Educerie · IB Diploma · Mathematics: analysis and approaches
Topic 1 Number and algebra · 1.3 Geometric sequences and series
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Recognise a geometric sequence and find its common ratio | SL, HL | "Write down the common ratio" (1 mark) |
| Use uₙ = u₁rⁿ⁻¹ to find a term, or r and u₁ from two terms, including both signs of r when the gap is even | SL, HL | "u₂ = 6 and u₅ = 162. Find r and u₁" (3 to 4 marks) |
| Use Sₙ = u₁(rⁿ − 1)/(r − 1) in either form, for any r ≠ 1 | SL, HL | "Find the sum of the first 8 terms" (2 to 3 marks) |
| Read, write and evaluate sigma notation for geometric series | SL, HL | "Find ∑ 5 × 2ᵏ from k = 1 to 4" (3 marks) |
| Find the least n for which a term or a sum passes a value, using a GDC table | SL, HL | Paper 2: "Find the first year in which…" (3 marks) |
| Turn a percentage increase or decrease into a ratio and model a context | SL, HL | Paper 2: salary rises of 3% a year, a population falling 2.4% a year, a disease spreading |
| Compare an arithmetic and a geometric model | SL, HL | Paper 2 section B: "Which offer gives more over 12 years?" |
Before you start
You need 1.2: the ideas of term, position, uₙ, Sₙ and sigma notation are the same here, and the geometric formulas are best learnt side by side with the arithmetic ones. You also need powers, including negative bases, (−3)⁵ = −243, and percentages as multipliers: a 4% rise multiplies by 1.04.
1The idea in one paragraph
In a geometric sequence you get from each term to the next by multiplying by the same number every time, the common ratio, r. Where an arithmetic sequence climbs by equal steps, a geometric one grows (or shrinks) by equal percentages, so it starts slowly and then runs away, or dies away towards zero. The nth term is the first term multiplied by r a total of n − 1 times, and the sum of the first n terms comes from one neat trick: multiply the whole series by r and subtract, so nearly every term cancels. Salaries that rise by a percentage, populations, disease outbreaks and (in 1.4) money earning compound interest are all geometric.
2The common ratio, and what it does
A sequence is geometric when each term divided by the one before gives the same number:
r = u₂ ÷ u₁ = u₃ ÷ u₂ = … and the sequence is geometric only if these are all equal.
So 3, 6, 12, 24, … has r = 2, and 80, 60, 45, … has r = 60 ÷ 80 = 0.75. To test a sequence, divide every term by the one before. 2, 6, 18, 50 is not geometric: the ratios are 3, 3 and 2.77….
No term can be zero, and r cannot be zero, because you cannot divide by zero and a zero term would make every later term zero. Beyond that, the value of r decides the shape. Figure 1 shows the three behaviours.
| r | behaviour | example |
|---|---|---|
| r > 1 | grows, faster and faster | 3, 6, 12, 24, … |
| 0 < r < 1 | shrinks towards zero, never reaching it | 80, 60, 45, 33.75, … |
| r < 0 | the sign flips every term | 1, −2, 4, −8, … |
(r = 1 gives a constant sequence, which is dull but legal: 5, 5, 5, …)
Percentages become ratios. A rise of p% multiplies by 1 + p/100, and a fall of p% multiplies by 1 − p/100. A salary rising by 3% a year has r = 1.03; a population falling by 2.4% a year has r = 0.976. Never use r = 0.03 or r = 0.024: that would keep only 3% or 2.4% of the value each year.
3The nth term
To reach u₆ from u₁ you multiply by r five times, for the same reason that arithmetic sequences have n − 1 steps: six terms, five gaps.
uₙ = u₁rⁿ⁻¹ (in the formula booklet)
Find a term. For 3, 6, 12, …, u₁₀ = 3 × 2⁹ = 3 × 512 = 1536.
With a ratio below 1. For 80, 60, 45, …, u₈ = 80 × 0.75⁷ = 10.678… = 10.7 (3 s.f.). On Paper 1, a fraction is easier: 0.75 = 3/4, and questions will use numbers that work.
With a negative ratio. For 1, −2, 4, …, u₇ = 1 × (−2)⁶ = 64, and u₈ = (−2)⁷ = −128. The brackets matter: −2⁶ means −(2⁶) = −64, which is wrong. An even power of a negative ratio gives a positive term.
Find r and u₁ from two terms. A geometric sequence has u₃ = 18 and u₆ = 486. From u₃ to u₆ is three multiplications by r, as Figure 2 shows.
Divide, do not subtract: in a geometric sequence the terms are linked by multiplication, so division is what cancels u₁.
An even gap gives two answers. If u₂ = 12 and u₄ = 108, then r² = 108 ÷ 12 = 9, so r = 3 or r = −3. Both work: 4, 12, 36, 108, … and −4, 12, −36, 108, …. A question that says "all terms are positive" or "r > 0" is telling you to reject −3; if it says nothing, give both, and find u₁ for each. Missing the negative root is one of the most common lost marks in this subtopic.
Least n for a term to pass a value. For 80, 60, 45, …, which is the first term below 1? On Paper 2, put uₙ = 80 × 0.75ⁿ⁻¹ into the GDC's table and read down:
| n | 15 | 16 | 17 |
|---|---|---|---|
| uₙ | 1.43 | 1.07 | 0.802 |
The first term below 1 is u₁₇. Write the two values either side of the boundary; they are the evidence. (Once you have logarithms from 1.7, you can also solve 80 × 0.75ⁿ⁻¹ < 1 algebraically.)
4Geometric series and the sum formula
Sₙ is the sum of the first n terms, as before. Here is where the formula comes from. Write Sₙ out, then write r × Sₙ underneath, shifted one place, as in Figure 3.
Every term of r × Sₙ except the last is already in Sₙ. Subtract, and all the middle terms vanish:
Sₙ = u₁(rⁿ − 1)/(r − 1) = u₁(1 − rⁿ)/(1 − r), r ≠ 1 (both in the formula booklet)
The two forms are the same fraction with top and bottom multiplied by −1. When r > 1 the first keeps both brackets positive; when 0 < r < 1 the second does, and fewer negatives means fewer sign slips. Knowing where the formula comes from is also the best protection against misremembering it.
r > 1. Sum of the first 8 terms of 5 + 15 + 45 + …
0 < r < 1. Sum of the first 10 terms of 64 + 48 + 36 + …, where r = 0.75.
r < 0. Sum of the first 7 terms of 2 − 6 + 18 − …, where r = −3.
Brackets round the negative ratio, every time: (−3)⁷ = −2187, so 1 − (−2187) = 2188.
r = 1 makes the formula divide by zero. There the series is u₁ + u₁ + … + u₁ = nu₁, so you do not need a formula at all.
Least n for a sum to pass a value. How many terms of 5 + 15 + 45 + … are needed for the sum to exceed one million?
A GDC table of Sₙ against n finds this in seconds. Show the two sums that straddle the target, and give n as a whole number.
5Sigma notation for geometric series
A rule with the counter in the exponent gives a geometric series. Because r already stands for the common ratio, use a different counter, such as k, to avoid confusion.
∑ 3 × 2ᵏ⁻¹, from k = 1 to 10, = 3 + 6 + 12 + … + 1536.
To evaluate, find three things: the first term (put in the lower limit), the ratio (the number being raised to the power of k), and the number of terms (upper − lower + 1).
The first term is 12, not 4. The rule is evaluated at k = 1, and 3¹ = 3.
A lower limit of zero adds one term: ∑ 2ᵏ from k = 0 to 7 is 1 + 2 + 4 + … + 128, which has 8 terms and sums to 1(2⁸ − 1)/(2 − 1) = 255.
6Using technology
On Paper 2, the GDC's sequence tools list the terms and the sums. A table of uₙ or Sₙ against n is the fastest way to answer any "first n such that…" question. A spreadsheet does the same with one formula, such as a cell equal to 1.03 times the cell above, filled down.
As with arithmetic sequences, the guide says that if you use technology you must still identify the first term and the ratio. Write u₁, r, n and the formula before the calculator's answer. That line carries the method mark, and it means an arithmetic slip on the keypad does not cost you everything.
7Applications: salaries, populations and disease
Salary increase. Amira starts a job on €32,000 and is promised a 3% rise every year. Her salaries form a geometric sequence with u₁ = 32 000 and r = 1.03.
Salary decrease. A sales bonus of €6,000 in the first year falls by 10% each year after that, so r = 0.9.
Population growth. A city of 240,000 grows by 1.8% a year. Take care with the positions: if u₁ is the population now, the population after 10 years is u₁₁ = 240 000 × 1.018¹⁰ ≈ 286 873. The exponent is the number of years that have passed. Decide what u₁ means, write it down, and count from there.
The spread of a disease. In the early stage of an outbreak, each week's new cases can be close to a fixed multiple of the week before. Figure 4 uses invented figures: 40 new cases in week 1, rising by 25% a week, so r = 1.25. Panel (a) shows the terms, the new cases each week. Panel (b) shows the sums, the running total.
The question decides which you need: "how many new cases in week 8" is a term, "how many cases altogether" is a sum. And comment on the model when asked. A geometric model of an outbreak cannot run for ever: the population is finite, people recover or are vaccinated, and behaviour changes, so the ratio falls over time. The same is true of any population. Geometric models describe the early stage well and the long run badly.
Geometric sequences also appear in physics, where the activity of a radioactive sample falls by the same fraction in each equal time interval. And uₙ = u₁rⁿ⁻¹ is an exponential function of n, which is why this subtopic links to exponential models in Topic 2 and to exponential regression in Topic 4.
8Arithmetic or geometric?
Figure 5 compares two invented pay offers, both starting at €30,000. Offer A adds €1,200 a year (arithmetic). Offer B adds 3.5% a year (geometric).
At first the arithmetic offer is ahead: in year 2, €1,200 is more than 3.5% of €30,000, which is €1,050. But the geometric offer's rise grows every year, because it is a percentage of a larger salary. By year 10, B pays €40,886.92 against A's €40,800, and after that it pulls away. Over the full 20 years, A pays €828,000 in total and B pays about €848,390.
That is the general lesson. A geometric sequence with r > 1 always overtakes an arithmetic one in the end, however generous the arithmetic step, because equal percentages of a growing amount keep getting bigger. The question is when, and that is what a GDC table answers.
| Arithmetic | Geometric | |
|---|---|---|
| each step | add d | multiply by r |
| test | differences equal | ratios equal |
| nth term | u₁ + (n − 1)d | u₁rⁿ⁻¹ |
| sum | n/2 (2u₁ + (n − 1)d) | u₁(rⁿ − 1)/(r − 1) |
| graph against n | straight line | curve, bending away |
| context | a fixed amount each time | a fixed percentage each time |
Figure 6 shows how extreme geometric growth becomes. A well-known legend about the invention of chess, attached in some tellings to a sage named Sissa ibn Dahir, has the inventor ask for one grain of wheat on the first square of the board, two on the second, four on the third, doubling on each of the 64 squares. The total is a geometric series with u₁ = 1, r = 2 and n = 64.
Ten squares need only 1023 grains. Sixty-four need a number with twenty digits. On an axis where each gridline is a power of ten, as in Figure 6, geometric growth plots as a straight line, and that is the idea behind the logarithms of 1.5.
(What happens when r is between −1 and 1 and you keep adding terms for ever? The sum settles on a finite value. That is the infinite geometric series of 1.8.)
9Where marks are lost
Using n instead of n − 1 in the exponent. uₙ = u₁rⁿ⁻¹. The first term has r⁰.
Subtracting instead of dividing to find r. From two terms, divide one equation by the other so that u₁ cancels. Subtracting leaves u₁ tangled up with r.
Missing the negative root. r² = 9 gives r = ±3. Give both unless the question rules one out, and say why you reject one if you do.
Losing the brackets round a negative ratio. (−2)⁶ = 64 but −2⁶ = −64. For odd powers the two happen to agree, which hides the habit until an even power catches you. Bracket a negative r on paper and on the GDC.
Writing a percentage as the ratio. A 3% rise is r = 1.03, not 0.03 or 3. A 10% fall is r = 0.9, not 0.1 or −0.1.
Being one out on the position. "After 10 years" is u₁₁ if u₁ is the starting value. Write down what u₁ is before you choose n.
Confusing a term with a sum. "The salary in year 10" is u₁₀; "the total earned in 10 years" is S₁₀.
Starting a sigma sum at the wrong term. ∑ 4 × 3ᵏ from k = 1 starts at 12, not 4. Put the lower limit into the rule.
10Work it right
- Check the ratios are equal, then write down u₁ and r. Turn any percentage into a multiplier first.
- Decide what u₁ stands for in a context, and so which n the question needs.
- Underline whether the question wants a term (uₙ) or a sum (Sₙ).
- Given two terms, divide to cancel u₁, count the gaps between the positions, and look for a second value of r when that power is even.
- For a sum, use u₁(rⁿ − 1)/(r − 1) when r > 1 and u₁(1 − rⁿ)/(1 − r) when 0 < r < 1. Bracket a negative r.
- For "least n", use a GDC table, and write the two values either side of the target.
- In context, round sensibly (people to whole numbers, money to the cent or as instructed) and comment on whether the model can go on for ever.
11Try it
Marks in brackets. Questions 1 to 3 are Paper 1 style: no calculator. Questions 4 and 5 are Paper 2 style.
Q1. A geometric sequence has first three terms 2, −6, 18.
(a) Write down the common ratio. 1 mark
(b) Find u₆. 2 marks
(c) Find the sum of the first six terms. 2 marks
Q2. In a geometric sequence, u₂ = 6 and u₅ = 162.
(a) Find the common ratio and the first term. 3 marks
(b) Find S₅. 2 marks
Q3. Find the value of ∑ 5 × 2ᵏ, from k = 1 to 4. 3 marks
Q4. The population of a town (invented data) is 18,500 at the start of 2025. It is expected to fall by 2.4% each year.
(a) Find the expected population at the start of 2035. 2 marks
(b) Find the first year at the start of which the population is expected to be below 14,000. 3 marks
Q5. Jonas is offered a job with a salary of €38,000 in the first year, rising by 2.5% each year.
(a) Find his salary in the 12th year. 2 marks
(b) Find the total he earns over the first 12 years. 2 marks
A second employer offers €38,000 in the first year, rising by €1,000 each year.
(c) Determine which offer pays more in total over the first 12 years, and by how much. 3 marks
12In one breath
A geometric sequence multiplies by the same common ratio r at every step, found by dividing a term by the one before; r > 1 grows faster and faster, 0 < r < 1 dies away towards zero, r < 0 flips sign, and a p% change is the multiplier 1 ± p/100. The nth term is uₙ = u₁rⁿ⁻¹. Given two terms, divide to cancel u₁ and count the gaps; an even gap gives r = ± something, and both must be considered. The sum Sₙ = u₁(rⁿ − 1)/(r − 1) = u₁(1 − rⁿ)/(1 − r) comes from multiplying by r and subtracting; use the form that keeps the brackets positive, bracket a negative r, and remember r = 1 just gives nu₁. In sigma notation the counter sits in the exponent, so put in the lower limit to get the first term. Use a GDC table for "least n" and show the values either side. Salaries, populations and outbreaks are geometric while the percentage stays fixed, and a geometric sequence with r > 1 always overtakes an arithmetic one in the end.
Answers
Q1. (a) r = −6 ÷ 2 = −3. (b) u₆ = 2 × (−3)⁵ = 2 × (−243) = −486. (c)
(a) A1. (b) M1 for 2 × (−3)⁵, A1 for −486. (c) M1 for a correct sum formula with r = −3, A1 for −364. Writing −3⁵ as 243 loses the A1 in (b); follow through from (a) throughout.
Q2. (a)
(b) S₅ = 2(3⁵ − 1)/(3 − 1) = 2 × 242 ÷ 2 = 242. (a) M1 for dividing to eliminate u₁ (r³ = 27), A1 for r = 3, A1 for u₁ = 2. There is only one real cube root, so no second value arises. (b) M1 for substituting into a sum formula, A1 for 242. Follow through from (a).
Q3. The first term is 5 × 2¹ = 10, the ratio is 2, and there are 4 terms. Sum = 10(2⁴ − 1)/(2 − 1) = 10 × 15 = 150. Check: 10 + 20 + 40 + 80 = 150. A1 for u₁ = 10 and r = 2, M1 for a sum formula with n = 4 (or for listing and adding the four terms), A1 for 150. Taking the first term as 5 gives 75 and scores M1 only.
Q4. u₁ = 18 500 (the start of 2025) and r = 1 − 0.024 = 0.976. (a) The start of 2035 is ten years later, so it is u₁₁ = 18 500 × 0.976¹⁰ = 14 510.08… ≈ 14 510 people. (b)
The population first falls below 14,000 at the start of 2037. (a) M1 for 18 500 × 0.976¹⁰ (r = 0.976 and ten years of change), A1 for 14 510 (accept 14 500). (b) M1 for setting up the inequality or a table of values, A1 for the two values either side of 14,000, A1 for 2037. Using r = 0.024 scores zero for the whole question. An answer of 2036 from reading the position as the year scores M1 A1 A0.
Q5. u₁ = 38 000 and r = 1.025. (a) u₁₂ = 38 000 × 1.025¹¹ = 49 859.29… ≈ €49,859.29. (b) S₁₂ = 38 000(1.025¹² − 1)/(1.025 − 1) = €524,231.01 (nearest cent). (c) The second offer is arithmetic with u₁ = 38 000 and d = 1000, so S₁₂ = 12/2 × (2 × 38 000 + 11 × 1000) = 6 × 87 000 = €522,000. The first (geometric) offer pays more, by 524 231.01 − 522 000 = €2,231.01. (a) M1 for u₁ × r¹¹, A1. (b) M1 for a geometric sum formula with n = 12, A1. (c) M1 for an arithmetic sum with d = 1000 and n = 12, A1 for €522,000, R1 for a conclusion naming the geometric offer with the difference (follow through from their (b)). Accept answers to the nearest euro throughout.
Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 1.3 Geometric sequences and series. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.