on a fixed original principal P at annual rate i, the annual interest is always Pi, so the balance after t years is P + tPi. These are models under the stated assumptions; the index should count completed time periods.

Example — Depreciation

Problem
A machine initially valued at ₹15625 loses 20% of its current value annually. Find its value after five years.

  1. 1.Each update retains 80%, so the yearly factor is 0.8 = 4/5.
  2. 2.After five years the value is 15625(4/5)⁵.
  3. 3.Since 15625 = 5⁶, the expression simplifies to 5 × 4⁵.
  4. 4.Thus the value is 5 × 1024 = ₹5120.
  5. 5.Subtracting 20% of the original value five times would instead model a fixed yearly subtraction and would not match the stated current-value rule.
Example — Simple interest is an additive pattern

Problem
An initial ₹10000 balance earns simple interest of 5% annually. Find the balances after 15 and 20 completed years.

  1. 1.The yearly interest is 10000 × 0.05 = ₹500, calculated on the original principal each year.
  2. 2.After t completed years the balance is 10000 + 500t.
  3. 3.At t = 15, the balance is ₹17500; at t = 20, it is ₹20000.
  4. 4.The balances at times 0, 1, 2, … form an A.P. with difference 500.
  5. 5.Do not add annual balance observations together to obtain the final balance. If a phrase such as “15th year” is ambiguous, state the time convention before substituting.

Instalment problems can have a fixed principal payment plus interest on the unpaid principal. After each principal payment, the unpaid amount decreases by the same amount. Therefore the interest charges form an arithmetic sequence even though the story involves a percentage rate. The rate is being applied to a changing unpaid balance, not compounded onto the debt.

Example — Fixed principal instalments and declining interest

Problem
A tractor costs ₹12000. A buyer pays ₹6000 immediately, then repays the remaining principal in annual ₹500 instalments plus 12% interest on the unpaid balance at the start of each year. Find the total paid.

  1. 1.The financed principal is ₹6000 and 6000/500 = 12 annual principal instalments are needed.
  2. 2.The successive starting balances are 6000, 5500, 5000, …, 500. Interest is charged before that year’s principal instalment.
  3. 3.The interest charges are 720, 660, 600, …, 60, an A.P. with twelve terms.
  4. 4.Pair the first and last charges: 720 + 60 = 780. Every opposite pair gives 780, and there are six such pairs.
  5. 5.Total interest is 6 × 780 = ₹4680.
  6. 6.Total paid is ₹6000 immediately + ₹6000 in principal instalments + ₹4680 interest = ₹16680.
  7. 7.If interest were charged after each year’s principal payment, the timing and answer would change; the calculation above explicitly fixes the intended convention.
Example — A second instalment model

Problem
A scooter costs ₹22000. After a ₹4000 immediate payment, the remaining principal is repaid in annual ₹1000 instalments, with 10% interest on each starting unpaid balance. Find the total paid.

  1. 1.The unpaid principal is ₹18000, so there are 18 principal instalments.
  2. 2.Starting balances are 18000, 17000, …, 1000, giving interest charges 1800, 1700, …, 100.
  3. 3.Pairing opposite interest charges gives 1800 + 100 = 1900 per pair, with 9 pairs.
  4. 4.Total interest is 9 × 1900 = ₹17100.
  5. 5.The full amount paid is 22000 + 17100 = ₹39100.
  6. 6.These charges form an A.P. because each yearly principal reduction is the same ₹1000.

Branching and changing workforce models

A branching process repeats a multiplicative rule, while a workforce losing a fixed number daily follows an additive rule. In either case, decide whether the requested quantity is one stage or the total over stages. A clear time or stage table often reveals which series is appropriate.

Example — A specified generation of letters

Problem
A letter reaches four recipients. Each recipient sends four copies to new recipients, and the rule continues. At ₹0.50 per letter, what is the postage cost of the eighth batch alone?

  1. 1.Define batch 1 as the four letters sent by the original sender.
  2. 2.The batch counts are 4, 16, 64, …, with first term 4 and ratio 4.
  3. 3.Batch 8 contains 4 × 4⁷ = 4⁸ = 65536 letters.
  4. 4.The cost of that batch alone is 65536 × ₹0.50 = ₹32768.
  5. 5.A total through batch 8 would require a geometric sum. This calculation concerns only the eighth batch and assumes all recipients are distinct and continue the chain.

For an arithmetic list with first value u and last value v, opposite pairs all sum to u + v. If there are n terms, adding the list forward and backward gives 2S = n(u + v), so S = n(u + v)/2. The argument also works when n is odd: the central entry pairs with itself. We need only this short recall to sum daily worker counts.

Example — Worker-days and a delayed completion

Problem
150 workers were expected to complete a job in D days. Four leave before each successive day after day 1, and the job takes eight extra days. Assuming equal fixed daily productivity per worker, find the actual completion time.

  1. 1.Let the actual number of days be N = D + 8. Daily worker counts are 150, 146, 142, …, 150 − 4(N − 1).
  2. 2.The planned work is 150D worker-days. The actual work is N[150 + 150 − 4(N − 1)]/2 = N(152 − 2N).
  3. 3.Equate them using D = N − 8: N(152 − 2N) = 150(N − 8).
  4. 4.Simplifying gives N² − N − 600 = 0, or (N − 25)(N + 24) = 0.
  5. 5.Reject the negative time N = −24. Thus N = 25 days and D = 17 days.
  6. 6.On day 25, 150 − 4(24) = 54 workers remain, so the daily counts stay positive.
  7. 7.Check the work: 25(150 + 54)/2 = 2550 worker-days, equal to 150 × 17.

Check your understanding

These questions test whether you can identify the mathematical structure before carrying out a calculation. In a proof, restrictions and cancellation steps matter as much as the final expression.

Quiz

Quick check

The third and seventh terms are u and w, and the fifth is v. Which relationship follows?

Quick check

Two G.P.s have ratios 3 and 1/6. What ratio do their corresponding products have?

Quick check

For a finite G.P. with endpoint values a and b, which is always valid?

Quick check

A sum of real squares is at most zero. What must be true?

Quick check

If f(x + y) = f(x)f(y) and f(1) = 2 on positive integers, what is f(4)?

Quick check

A value loses 15% of its current value in each period. What factor applies each period?

Quick check

For first and next blocks of n terms, T = rⁿS. When is S/T = 1/rⁿ valid?

Quick check

Positive roots have A.M. 6 and G.M. 4. Which monic quadratic has those roots?

Practice Problems

Practice Problems
  1. The fourth, ninth and fourteenth G.P. terms are x, y, z. Prove y² = xz in two ways.
  2. Find the sum of corresponding products of 3, 6, 12, 24 and 16, 8, 4, 2. Explain the rR = 1 case.
  3. For five nonzero G.P. terms with first value 2 and last value 162, verify the squared endpoint-product identity for every real ratio allowed by those endpoints.
  4. A G.P. has first-three-term sum 13 and next-three-term sum 104. Find the first term and ratio.
  5. Derive the sum of 5 + 55 + 555 + … through n terms by expressing each term with a power of 10.
  6. Positive roots have arithmetic mean 7 and geometric mean 5. Find their quadratic and explain why both roots are positive.
  7. Compare ₹1000 growing at 10% simple interest and 10% annual compounding after three years, identifying the two update rules.
  8. Rewrite (u² + v²)r² − 2(uv + vw)r + (v² + w²) as a sum of squares and explain what equality implies.
  9. A debt of ₹2400 is repaid in four equal annual principal instalments plus 5% interest on each starting balance. Find the total interest, stating the payment timing.

Key Takeaways

Key Takeaways

• Translate each condition into an equation before choosing a solution method. • Equally spaced G.P. terms form a G.P. with a ratio that is a power of the original ratio. • Corresponding products have ratio rR, while reciprocal terms have ratio 1/r. • Endpoint pairing gives P² = (ab)ⁿ and helps connect term sums to reciprocal sums. • A shifted block sum is multiplied by rⁿ; quotient statements need nonzero sums. • A sum of nonnegative squares can be zero only when every square is zero. • A.M. and G.M. determine a sum and product, which also determine a quadratic with those numbers as roots. • Fixed additions create arithmetic patterns; fixed current-value percentages create geometric patterns. • Check signs, index domains, timing assumptions and all candidate solutions against the original problem.

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Geometric Progression and Means

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Chapter Summary and Practice