Sum of infinite geometric sequences AI HL

IB Mathematics: Applications and Interpretation · AI HL

Lesson 1.11 · Sums of Infinite Geometric Sequences

Discover when infinitely many terms have a finite sum and apply the result to repeating patterns.

Faith connection · God’s love

An infinite geometric series invites us to think carefully about infinity: infinitely many terms can sometimes add to a finite number. The Bible says, “For I am persuaded, that neither death, nor life … shall be able to separate us from the love of God, which is in Christ Jesus our Lord” (Romans 8:38–39, KJV). Mathematics gives us a precise way to study one kind of infinity. God’s enduring love is known through Jesus. Come to Jesus, trust Him as your Savior, and follow Him as Lord.

Learning objectives

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

identify the first term and common ratio of a geometric series;
decide whether an infinite geometric series has a finite sum;
calculate the sum to infinity and explain its connection to a limit;
solve problems involving unknown terms or ratios;
model bouncing objects, repeating decimals and other repeated processes.

1 · From a geometric sequence to a series

A geometric sequence consists of terms obtained by multiplying by a fixed common ratio $r$ . A geometric series adds those terms. If the first term is $a$ , the series begins:

$a+ar+ar^2+ar^3+\cdots$

For example, the terms $12,\ 6,\ 3,\ 1.5,\ldots$ have first term $a=12$ and common ratio $r=\frac12$ . Their series is $12+6+3+1.5+\cdots$ .

Worked example · Identifying the ratio

Find $a$ and $r$ for the infinite series below.

$40-20+10-5+\cdots$

The first term is $40$ . Divide any term by the preceding term to find the ratio:

$a=40,\qquad r=\frac{-20}{40}=\boxed{-\frac12}.$

The negative ratio causes the signs to alternate.

2 · When does a sum to infinity exist?

Imagine adding more and more terms. A finite sum to infinity exists when these partial sums approach a fixed value. For a geometric series, this happens when the magnitude of the common ratio is less than 1.

$\boxed{|r|<1}$

Equivalently, $-1<r<1$ . The terms become smaller in magnitude, and $r^n$ approaches zero as $n$ increases.

Worked example · Testing for convergence

Decide which series have a finite sum to infinity.

A: $6+3+1.5+\cdots$ has $r=\frac12$ . Since $|r|<1$ , it has a finite sum.

B: $4-2+1-\frac12+\cdots$ has $r=-\frac12$ . Since $|r|<1$ , it also has a finite sum.

C: $3+6+12+\cdots$ has $r=2$ . Its partial sums continue to grow, so it has no finite sum to infinity.

Important: Having infinitely many terms does not automatically mean that their sum is infinite. The size of the terms and the common ratio matter.

3 · Deriving the sum to infinity

The sum of the first $n$ terms of a geometric series is:

$S_n=\frac{a(1-r^n)}{1-r},\qquad r\ne1.$

When $|r|<1$ , the value of $r^n$ approaches zero as $n$ becomes arbitrarily large. The limiting sum is therefore:

$S_{\infty} =\lim_{n\to\infty}\frac{a(1-r^n)}{1-r} =\boxed{\frac{a}{1-r}},\qquad |r|<1.$

Worked example · A positive ratio

Find the sum to infinity of $12+6+3+1.5+\cdots$ .

Here $a=12$ and $r=\frac12$ . The ratio satisfies $|r|<1$ .

$S_{\infty} =\frac{12}{1-\frac12} =\boxed{24}.$

Worked example · An alternating series

Find the sum to infinity of $40-20+10-5+\cdots$ .

Here $a=40$ and $r=-\frac12$ .

$S_{\infty} =\frac{40}{1-(-\frac12)} =\frac{40}{\frac32} =\boxed{\frac{80}{3}}.$

Common mistake: Check $|r|<1$ before using the formula. Substituting $r=2$ into $a/(1-r)$ does not give the sum of a series with ratio 2.

4 · Finding an unknown common ratio

If the first term and sum to infinity are known, substitute them into $S_{\infty}=a/(1-r)$ and solve for $r$ . Then check that the result lies between $-1$ and $1$ .

Worked example · Working backwards

An infinite geometric series has first term 12 and sum 40. Find its common ratio and second term.

$40=\frac{12}{1-r} \quad\Longrightarrow\quad 1-r=\frac{12}{40}=0.3.$

$\boxed{r=0.7}.$

The second term is $ar=12(0.7)=\boxed{8.4}$ . Since $|0.7|<1$ , the sum is valid.

5 · How close is a partial sum to the infinite sum?

The difference between the infinite sum and the sum of the first $n$ terms is called the remainder.

$S_{\infty}-S_n =\frac{a}{1-r}-\frac{a(1-r^n)}{1-r} =\boxed{\frac{ar^n}{1-r}}.$

When $a>0$ and $0<r<1$ , the proportion of the infinite sum contained in the first $n$ terms is:

$\frac{S_n}{S_{\infty}}=1-r^n.$

Worked example · Reaching 95% of the sum

How many terms of the series $10+8+6.4+\cdots$ are needed to obtain at least 95% of its sum to infinity?

The ratio is $r=0.8$ . We require:

$1-0.8^n\ge0.95 \quad\Longrightarrow\quad 0.8^n\le0.05.$

Solving with logarithms, or checking consecutive whole-number values on a calculator, gives:

$n\ge\frac{\log(0.05)}{\log(0.8)} \approx13.43.$

The smallest whole number of terms is therefore $\boxed{14}$ .

IB-style Paper 1 · Skill practice

Show your method and state when a sum to infinity does not exist.

Question 1 · [3 marks]

Find the sum to infinity of the series below.

$18+12+8+\cdots$

Show solution

The first term is $a=18$ and the common ratio is $r=12/18=2/3$ . Since $|r|<1$ :

$S_{\infty} =\frac{18}{1-\frac23} =\boxed{54}.$

Question 2 · [3 marks]

Determine whether the series below has a finite sum to infinity. Give a reason.

$5-7.5+11.25-\cdots$

Show solution

$r=\frac{-7.5}{5}=-1.5.$

Because $|r|=1.5>1$ , the terms do not approach zero. The series has no finite sum to infinity.

Question 3 · [4 marks]

An infinite geometric series has first term 15 and sum to infinity 60. Find its common ratio and third term.

Show solution

$60=\frac{15}{1-r} \quad\Longrightarrow\quad 1-r=\frac14 \quad\Longrightarrow\quad \boxed{r=\frac34}.$

$u_3=ar^2 =15\left(\frac34\right)^2 =\boxed{\frac{135}{16}}.$

Question 4 · [4 marks]

The first two terms of an infinite geometric series are 8 and $-2$ . Find the sum to infinity.

Show solution

$a=8,\qquad r=\frac{-2}{8}=-\frac14.$

As $|r|<1$ :

$S_{\infty} =\frac{8}{1-(-\frac14)} =\frac{8}{\frac54} =\boxed{\frac{32}{5}}.$

Question 5 · [5 marks]

A geometric series has first term 3 and common ratio $\frac12$ . Find the sum of its first five terms and the sum of all the remaining terms.

Show solution

$S_5 =\frac{3(1-(\frac12)^5)}{1-\frac12} =\boxed{\frac{93}{16}}.$

The infinite sum is $3/(1-\frac12)=6$ . Therefore the remaining terms sum to:

$S_{\infty}-S_5 =6-\frac{93}{16} =\boxed{\frac3{16}}.$

Question 6 · [5 marks]

A series has first term $a$ and common ratio $r$ , where $0<r<1$ . Its sum to infinity is 20, and the sum of its first two terms is 15. Find $a$ and $r$ .

Show solution

Since $S_{\infty}=20$ , we have $a=20(1-r)$ . Also, $S_2=a(1+r)=15$ . Substitute:

$20(1-r)(1+r)=15 \quad\Longrightarrow\quad 1-r^2=\frac34.$

$r^2=\frac14.$

The question states $0<r<1$ , so $r=\frac12$ . Hence:

$\boxed{r=\frac12,\qquad a=20\left(1-\frac12\right)=10}.$

Applications · Advanced IB-style questions

Application 1 · A bouncing ball [7 marks]

A ball is dropped from a height of 2.4 m. After each impact, it rises to 64% of its previous height. Assume that this pattern continues indefinitely.

(a) Find the height reached after the third bounce. (b) Find the total vertical distance travelled by the ball.

Show solution

(a) The height after bounce $n$ is $2.4(0.64)^n$ . Thus:

$h_3=2.4(0.64)^3 =0.6291456 \approx\boxed{0.629\text{ m}}.$

(b) The initial fall occurs once. Each bounce height is then travelled twice, on the way up and down:

$D=2.4+2\bigl(2.4(0.64)+2.4(0.64)^2+\cdots\bigr).$

$D=2.4+2\left(\frac{2.4(0.64)}{1-0.64}\right) =2.4+8.533\ldots \approx\boxed{10.9\text{ m}}.$

The result is given to three significant figures.

Application 2 · Repeating decimal [6 marks]

Express the repeating decimal $0.272727\ldots$ as a fraction in its simplest form. Use an infinite geometric series.

Show solution

Write each repeated block as a term:

$0.272727\ldots =\frac{27}{100} +\frac{27}{10000} +\frac{27}{1000000} +\cdots$

This is a geometric series with $a=27/100$ and $r=1/100$ . Therefore:

$S_{\infty} =\frac{\frac{27}{100}}{1-\frac1{100}} =\frac{27}{99} =\boxed{\frac3{11}}.$

Application 3 · Repeated reduction [7 marks]

A machine removes 40% of the material remaining in a container during each cycle. Initially, the container holds 75 kg.

(a) Find the amount removed in the fourth cycle. (b) Find the total amount removed over all cycles. (c) Explain why your answer to part (b) is reasonable.

Show solution

The amounts removed form a geometric series. The first amount is $75(0.4)=30$ kg, and 60% remains after each cycle, so $r=0.6$ .

(a)

$u_4=30(0.6)^3=\boxed{6.48\text{ kg}}.$

(b)

$S_{\infty} =\frac{30}{1-0.6} =\boxed{75\text{ kg}}.$

(c) The total removed cannot exceed the original 75 kg. In this mathematical model, the amount remaining approaches zero, so the total removed approaches 75 kg.

Application 4 · Fractal pattern [8 marks]

A design starts with a region of area 1 m $^2$ . During stage 1, regions with total area $\frac14$ m $^2$ are coloured. In each subsequent stage, the newly coloured area is $\frac34$ of the area newly coloured in the preceding stage. No region is coloured twice.

(a) Find the total area coloured after four stages. (b) Find the limiting coloured area. (c) What area remains uncoloured in the limit?

Show solution

The newly coloured areas form a geometric series with $a=\frac14$ and $r=\frac34$ .

(a)

$S_4 =\frac{\frac14\left(1-(\frac34)^4\right)}{1-\frac34} =1-\frac{81}{256} =\boxed{\frac{175}{256}\text{ m}^2}.$

(b) Since $|r|<1$ :

$S_{\infty} =\frac{\frac14}{1-\frac34} =\boxed{1\text{ m}^2}.$

(c) The uncoloured area approaches $1-1=\boxed{0\text{ m}^2}$ .

Connections and reflection

Limits: The infinite sum is the value approached by the partial sums $S_n$ as $n$ grows.

Fractals: Repeated geometric changes can produce an infinite sequence of smaller areas, as in Application 4.

Markov chains: In later work, differences between a changing distribution and its long-term distribution can sometimes shrink by a constant factor each step.

TOK question: Is it possible to know something about infinity even though we cannot directly experience infinitely many steps? What role do definitions and proofs play in our answer?

Quick recap

An infinite geometric series has a first term $a$ and common ratio $r$ :

$a+ar+ar^2+ar^3+\cdots$

It has a finite sum precisely when $|r|<1$ . In that case:

$\boxed{S_{\infty}=\frac{a}{1-r}}$

Exam habit: Identify $a$ and $r$ , check $|r|<1$ , and decide which distances or amounts must be counted more than once in an application.

Lesson 1.11 · Sums of Infinite Geometric Sequences

Faith connection · God’s love

Learning objectives

1 · From a geometric sequence to a series

2 · When does a sum to infinity exist?

3 · Deriving the sum to infinity

4 · Finding an unknown common ratio

5 · How close is a partial sum to the infinite sum?

IB-style Paper 1 · Skill practice

Applications · Advanced IB-style questions

Connections and reflection

Quick recap

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IB Mathematics: Applications and Interpretation HL 6 Sections80 LessonsLifetime

Curriculum

Sum of infinite geometric sequences AI HL

Faith connection · God’s love

Learning objectives

1 · From a geometric sequence to a series

2 · When does a sum to infinity exist?

3 · Deriving the sum to infinity

4 · Finding an unknown common ratio

5 · How close is a partial sum to the infinite sum?

IB-style Paper 1 · Skill practice

Applications · Advanced IB-style questions

Connections and reflection

Quick recap

Related

Leave A Reply Cancel reply

Company

Links

Support

Recommend

alphy school publishers

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