Geometric sequences and series AI HL

IB Mathematics AI HL · Unit 1

Number and Algebra · Lesson 1.3

Geometric Sequences and Series

Explore repeated multiplication, growth, decay, and the total produced by a changing pattern.

Unit 1 theme · God’s love

✦ Wisdom

Some geometric patterns grow beyond what we first expect. When the Bible speaks of God’s love, it also directs our attention beyond what we can measure. Ephesians 3:18–19 invites us to consider the breadth, length, height, and depth of Christ’s love.

God’s love is made known to us in Jesus Christ. You are invited to come to Jesus with your questions, your burdens, and your need for forgiveness, and to trust him as your Savior and Lord.

🎯 Learning objectives

By the end of this lesson, you should be able to:

identify a geometric sequence and its common ratio;
find any term using the nth-term formula;
find the sum of the first n terms;
interpret sigma notation for geometric series;
model repeated percentage increase and decrease;
compare arithmetic and geometric models; and
explain the assumptions and limitations of a model.

01 · Recognising a geometric sequence

A geometric sequence is formed by multiplying each term by the same number. This number is the common ratio, written

$r.$

$r=\frac{\text{a term}}{\text{the preceding term}}$

Worked example

Consider

$6,\ 18,\ 54,\ 162,\ldots$

$\frac{18}{6}=3,\qquad \frac{54}{18}=3,\qquad \frac{162}{54}=3.$

The common ratio is

$\boxed{r=3}.$

Each term is three times the previous term.

Compare: An arithmetic sequence has a constant difference. A geometric sequence has a constant ratio.

02 · Growth, decay, and alternating signs

The value of the ratio tells us how the sequence behaves:

If
$r>1,$

positive terms increase.
If
$0<r<1,$

positive terms decrease towards zero.
If
$r<0,$

the signs alternate.

Worked example

In the sequence

$80,\ 40,\ 20,\ 10,\ldots,$

$r=\frac{40}{80}=\frac12.$

Each term is half the preceding term. This is a model of 50% decrease per step.

Percentage language: A 10% increase means

$r=1.10.$

A 10% decrease means

$r=0.90.$

03 · Finding the nth term

Starting with the first term

$u_1,$

we multiply by

$r$

a total of

$n-1$

times to reach

$u_n.$

$\boxed{u_n=u_1r^{\,n-1}}$

Worked example

A sequence begins

$5,\ 10,\ 20,\ldots$

Find the eighth term.

Here

$u_1=5$

and

$r=2.$

Therefore,

$u_8=5(2)^{8-1}=5(128)=\boxed{640}.$

Watch the index: The exponent for the eighth term is 7, not 8.

04 · Finding the ratio from two terms

If two terms are known, substitute each into

$u_n=u_1r^{n-1}.$

When solving an even power of

$r,$

check whether both a positive and a negative ratio are possible.

Worked example

A geometric sequence has

$u_1=5$

and

$u_6=\frac{32}{625}.$

Find the common ratio.

$\frac{32}{625}=5r^5$

$r^5=\frac{32}{3125} =\left(\frac25\right)^5.$

Since the fifth power preserves the sign,

$\boxed{r=\frac25}.$

The second term is

$u_2=5\left(\frac25\right)=2.$

05 · The sum of the first n terms

An arithmetic or geometric sequence is a list of terms. A series is their sum.

For a geometric sequence with

$r\ne1,$

$\boxed{S_n=u_1\frac{1-r^n}{1-r}}$

The equivalent form

$S_n=u_1\frac{r^n-1}{r-1}$

is often convenient when

$r>1.$

Both forms give the same result.

Worked example

Find the sum of the first six terms of

$3,\ 6,\ 12,\ldots$

$S_6=3\frac{2^6-1}{2-1} =3(63)=\boxed{189}.$

Special case: If

$r=1,$

every term equals

$u_1,$

$S_n=nu_1.$

06 · Sigma notation

The capital Greek letter

$\Sigma$

means “sum.” The number below it gives the starting index and the number above it gives the final index.

Worked example

Evaluate

$\displaystyle\sum_{k=1}^{5}3(2)^{k-1}.$

The terms are

$3,\ 6,\ 12,\ 24,\ 48.$

Therefore,

$\sum_{k=1}^{5}3(2)^{k-1} =3\frac{2^5-1}{2-1} =\boxed{93}.$

07 · Repeated percentage change

If a quantity increases by

$p\%$

each period, multiply it by

$1+\frac{p}{100}$

each time. If it decreases by

$p\%,$

multiply by

$1-\frac{p}{100}.$

Worked example

A business earns €200 in its first week. Its weekly earnings increase by 8% each week. Estimate its earnings in week 10.

$u_{10}=200(1.08)^9\approx\boxed{€400}.$

There are nine increases between week 1 and week 10. The answer is rounded to the nearest euro.

Interpretation: A 10% increase every week is not the same as adding 10% of the first week’s amount every week. One produces a geometric pattern; the other produces an arithmetic pattern.

08 · Decreasing series and a limiting total

When

$0<r<1,$

a geometric series continues to grow, but each new term is smaller. Its sums can approach a fixed value. For

$|r|<1,$

the limiting sum is

$S_\infty=\frac{u_1}{1-r}.$

Worked example

A balloon rises 450 m in the first minute. During each following minute, it rises 82% of the distance risen in the preceding minute. What height does this model approach?

$S_\infty=\frac{450}{1-0.82} =\frac{450}{0.18} =\boxed{2500\text{ m}}.$

Under this model, the balloon remains below 2500 m after every finite number of minutes.

Model limitation: A real balloon’s motion depends on conditions such as wind, air density, and temperature. A fixed ratio is an assumption, not a physical law.

🧮 Technology and examination tip

A spreadsheet or graphing calculator can generate geometric terms and their running totals. In an IB examination, identify the first term, common ratio, and term number before entering a formula. Check whether “week 1” is the first term: reaching week 11 involves ten changes.

✏️ IB-style Paper 1 practice

A graphic display calculator may be used. Show appropriate working.

The first three terms of a geometric sequence are

$4,\ 6,\ 9.$

(a) Find the common ratio. [1 mark]

(b) Find the eighth term. [2 marks]

(c) Find the sum of the first eight terms. [3 marks]
A geometric sequence has first term 81 and fifth term 1.
Its common ratio is positive.

(a) Find the common ratio. [2 marks]

(b) Find the second term. [1 mark]

(c) Find the sum of the first five terms. [3 marks]
Evaluate

$\displaystyle\sum_{k=3}^{8}7(2)^{k-3}.$

State the number of terms in the sum. [4 marks]
A geometric sequence has

$u_1=5$

and

$r=\frac25.$

Find the smallest positive integer

$k$

for which

$S_k>8.$

[4 marks]
The first term of a geometric sequence is 64 and its common ratio is

$-\frac12.$

(a) Write down its first four terms. [2 marks]

(b) Find

$u_7.$

[2 marks]

(c) Find

$S_8.$

[3 marks]

Reveal Paper 1 solutions

(a)

$r=\frac64=\frac32=1.5.$

(b)

$u_8=4(1.5)^7=68.34375.$

(c)

$S_8=4\frac{1.5^8-1}{1.5-1} =197.03125\approx\boxed{197}.$
(a)

$1=81r^4,$

so

$r^4=\frac1{81}.$

The ratio is positive, hence

$r=\frac13.$

(b)

$u_2=81\left(\frac13\right)=27.$

(c) The terms are

$81,\ 27,\ 9,\ 3,\ 1,$

giving

$\boxed{S_5=121}.$
There are

$8-3+1=6$

terms. The first is 7 and the ratio is 2:

$\sum_{k=3}^{8}7(2)^{k-3} =7\frac{2^6-1}{2-1} =\boxed{441}.$
$S_3=5+2+0.8=7.8<8,$

whereas

$S_4=7.8+0.32=8.12>8.$

Therefore,

$\boxed{k=4}.$
(a)

$64,\ -32,\ 16,\ -8.$

(b)

$u_7=64\left(-\frac12\right)^6=1.$

(c)

$S_8=64\frac{1-(-1/2)^8}{1-(-1/2)} =\boxed{42.5}.$

🌍 Applications and advanced challenges

Linking two kinds of sequence. The second term of an arithmetic
sequence is 30 and its fifth term is 90.

(a) Find the common difference and first term of the arithmetic sequence.
[3 marks]

(b) Its first, second, and fifth terms become the first three terms of a
geometric sequence. Find the seventh term of the geometric sequence.
[4 marks]
Two businesses. A café earns 60 USD profit in its first week, and its weekly profit increases by 10USD each week. A tea shop also earns
60USD in its first week, but its weekly profit increases by 10% each week.

(a) Find each business’s profit in week 11.
[4 marks]

(b) Find each business’s total profit over its first 12 weeks.
[4 marks]

(c) Find the first week $m$ at the end of which the tea shop’s
cumulative profit exceeds the café’s cumulative profit.
[4 marks]
Balloon model. A balloon rises 450 m during its first minute.
During each later minute it rises 82% of the distance it rose during the
previous minute.

(a) Find its height after 10 complete minutes.
[3 marks]

(b) According to the model, will it ever reach a height of 2520 m?
Justify your answer. [3 marks]

(c) Suggest one limitation of this model. [1 mark]
Population model. A town has 12 000 residents at the
beginning of 2026. Its population is modelled as increasing by 4%
each year.

(a) Write a formula for its modelled population

$P_n$

at the
beginning of the year

$2026+n.$

[2 marks]

(b) Predict the population at the beginning of 2031, to the
nearest person. [2 marks]

(c) State one reason the prediction may be inaccurate.
[1 mark]
Advanced threshold problem. A video receives 200 views
on its first day. The number of views each day is modelled as 20%
greater than on the previous day.

(a) Write an expression for the total modelled views over the
first

$n$

days. [3 marks]

(b) Find the first day by the end of which the cumulative total
exceeds 5000 views. [4 marks]

(c) Explain why the model is unlikely to remain valid indefinitely.
[1 mark]

Reveal application solutions

(a)

$u_2=u_1+d=30$

and

$u_5=u_1+4d=90.$

Subtracting gives

$3d=60,$

so

$\boxed{d=20}$

and

$\boxed{u_1=10}.$

(b) The three selected arithmetic terms are

$10,\ 30,\ 90.$

They form a geometric sequence with

$r=3.$

Its seventh term is

$u_7=10(3)^6=\boxed{7290}.$
(a) Café:

$u_{11}=60+10(10)=\boxed{\$160}.$

Tea shop:

$u_{11}=60(1.10)^{10} \approx\boxed{\$155.62}.$

(b) Café:

$S_{12}=\frac{12}{2}(60+170)=\boxed{\$1380}.$

Tea shop:

$S_{12}=60\frac{1.10^{12}-1}{0.10} \approx\boxed{\$1283.06}.$

(c) Compare

$C_m=\frac{m}{2}[120+10(m-1)]$

with

$T_m=600(1.10^m-1).$

At

$m=16,$

the tea shop’s cumulative total is still below
the café’s. At

$m=17,$

$T_{17}\approx\$2432.68>C_{17}=\$2380.$

Thus the first such week is

$\boxed{m=17}.$
(a) The distances risen each minute form a geometric sequence:

$S_{10}=450\frac{1-0.82^{10}}{1-0.82} \approx\boxed{2156\text{ m}}.$

(b) The limiting height is

$S_\infty=\frac{450}{1-0.82}=2500\text{ m}.$

Since

$2500<2520,$

the balloon will not
reach 2520 m under this model.

(c) For example, wind and changing air conditions mean that
the rise is unlikely to remain exactly 82% of the previous
minute’s rise.
(a)

$\boxed{P_n=12000(1.04)^n}.$

(b) At the beginning of 2031,

$n=5:$

$P_5=12000(1.04)^5\approx\boxed{14\,600\text{ people}}.$

(c) Births, deaths, or migration may change, so annual growth
need not remain at 4%.
(a)

$\boxed{S_n=200\frac{1.2^n-1}{0.2} =1000(1.2^n-1)}.$

(b) We need

$1000(1.2^n-1)>5000,$

or

$1.2^n>6.$

A calculator gives

$S_9\approx4159.78<5000,$

while

$S_{10}\approx5191.74>5000.$

The first such day is

$\boxed{\text{day 10}}.$

(c) For example, the potential audience is limited; daily
views cannot keep increasing by 20% forever.

✓ Lesson recap

A geometric sequence has a constant ratio
$r.$
The nth term is
$u_n=u_1r^{n-1}.$
For
$r\ne1,$

the sum is

$S_n=u_1\frac{1-r^n}{1-r}.$
Repeated percentage change produces a geometric model.
Use the model to calculate, then consider whether its assumptions
remain realistic.

Geometric Sequences and Series

✦ Wisdom

🎯 Learning objectives

01 · Recognising a geometric sequence

02 · Growth, decay, and alternating signs

03 · Finding the nth term

04 · Finding the ratio from two terms

05 · The sum of the first n terms

06 · Sigma notation

07 · Repeated percentage change

08 · Decreasing series and a limiting total

🧮 Technology and examination tip

✏️ IB-style Paper 1 practice

🌍 Applications and advanced challenges

✓ Lesson recap

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IB Mathematics: Applications and Interpretation HL 6 Sections80 LessonsLifetime

Curriculum

Geometric sequences and series AI HL

✦ Wisdom

🎯 Learning objectives

01 · Recognising a geometric sequence

02 · Growth, decay, and alternating signs

03 · Finding the nth term

04 · Finding the ratio from two terms

05 · The sum of the first n terms

06 · Sigma notation

07 · Repeated percentage change

08 · Decreasing series and a limiting total

🧮 Technology and examination tip

✏️ IB-style Paper 1 practice

🌍 Applications and advanced challenges

✓ Lesson recap

Related

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