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Lesson 2 of 4

Sequences and Series · Lesson 2 of 4

Geometric Progression and Means

“Follow repeated multiplication to find distant terms, finite sums and geometric means.”

Learning Objectives

• Recognise a geometric progression and interpret positive, negative and fractional ratios. • Derive and apply the general-term and finite-sum formulas, including r = 1. • Recover a first term, a ratio or a term count from given information. • Transform a related series into geometric sums and model repeated proportional change. • Insert geometric means and distinguish positive means from signed G.P. branches. • Prove A.M. ≥ G.M. for positive numbers and recover numbers from their means.

Repeated multiplication creates a progression

In 2, 4, 8, 16, …, each value is twice the preceding value. In 81, 27, 9, 3, …, each is one-third of the preceding value. These lists change at different speeds and in different directions, yet both follow one repeated multiplication rule. That common structure is a geometric progression.

Definition
Geometric progression

A geometric progression, or G.P., is a sequence of nonzero terms in which the ratio of each term to its immediately preceding term is a fixed nonzero number r. This number is called the common ratio.

To find r, divide a later term by the term immediately before it. Check that the same value works for all displayed pairs. The list 2, 4, 7, … is not a G.P.: the first quotient is 2 but the next is 7/4. A single quotient gives a candidate ratio, not enough evidence that a longer given list follows it.

Constant-ratio conditionLaTeX
All terms are nonzero, so the quotient is defined. The ratio r is nonzero under this chapter’s definition. The first term is denoted by a.

A negative common ratio reverses the sign at every step. For 1/9, −1/27, 1/81, −1/243, …, dividing consecutive terms gives −1/3. The magnitudes shrink because the magnitude of the ratio is less than 1. For 0.01, 0.0001, 0.000001, …, the common ratio is 0.01, not 0.1: multiplying by 0.01 moves the decimal point two places.

RatioEffect for a positive first termExample
r > 1Positive terms grow.2, 6, 18, 54, …
0 < r < 1Positive terms shrink.8, 4, 2, 1, …
r = 1The term remains unchanged.5, 5, 5, 5, …
r < 0Signs alternate; magnitudes depend on |r|.3, −6, 12, −24, …
Common mistake

A G.P. uses a constant ratio, not a constant difference. Also, “r > 1 means increasing” needs a sign condition: for a negative first term, multiplying by 2 makes the values more negative. Under the nonzero-term definition, a list that reaches zero is not a G.P.

Why the general term contains n − 1

The first term is already present before any multiplication by r occurs. Reaching the second term takes one multiplication; reaching the third takes two. Count the gaps from the first position to the required position, rather than counting the positions themselves.

Starting from a, the terms are a, ar, ar², ar³, … . Their exponents are 0, 1, 2, 3, …, one less than their positions. Therefore the term at position n is the first term multiplied by r exactly n − 1 times.

General term of a G.P.LaTeX
a is the first term, r is the common ratio and n is a positive integer position. For n = 1 the formula gives ar⁰ = a, as it must.
Count the gaps, not the positionsaarar²ar³× r →× r →× r →position 1position 2position 3position 4Position n requires n − 1 multiplications
Positions and multiplication steps— There are three multiplication steps from the first term to the fourth term. That is why the exponent at position 4 is 3.
Example — A distant term and a general term

Problem
Find the tenth and nth terms of 5, 25, 125, … .

  1. 1.The first term is a = 5 and the ratio is r = 25/5 = 5.
  2. 2.a₁₀ = 5 × 5⁹ = 5¹⁰ = 9,765,625.
  3. 3.For an arbitrary position n, aₙ = 5 × 5ⁿ⁻¹ = 5ⁿ.
  4. 4.Checking n = 1 and n = 2 gives 5 and 25, so the index convention agrees with the list.
Example — Finding a position

Problem
Which term of 2, 8, 32, … equals 131072?

  1. 1.Here a = 2 and r = 4. Set 2 × 4ⁿ⁻¹ = 131072.
  2. 2.Divide by 2: 4ⁿ⁻¹ = 65536.
  3. 3.Since 65536 = 4⁸, compare exponents: n − 1 = 8.
  4. 4.Thus n = 9. The value is the ninth term, after eight multiplication steps.
  5. 5.A position must be a positive integer. If the equation produced no such integer, the proposed value would not occur in the sequence.

When two terms are known, writing two general-term equations often lets us eliminate a. Dividing a later-term equation by an earlier-term equation cancels the first term and subtracts the exponents. The exponent left over is the gap between the two positions.

Comparing two termsLaTeX
The subscripts p and q denote positions. This relationship follows by dividing ar^(q−1) by ar^(p−1); it does not assume the terms are consecutive.
Example — Recovering the ratio and first term

Problem
The third term of a G.P. is 24 and the sixth is 192. Find the tenth term.

  1. 1.Write ar² = 24 and ar⁵ = 192.
  2. 2.Divide the second equation by the first: r³ = 192/24 = 8.
  3. 3.The real solution is r = 2. Substitute into ar² = 24 to get 4a = 24, so a = 6.
  4. 4.Then a₁₀ = 6 × 2⁹ = 6 × 512 = 3072.
  5. 5.Check the given data: 6 × 2² = 24 and 6 × 2⁵ = 192.
Common mistake

If an even power of r is known, both positive and negative real ratios may fit. For example, r² = 4 permits r = 2 and r = −2. Use other conditions to choose between them; do not discard a valid sign merely because the displayed term is positive.

Deriving a finite geometric sum

Adding a few terms directly is easy, but adding hundreds separately is inefficient. The useful observation is that multiplying the entire sum by r shifts each term to the next term in the progression. Subtraction then cancels all the interior terms.

Write Sₙ = a + ar + ar² + … + arⁿ⁻¹. Multiplication by r gives rSₙ = ar + ar² + … + arⁿ. In the second expression the initial a has disappeared and the new final arⁿ has appeared. All the shared terms cancel when the second line is subtracted from the first.

Subtract the shifted sumLaTeX
The cancellation is the reason the formula works. We may divide by 1 − r only when r ≠ 1.
Sum of the first n termsLaTeX
Both expressions are identical: the numerator and denominator have each been multiplied by −1. Choose whichever makes the calculation easier.

For r = 1, every term is a, so the sum is a added n times. The fraction formula would have denominator zero and cannot be used. Handling this case separately is part of using the sum formula correctly, not an optional shortcut.

Constant progressionLaTeX
This follows directly from n equal terms. The first term is still nonzero under the definition used here.

If l denotes the last included term, then l = arⁿ⁻¹ and lr = arⁿ. Substituting this into the sum formula gives Sₙ = (a − lr)/(1 − r). This version can help when the first term, last term and ratio are given but n is not yet known. It still requires r ≠ 1.

Using the last termLaTeX
l is the nth term, not the term immediately after it. Multiplying l by r gives the extra term in the shifted sum.
Example — A fractional common ratio

Problem
Find Sₙ and S₅ for 1 + 2/3 + 4/9 + … .

  1. 1.The first term is 1 and the ratio is 2/3.
  2. 2.Sₙ = [1 − (2/3)ⁿ]/[1 − 2/3] = 3[1 − (2/3)ⁿ].
  3. 3.For n = 5, (2/3)⁵ = 32/243.
  4. 4.Thus S₅ = 3(1 − 32/243) = 3 × 211/243 = 211/81.
  5. 5.There are five terms; the last is (2/3)⁴, but the sum formula contains (2/3)⁵ because it comes from the shifted sum.
Example — The separate r = 1 case

Problem
Find the sum of twelve terms of 7, 7, 7, … .

  1. 1.The common ratio is 7/7 = 1.
  2. 2.Use S₁₂ = 12 × 7 = 84.
  3. 3.Substituting r = 1 into a(1 − rⁿ)/(1 − r) would produce 0/0, which is undefined.
  4. 4.The direct constant-term calculation supplies the correct result.
Example — A sum with alternating signs

Problem
Find the sum of the first five terms of 3, −6, 12, −24, … .

  1. 1.Here a = 3 and r = −2.
  2. 2.S₅ = 3[1 − (−2)⁵]/[1 − (−2)].
  3. 3.Since (−2)⁵ = −32, the numerator is 3(1 + 32) = 99 and the denominator is 3.
  4. 4.Therefore S₅ = 33.
  5. 5.Directly, 3 − 6 + 12 − 24 + 48 = 33, confirming the sign handling.
Example — Recovering a term count from a sum

Problem
How many terms of 3, 3/2, 3/4, … have sum 3069/512?

  1. 1.Here a = 3 and r = 1/2, so Sₙ = 6(1 − 1/2ⁿ).
  2. 2.Set 3069/512 = 6(1 − 1/2ⁿ) and divide by 6.
  3. 3.This gives 3069/3072 = 1 − 1/2ⁿ.
  4. 4.Hence 1/2ⁿ = (3072 − 3069)/3072 = 3/3072 = 1/1024.
  5. 5.Since 1024 = 2¹⁰, n = 10. Ten terms are required.
  6. 6.The sum is less than 6, consistent with the expression 6(1 − 1/2ⁿ) for a finite positive n.
Common mistake

Do not confuse the last-term exponent n − 1 with the sum-formula exponent n. Also, finite sums are valid for negative ratios and ratios of magnitude greater than 1. We are not assuming that an infinite geometric series has a finite sum.

Choosing convenient unknown terms

Unknown terms can be written in several equivalent ways. For three consecutive nonzero terms, a/r, a, ar makes the central value easy to identify because their product is a³. In this representation a is the middle term, so it is important to say that explicitly.

Example — Three terms from a sum and a product

Problem
Three consecutive G.P. terms have sum 13/12 and product −1. Find the possible ratios and terms.

  1. 1.Represent the terms as m/r, m, mr, with m the middle term and r ≠ 0.
  2. 2.Their product is m³ = −1, so the real middle term is m = −1.
  3. 3.The sum equation becomes −1/r − 1 − r = 13/12.
  4. 4.Multiply by 12r and rearrange: 12r² + 25r + 12 = 0.
  5. 5.Factor: (3r + 4)(4r + 3) = 0. Therefore r = −4/3 or r = −3/4.
  6. 6.For r = −3/4, the terms are 4/3, −1, 3/4. For r = −4/3, they are 3/4, −1, 4/3.
  7. 7.Both have sum 13/12 and product −1. The two ratios correspond to reversing the order of the terms.

A series can contain a hidden G.P.

A list of repeated digits, such as 7, 77, 777, …, does not have a constant ratio. Nevertheless, each entry can be expressed using a power of 10. This turns its sum into a geometric sum together with a simple correction.

The k-digit number made entirely of sevens is 7(10ᵏ − 1)/9. To see why, observe that 10ᵏ − 1 is a string of k nines; dividing by 9 creates k ones and multiplying by 7 creates k sevens. We then add these expressions for k = 1 through n.

Example — Repeated sevens

Problem
Find 7 + 77 + 777 + … through the term containing n sevens.

  1. 1.The kth term is (7/9)(10ᵏ − 1).
  2. 2.Therefore Sₙ = (7/9)[(10 + 10² + … + 10ⁿ) − n].
  3. 3.The power-of-ten sum has first term 10, ratio 10 and n terms. Its sum is 10(10ⁿ − 1)/9.
  4. 4.Thus Sₙ = (7/9)[10(10ⁿ − 1)/9 − n].
  5. 5.For n = 2, this gives (7/9)(110 − 2) = 84, which equals 7 + 77.
  6. 6.We used a G.P. inside the expression; we did not claim that the original repeated-digit sequence is itself a G.P.

Counting repeated proportional change

A geometric model applies when each stage multiplies the previous stage by the same factor. Distinguish the value at one stage from the total across stages. Also decide whether the original observation is indexed at time zero or at position one before applying a formula.

Example — One generation and all generations

Problem
In the idealised ancestor model, find the total number of ancestor positions across ten preceding generations.

  1. 1.The generation counts are 2, 4, 8, …, 1024, a G.P. with a = 2, r = 2 and n = 10.
  2. 2.The tenth-generation count alone is a₁₀ = 2 × 2⁹ = 1024.
  3. 3.The requested total is S₁₀ = 2(2¹⁰ − 1)/(2 − 1).
  4. 4.Therefore S₁₀ = 2(1024 − 1) = 2046 positions.
  5. 5.The distinction between 1024 and 2046 is the distinction between one term and the sum of ten terms.
Example — Doubling from time zero

Problem
A culture initially contains 30 bacteria and doubles every hour. Find the populations after 2, 4 and n hours.

  1. 1.Treat the original observation as time t = 0: N₀ = 30.
  2. 2.After one hour, N₁ = 30 × 2; after two hours, N₂ = 30 × 2² = 120.
  3. 3.After four hours, N₄ = 30 × 2⁴ = 480.
  4. 4.After n hours, Nₙ = 30 × 2ⁿ.
  5. 5.If we list the initial population as the first sequence term, then the population after n hours is at position n + 1. This explains why the time formula uses exponent n rather than n − 1.

An annual proportional increase works the same way. If an amount gains 10% each year, its new value is the old value plus 0.10 times that old value: together these are 1.10 times the old value. Repeating this update gives a geometric sequence of balances. This is a mathematical model with the stated rate held fixed.

Example — Compounding a fixed percentage

Problem
Find the amount after ten years for an initial ₹500 growing by 10% per year, compounded annually.

  1. 1.One yearly update multiplies the amount by 1 + 10/100 = 1.1.
  2. 2.After ten updates, A₁₀ = 500(1.1)¹⁰.
  3. 3.Since (1.1)¹⁰ = 2.5937424601, A₁₀ = 1296.87123005.
  4. 4.The amount is approximately ₹1296.87, with rounding done at the end.
  5. 5.The exponent counts ten completed annual updates; adding the balances from separate years would answer a different question.

A geometric mean balances a ratio

An arithmetic mean balances equal differences. A geometric mean balances equal ratios. For positive numbers a and b, placing G between them as a, G, b requires the same multiplication factor in both steps.

The equal-ratio condition is G/a = b/G. Multiplying by aG gives G² = ab. Two algebraic square roots satisfy this equation, but the geometric mean of two positive numbers is defined to be the positive root. That definition distinguishes the usual mean from a signed middle term in a G.P.

Geometric mean of two positive numbersLaTeX
The positive square root is chosen. For example, between 2 and 8 the geometric mean is 4, since 2, 4, 8 have ratio 2.
Example — Mean versus signed middle term

Problem
Find the geometric mean of 4 and 36, and all real middle terms that make 4, x, 36 a G.P.

  1. 1.The geometric mean is √(4 × 36) = √144 = 12.
  2. 2.For the signed G.P. question, x² = 144 permits x = 12 and x = −12.
  3. 3.The positive branch 4, 12, 36 has ratio 3.
  4. 4.The negative branch 4, −12, 36 has ratio −3.
  5. 5.Only 12 is the geometric mean under the positive-number definition; both values answer the signed G.P. question.

Inserting several geometric means

To insert n values between two positive endpoints, count the complete list and then count its gaps. The endpoints plus the inserted values make n + 2 terms, so there are n + 1 ratio steps from the first endpoint to the last. This count determines the power of r.

Write the list as a, G₁, G₂, …, Gₙ, b. The endpoint equation is b = arⁿ⁺¹. For the positive geometric means we take the positive root r = (b/a)^(1/(n + 1)). The kth inserted mean is then Gₖ = arᵏ, because it is k steps after a. This rule works whether the positive endpoints increase or decrease.

Positive inserted geometric meansLaTeX
Here n counts inserted means, not all terms. The endpoint b occupies position n + 2. Both a and b are positive.
Example — Three inserted means, with a signed branch

Problem
Insert three numbers between 1 and 256 so that the complete list is a G.P.

  1. 1.There are five terms and four gaps, so 256 = 1 × r⁴.
  2. 2.The real ratios are r = 4 and r = −4.
  3. 3.For r = 4, the list is 1, 4, 16, 64, 256. The three positive means are 4, 16, 64.
  4. 4.For r = −4, the list is 1, −4, 16, −64, 256. These also satisfy the signed G.P. requirement.
  5. 5.If the request specifically asks for positive geometric means, retain only the positive branch. If it asks for all real insertions, show both.
  6. 6.An even exponent in the endpoint equation allows the negative-ratio branch here; an odd exponent with positive endpoints would force a positive real ratio.
Common mistake

“Between” here means between the endpoints in the written list, not necessarily between them in numerical value. Inserting n values requires n + 1 multiplications. Using exponent n would put the final endpoint one position too early.

Why the arithmetic mean is at least the geometric mean

For two positive numbers, the arithmetic mean is their sum divided by 2 and the geometric mean is the positive square root of their product. Comparing them is easier if we turn their difference into a square. A real square can never be negative, so it gives both an inequality and its equality condition.

Let A = (a + b)/2 and G = √ab. Because a and b are positive, their square roots are real and √a√b = √ab. The numerator a + b − 2√ab equals (√a − √b)². Dividing this nonnegative square by 2 gives A − G ≥ 0.

Arithmetic mean–geometric mean inequalityLaTeX
Equality holds exactly when √a = √b, or a = b. The stated setting is two positive real numbers.

For 4 and 16, A = 10 and G = 8, so A exceeds G. For 9 and 9, both means are 9 and equality holds. Equal endpoints have no imbalance to average out. Unequal endpoints make the square in the proof positive, giving a strict inequality.

Example — Recovering numbers from their means

Problem
Two positive numbers have arithmetic mean 10 and geometric mean 8. Find them.

  1. 1.The arithmetic mean gives a + b = 20. The geometric mean gives ab = 8² = 64.
  2. 2.Use (a − b)² = (a + b)² − 4ab = 400 − 256 = 144.
  3. 3.Therefore a − b = 12 or −12.
  4. 4.Combining a + b = 20 with a − b = 12 gives a = 16 and b = 4. The other sign reverses their order.
  5. 5.Check: (16 + 4)/2 = 10 and √(16 × 4) = 8. The numbers are 4 and 16.

More generally, if A and G are given positive means, write the unknown numbers as A + d and A − d. Their product is A² − d² = G², so d² = A² − G². This explains both the formula below and the requirement A ≥ G. If A < G, there is no real pair of positive numbers with those means.

Numbers from their two meansLaTeX
The two signs give the two numbers. For G > 0 and A ≥ G, the smaller value is positive because √(A² − G²) < A.
Example — Checking whether means are possible

Problem
Can positive numbers have A = 5 and G = 6? What happens if A = G = 5?

  1. 1.The inequality requires A ≥ G. The proposed values 5 and 6 violate that condition, so no such positive pair exists.
  2. 2.For A = G = 5, the square root √(A² − G²) is zero.
  3. 3.Both numbers are 5, agreeing with the equality condition a = b.

Check your understanding

Decide which quantity is being requested before calculating: a term, a sum, a ratio or a mean. The sign and index conditions are part of the reasoning.

Quiz

Quick check

For 3, −6, 12, …, what are r and a₄?

Quick check

Why is the exponent in aₙ = arⁿ⁻¹ one less than n?

Quick check

What is the sum of eight terms of 5, 5, 5, …?

Quick check

If a₃ = 12 and a₆ = 96 in a real G.P., what is r?

Quick check

How many ratio steps connect the endpoints when four means are inserted?

Quick check

What is the geometric mean of 9 and 25?

Quick check

For positive a and b, when does A.M. = G.M.?

Quick check

An initial population of 20 triples hourly. What is it after three hours?

Practice Problems

Practice Problems
  1. For the G.P. 5/2, 5/4, 5/8, …, find its nth and twentieth terms.
  2. Find both real ratios if a₂ = 6 and a₄ = 24, and give the corresponding first terms.
  3. Derive the finite-sum formula by writing the original and shifted sums, then explain why r = 1 is separate.
  4. Find how many terms of 4, 12, 36, … have sum 484.
  5. Insert two positive geometric means between 3 and 81, checking every consecutive ratio.
  6. Find positive numbers with A.M. 13 and G.M. 12, and verify the result.
  7. A value decreases by 20% in each stage. Write its value after five stages in terms of its initial value V.
  8. Express the sum 8 + 88 + 888 + … through n terms using a geometric sum.

Key Takeaways

Key Takeaways

• A G.P. has nonzero terms and a fixed nonzero ratio; negative ratios alternate signs. • The nth term is arⁿ⁻¹ because n − 1 multiplications follow the first term. • For r ≠ 1, Sₙ = a(1 − rⁿ)/(1 − r); for r = 1, Sₙ = na. • Dividing term equations cancels the first term and exposes a power of the ratio. • n inserted means create n + 1 gaps between the endpoints. • The geometric mean of positive a and b is √ab; signed G.P. insertions can give extra branches. • For positive numbers, A.M. ≥ G.M., with equality exactly when the numbers are equal. • A time-zero observation needs one update per elapsed period; model its exponent accordingly.

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Sequences and Series

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Mixed Applications