Maths Quest 11 Chapter 5

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5

5A
5B
5C
5D
5E
5F

Describing sequences
Arithmetic sequences
Arithmetic series
Geometric sequences
Geometric series
Applications of sequences and series

Sequences
and series
areaS oF STudY

• Sequences and series as maps between the
natural numbers and real numbers, and the use
of technology to generate sequences and series
and their graphs
• Sequences generated by recursion: arithmetic
(tn + 1 = tn + d), geometric (tn + 1 = rtn) and fixed

point iteration (for example, t1 = 2, tn + 1 = tn2;
t1 = 5, tn + 1 = 0.8tn(1 – tn))
• Practical applications of sequences and series,
such as financial arithmetic and population
modelling.

eBook plus
Digital doc

5a

describing sequences

10 Quick Questions

Sequences of numbers play an important part in our everyday life. For example, the following
sequence:
2.25, 2.37, 2.58, 2.57, 2.63, . . .
gives the end-of-day trading price (for 5 consecutive days) of a share in an electronics company.
It looks like the price is on the rise, but is it possible to accurately predict the future price per
share of the company?
The following sequence is more predictable:
10 000, 9000, 8100, . . .
This is the estimated
number of radioactive decays
of a medical compound each
minute after administration to a
patient. The compound is used
to diagnose tumours. In the
first minute, 10 000 radioactive
decays are predicted; during the
second minute, 9000, and so on.
Can you predict the next number
in the sequence? You’re correct if
you said 7290. Each successive
term here is 90% of, or
0.90 times, the previous term.

212

maths Quest 11 Standard General mathematics for the TI-nspire

Sequences are strings of numbers. They may be finite in number or infinite. Number
sequences may follow an easily recognisable pattern or they may not. A great deal of recent
mathematical work has gone into deciding whether certain strings follow a pattern (in
which case subsequent terms could be predicted) or whether they are random (in which case
subsequent terms cannot be predicted). This work forms the basis of chaos theory, speech
recognition software for com­puters, weather ­prediction and stock market fore­casting, to name
but a few uses. The list is almost ­endless.
Sequences which follow a pattern can be described a number of ­different ways. They may
be listed in sequential order, they may be described as a func­tional definition, or they may be
described in an i­terative definition.

1  Listing in sequential order
Consider the sequence of numbers t : {5, 7, 9, .  .  .}. The numbers in sequential order are firstly
5 then 7 and 9 with the indication that there are more numbers to follow. The symbol t is the
name of the sequence and the first three terms in the sequence shown are t1 = 5, t2 = 7 and t3 = 9.
The fourth term, t4 if the pattern were to continue, would be the number 11. In general, tn is the
nth term in the sequence. In this example, the next term is simply the previous term with the
number 2 added to it, with the first term being the number 5.
Another possible sequence is t : {5, 10, 20, 40, .  .  .}. In this case it appears that the next term is
twice the previous term. The fifth term here, if the pattern continued, would be t5 = 80. It can be
difficult to determine whether or not a pattern exists in some sequences. Can you find the next
term in the following sequence?
t : {1, 1, 2, 3, 5, 8, .  .  .}
Here the next term is the sum of the previous two terms, hence the next term would be 5 + 8
which is equal to 13, and so on. This sequence is called the Fibonacci sequence and is named
after its discoverer, Leon­ardo Fibonacci, a thirteenth century mathematician.
Here is another sequence; can you find the next term here?
t : {7, 11, 16, 22, 29, .  .  .}
In this sequence the difference between successive terms increases by 1 for each pair. The
first difference is 4, the next difference is 5 and so on. The sixth term is thus 37 which is 8 more
than 29.

2  Functional definition
A functional definition is expressed in the form:
tn = 2n - 7, n ∈ {1, 2, 3, 4, .  .  .}
Using this definition the nth term can be readily calculated. For this example t1 = 2 × 1 - 7 = -5,
t2 = 2 × 2 - 7 = -3, t3 = 2 × 3 - 7 = -1 and so on. We can readily calculate the 100th term,
t100 = 2 × 100 - 7 = 193, simply by substituting the value n = 100 into the expression for tn.
Look at the following example:
dn = 4.9n2, n ∈ {1, 2, 3, .  .  .}
For this example, in which the sequence is given the name d, d1 = 4.9 × 12 = 4.9,
d2 = 4.9 × 22 = 19.6. Listing the sequence would yield d : {4.9, 19.6, 44.1, 78.4, .  .  .}. The
10th term would be 4.9 × 102 = 490.
Here is another example:
cn = cos (np) + 1, n ∈ {1, 2, 3, .  .  .}
Here the sequence would be c: {0, 2, 0, 2, .  .  .}.

3  Iterative definition
An iterative definition is expressed in the form:
tn + 1 = 3tn - 2; t1 = 6

Chapter 5  Sequences and series

213

This definition looks complicated, but is actually straightforward. You may have already come
across this idea on a spreadsheet. The word iteration means the calculation of the next term from
the previous term using the same procedure. The symbol tn + 1 simply means the next term after
the term tn. In the above example the first term, t1, is 6 (this is given in the definition) and so the
next term, t2, is 3 × 6 - 2 = 16, and the following term is 3 × 16 - 2 = 46. In each and all cases
the next term is found by multiplying the previous term by 3 and then subtracting 2. We could
write the sequence out as a table:
n

Comment

tn
t1 = 6

Given in the definition

1

t2 = 3t1 - 2
  =3×6-2
  = 16

Using t1 to find the next term, t2

2

t3 = 3t2 - 2
  = 3 × 16 - 2
  = 46

Using t2 to find the next term, t3

3

t4 = 3t3 - 2
  = 3 × 46 - 2
  = 136

Using t3 to find the next term, t4

An example of this sequence using notation found in a spreadsheet would be:
A1 = 6 (the first term is equal to 6)
A2 = 3 × A1 - 2 (the next term is 3 times the previous term minus 2).
You could then apply the F i l l Do wn option in the E di t menu of the spreadsheet from cell A2
downwards to generate as many terms in the sequence as required. This would result in the next
cell down being three times the previous cell, less 2. The iterative definition finds a natural use
in a spreadsheet environment and consequently much use is made of it. A drawback is that you
cannot find the nth term directly as in the functional definition, but the advantage is that more
complicated systems can be successfully modelled using iterative descriptions and hence are
more interesting and relevant.
Worked exampLe 1

eBook plus
7
,
2

a Find the next three terms in the sequence, b: {14, 7, . . .}.
b Find the 4th, 8th and 12th term in the following sequence:

en = n2 - 3n, n ∈ {1, 2, 3, . . .}.
c Find the 2nd, 3rd and 5th term for the following sequence:
kn + 1 = 2kn + 1, k1 = -0.50.
ThInk

int-0874
Worked example 1

WrITe

a

In this example the sequence is listed and a
simple pattern is evident. From inspection, the
next term is half the previous term and so the
sequence would be 14, 7, 7 , 7 , 7 , 7 .

a

The next three terms are 7 , 7 , 7 .

b

1

b

en = n2 - 3n

4 8 16

2 4 8 16

214

Tutorial

This is an example of a functional
definition. The nth term of the sequence
is found simply by substitution into the
expression en = n2 - 3n.

maths Quest 11 Standard General mathematics for the TI-nspire

c

2

First, define the rule as follows.
On a Calculator page, press:
•  MENU b
•  1: Actions 1
•  1: Define 1
Complete the entry line as:
Define e(n) = n2 – 3 × n
Then press ENTER ·.
Find all required terms. For instance, to find
the 4th term, complete the entry line as:
e(4).
Then press ENTER ·.

3

Write the solution.
e4 = 4; e8 = 40; e12 = 108.
c
This is an example of an iterative
kn + 1 = 2kn + 1,
definition. We can find the 2nd, 3rd and
k1 = -0.50
5th terms for the sequence
kn + 1 = 2kn + 1, k1 = -0.50 by iteration.
Substitute k1 = -0.50 into the formula to find k2.
k2 = 2 × -0.50 + 1 = 0

1

2
3

Continue the process until the value of k5
is found.

4

Write the answer.

5

Alternatively, on a Lists & Spreadsheet
page, label column A: k.
To generate a sequence, press
•  MENU b
•  3: Data 3
•  1: Generate Sequence 1
Complete the entry lines as:
Formula: u(n) = 2 × u(n − 1) + 1
Initial Terms: −0.5
Max No. Terms: 5
Then press ENTER ·.

6

The terms of the sequence are now in
column A and can be read directly from
the screen.

7

Write the answer from the screen.

k3 = 2 × 0 + 1 = 1
k4 = 2 × 1 + 1 = 3
k5 = 2 × 3 + 1 = 7
Thus k2 = 0, k3 = 1 and k5 = 7.

Chapter 5  Sequences and series

215

Logistic equation
The logistic equation is a model of population growth. It gives the rule for determining the
population in any year, based on the population in the previous year. Since we need the previous
term in order to be able to generate the next term of the sequence, the logistic equation is an
example of an iterative definition. It is of the general form:
tn + 1 = atn(1 - tn),
where 0 < t0 < 1 and a is a constant.
Depending on the value of a, sequences generated by use of the logistic equation could be
convergent, divergent, or oscillating. A string of numbers that converges to (settles at) a certain
fixed value is called a convergent sequence. Sequence tn can converge to only one possible
number, x, called the limit of the sequence. This can be written as tn → x. (The symbol → is read
as ‘tends to’, or ‘approaches’.) A sequence whose terms grow further and further apart is called
divergent. That is, a sequence is divergent if tn → ∞, or tn → -∞ as n → ∞. Finally, a sequence
whose terms tend to fluctuate between two (or more) values is called oscillating. An oscillating
sequence is neither convergent nor divergent.
Worked Example 2

Given that a = 2 and t0 = 0.7, use the logistic equation to generate a sequence of 6 terms, and state
whether the sequence is convergent, divergent, or oscillating. If the sequence is convergent, state its
limit.
Think

Write

1

Write the logistic equation, replacing a with its given
value (that is, 2).

tn + 1 = atn(1 – tn)

= 2tn(1 – tn)

2

To find t1, substitute the value of t0 (that is, 0.7) in
place of tn and evaluate.

t1 = 2t0(1 – t0)

= 2 × 0.7 × (1 – 0.7)

= 0.42

3

To find the next term, t2, substitute the value of t1
(that is, 0.42) in place of tn and evaluate.

t2 = 2t1(1 – t1)

= 2 × 0.42 × (1 – 0.42)

= 0.4872

4

Continue the iterative process four more times, each
time substituting the value of the previous term into
the logistic equation to find the next term.

5

The terms of the sequence are growing closer and
closer to each other, finally settling at 0.5.

t3 = 2t2(1 – t2)
= 2 × 0.4872 × (1 – 0.4872)
= 0.499  672  3
t4 = 2t3(1 – t3)
= 2 × 0.499  672  3 × (1 – 0.499  672  3)
= 0.499  999  8
t5 = 2t4(1 – t4)
= 2 × 0.499  999  8 × (1 – 0.499  999  8)
= 0.5
t6 = 2t5(1 – t5)
= 2 × 0.5 × (1 – 0.5)
= 0.5
The sequence is convergent; the limit of
the sequence is 0.5.

Note that instead of saying ‘the limit of the sequence is 0.5’ in the pre­vious example, we could
simply write tn → 0.5.
216

Maths Quest 11 Standard General Mathematics for the TI-Nspire

REMEMBER

1. A sequence is a string of numbers or expressions which may follow a recognisable
pattern.
2. A sequence can be described in a number of ways.
(a) As a list — for example: tn: {1, 7, 2, 6, . . .}
(Note: t3 = 2)
(b) As a function — for example: tn = 2n - n2, n ∈ {1, 2, 3, . . .}
(Note: t5 = 2 × 5 - 52 = -15)
(c) As a recursive or iterative formula — for example: tn + 1 = 2tn - 3, t1 = 6
(Note: t2 = 2 × 6 - 3 = 9)
3. The logistic equation is a model of population growth and is an example of an iterative
definition. It is of the general form:
tn + 1 = atn(1 – tn),
where 0 < t0 < 1 and a is a constant.
4. A sequence that converges to (settles at) a certain fixed number, x (the limit of the
sequence) is called convergent. This can be written as tn → x. A sequence whose terms
grow further and further apart is called divergent. That is, a sequence is divergent if
tn → ∞, or tn → -∞ as n → ∞. A sequence whose terms fluctuate between two (or
more) values is called oscillating. An oscillating sequence is neither convergent nor
divergent.

Exercise

5A

Describing sequences
  1   WE1a  For each of the following sequences, find the next three terms.
a {1, 4, 7, …}
b {1, 0, -1, -2, …}
3 3
c {1, 4, 16, 64, …}
d {3, 2 , 4 , …}
e {2, -5, 8, -11, 14, …}
f {2, 5, 9, 14, 20, …}
g {3, 4, 7, 11, 18, …}
h {2a - 5b, a - 2b, b, -a + 4b, …}
i {1, 0, 1, 0, 1, …}
j {1.0, 1.1, 1.11, …}
k {1024, -512, 256, -128, …}
  2   WE1b  Find the first, fifth and tenth terms in the following sequences.
a tn = 2n - 5, n ∈ {1, 2, 3, …}
b tn = 4 × 3n - 2, n ∈ {1, 2, 3, …}
n
c tn =
, n ∈ {1, 2, 3, …}
d tn = 17 - 3.7n, n ∈ {1, 2, 3, …}
n +1
(3 − n)
n
 1
 1
e tn = 5 ×  2  , n ∈ {1, 2, 3, …}
f tn = 5 ×  2 
, n ∈ {1, 2, 3, …}
g tn = (-1)n + n, n ∈ {1, 2, 3, …}
i tn = n2 - n + 41, n ∈ {1, 2, 3, …}
k tn = ar n - 1, n ∈ {1, 2, 3, …}

h tn = 3n2-n, n ∈ {1, 2, 3, …}
j tn = a + (n - 1)d, n ∈ {1, 2, 3, …}

  3   WE1c  Using a spreadsheet, or other method, find the third, eighth and tenth terms in the
following sequences.
1
a un + 1 = un + 2, u1 = 3
b un + 1 = un - 2, u1 = 2
c un + 1 = 3un, u1 = 0.85
d un + 1 = 2un, u1 = 3
4
3
e un + 1 = 4 un, u1 = 3
f un + 1 = un - 7, u1 = 14
g un + 1 = -un + 2, u1 = 3
h un + 1 = un + (-1)nun, u1 = 3
−1
i un = 2un - 1, u1 = 4
j un + 1 = aun + a, u1 = a
k un + 2 = un + 1 + un, u1 = 1, u2 = 1
l un + 1 = -un2 - 2, u1 = 3

Chapter 5  Sequences and series

217

  4   WE2  Given the following values of a and t0, use the logistic equation to generate a sequence
of 6 terms. State whether the sequence is convergent, divergent, or oscillating. If the sequence
is convergent, state its limit.
a a = 0.8, t0 = 0.5
b a = 0.4, t0 = 0.6
c a = 1.1, t0 = 0.9
d a = 1.9, t0 = 0.4
e a = 2.1, t0 = 0.5
f a = 2.5, t0 = 0.3
g a = 3, t0 = 0.2
h a = 3.4, t0 = 0.7
i a = 4.2, t0 = 0.1
j a = 4.5, t0 = 0.8
  5

Study the pattern in each of the following sequences and where possible write the next
two terms in the sequence, describing the pattern that you use.
a 5, 6, 8, 11, …
b 4, 9, 12, 13, 12, 9, …
c 9, 8, 9, 0, …
d 6, 12, 12, 6, 11 , …
e 5, 8, 13, 21, …
f 1, 3, 7, 15, …
2
g 1, 3, 2, 4, 3, …

  6   MC  a Which of the following functional definitions could be used to describe a sequence
   {3, 1, -1, …}?
A tn = n - 2, n ∈ {1, 2, 3, …}
B tn = 2n - 5, n ∈ {1, 2, 3, …}
C tn = 5n - 2, n ∈ {1, 2, 3, …}
D tn = 5 - 2n, n ∈ {1, 2, 3, …}
E tn = 2(5 - n), n ∈ {1, 2, 3, …}
b Which of the following recursive definitions could be used to describe a sequence
{20, ‑10, 5, …}?
t
t
A  tn + 1 = tn - 30, t1 = 20 B tn + 1 = n , t1 = -20
C tn + 1 = tn - n , t1 = 20
2
2
−t
n
D  tn + 1 = tn - 10, t1 = 20 E tn + 1 = , t1 = 20
2
6 n2 − 12
c Which of the following sequences is generated by the definition tn =
,
2
n ∈ {1, 2, 3, …}?
A {-3, 6, 15, …}
B {-3, 6, -12, …}
C {-3, 6, 21, …}
D { 3, 6, 12, …}
E { 3, 6, 18, …}
  7 The first term in a sequence is −2. The second term is obtained by multiplying the value of
the first term by 3, then adding 1. The third term is obtained by multiplying the value of the
second term by 3, then adding 1. If this pattern continues, the equation which generates this
sequence is:
A tn = 3tn + 1 + 1 where t1 = −2
B tn + 1 = 3tn + 1 where t1 = −2

C 3tn + 1 = tn + 1 where t1 = 2
D 3tn = tn + 1 + 1 where t1 = −2

E tn + 1 + 1 = 3tn where t1 = 2
  8 A sequence follows the rule wn + 1 = 4wn + 2, where wn is the nth term and n = 1, 2, 3, 4 .  .  . .
The value of the second term, w2, is 10.
The value of the fourth term, w4, is:
A 18
B 20
C 42
D 170
E 200
[©VCAA 2004]
Exam tip   In this question, students were given the equation and
the value of the second term of the sequence generated by this
equation, and were asked to generate the fourth term. Only 39% of
students were able to perform this relatively routine task. While this
could have been done with the aid of technology, it was best treated as
a pencil and paper exercise.
[Assessment report 2004]

  9 Write the iterative definition for each of the following sequences.
a {7, 5, 3, 1, -1, …}
b {12, 6, 3, 1.5, …}
c {12, 12.6, 13.2, …}
d {2, 11, 56, 281, …}
e {4, -12, 36, …}
f {2, 4, 16, 256, …}
218

Maths Quest 11 Standard General Mathematics for the TI-Nspire

10 In the township of Grizabella, the population of stray
cats in any given year is given as pn + 1. This can be
calculated using the formula pn + 1 = 1.3pn (1 – pn),
where pn is the number of cats (in hundreds) in the
preceding year. If in 2005 there were 28 stray cats in
Grizabella township, calculate:
a the expected number of stray cats for 2006 and 2007
b the limiting number of stray cats that Grizabella
township can sus­tain.
11 In the neighbouring township of Macavity, the size of the
population of stray cats follows the logistic equation
pn + 1 = 0.3pn(1 – pn),
where pn + 1 and pn refer to the popu­lation size (in
hundreds) in any given year and in the preceding year
respectively. It is known that in 2005, there were 62 stray cats in the township. By generating
and examining the sequence of numbers using the above equation, decide what will happen
in the long run to the population size of stray cats in Macavity township. (That is, will the
population of cats keep increasing, decreasing, or settle at a particular value?)

5B

Arithmetic sequences
At a racetrack a new prototype racing car
unfortunately develops an oil leak. Each
second, a drop of oil hits the road. The
driver of the car puts her foot on the
accelerator and the car increases speed at
a steady rate as it hurtles down the
straight. The diagram below shows the
pattern of oil drops on the road with the
distances between the drops labelled.

10 metres

18 metres

26 metres

42 metres

34 metres

The sequence of distances travelled in metres each second is {10, 18, 26, 34, 42, …}.
The first term in the sequence, t1, is 10 and as you can see, each ­subsequent term is 8 more
than the previous term. This type of sequence is given a special name — an arithmetic
sequence.
An arithmetic sequence is a sequence where there is a common difference between any two
successive terms.
We can list the sequence in a table as in table A. From this table we can see that it is possible
to write a functional definition for the sequence in terms of the first term, 10, and the common
difference, 8, and thus:
tn = 10 + (n - 1) × 8
= 2 + 8n, n ∈ {1, 2, 3, …}

Chapter 5  Sequences and series

219

We can readily get a general formula for the nth term of an arithmetic sequence whose first
term is a and whose common difference is d (see table B).
Table A

Table B

n

tn

n

tn

tn

1

10 + 0 × 8

10

1

a+0×d

2

10 + 1 × 8

18

2

a+1×d

3

10 + 2 × 8

26

3

a+2×d

4

10 + 3 × 8

34

4

a+3×d

n

10 + (n - 1) × 8
= 10 + 8n - 8
= 2 + 8n

n

a + (n - 1) × d
= (a - d) + dn

2 + 8n

In general then:
The nth term of an arithmetic sequence is given by
tn = a + (n - 1) × d = (a - d) + nd, n ∈ {1, 2, 3, .  .  .}
where a is the first term and d is the common difference.
If we consider three successive terms in an arithmetic sequence, namely x, y and z, then since
y - x = the common difference, d, and z - y = d, it follows that:
z+x
y-x=z-y⇒y=
2
The middle term of any three consecutive terms in an arithmetic sequence is called an
arithmetic mean and is the average of the outer two.
z+x
That is, y =
for any 3 consecutive terms, x, y, z of the arithmetic sequence.
2
Worked Example 3

Show that the following sequences are arithmetic.
− 7 − 11
a 4 , 8 , -1, …
b x2 - 4x, 3x2 - 7x, 5x2 - 10x, …
Think
a

1

2

3

b

220

1

Write

To show that a sequence is arithmetic you
need to show that the difference between
any two successive terms is a constant.
Find the difference between the first and
the second terms.
Find the difference between the second
and the third terms.

Compare the differences and draw your
conclusion.
Find the difference between the first two
terms.

a t2 - t1



=



=



=

− 11
8

−7
4 

- 

− 11 + 14

8
3
8

t3 - t2


= -1 - 



=

− 11
8 

3
8

t2 - t1 = t3 - t2 =

3
8

∴ The sequence is arithmetic.
b t2 - t1




Maths Quest 11 Standard General Mathematics for the TI-Nspire

= 3x2 - 7x - (x2 - 4x)
= 2x2 - 3x

2

Find the difference between the second
and the third terms.

t3 - t2

= 5x2 - 10x - (3x2 - 7x)

= 2x2 - 3x

3

Compare the differences and draw your
conclusion.

t2 - t1 = t3 - t2 = 2x2 - 3x
∴ The sequence is arithmetic.

Worked Example 4

State which of the following are arithmetic sequences by finding the difference between successive
terms. For those which are arithmetic, find the next term in the sequence, t4, and consequently find
the functional definition for the nth term for the sequence, tn.
a t : {4, 9, 15, …}
b t : {-2, 1, 4, …}
Think
a

1

b

Write

To check that a sequence is arithmetic, see
if a common difference exists.

2

There is no common difference, as 5 ≠ 6.

1

To check that a sequence is arithmetic, see
if a common difference exists.

a  9-4=5

15 - 9 = 6
Since there is no common difference the
sequence is not arithmetic.

b 1 - -2 = 3

4 - 1 = 3

2

The common difference is 3.

The sequence is arithmetic with the common
difference d = 3.

3

The next term in the sequence, t4, can be
found by adding 3 to the previous term, t3.





t4 = t3 + 3
=4+3
=7

4

To find the functional definition, write the
formula for the nth term of the arithmetic
sequence.





tn = a + (n - 1) × d
= a + nd  - d
= (a - d ) + nd

5

Identify the values of a and d.

a = -2 and d = 3

6

Substitute a = -2 and d = 3 into the
formula and simplify.




tn = (-2 - 3) + n × 3
tn = 3n - 5

Worked Example 5

Find the missing terms in this arithmetic sequence: {41, a, 55, b, …}.
Think

Write

x+z
2

1

The first three successive terms are 41, a, 55. Write
the rule for the middle term of the three successive
terms of an arithmetic sequence.

For x, y, z: y =

2

Identify the variables.

x = 41; y = a; z = 55

3

Substitute the values of x, y and z into the formula in
step 1 and evaluate.

a=

41 + 55
2
= 48

Chapter 5  Sequences and series

221

4

Find the common difference. (The second term is now
known.)

d = t2 - t1
= 48 - 41
=7

5

Find the value of b by adding the common difference
to the preceding taerm.

b = 55 + 7
= 62

6

State your answer.

So a = 48, b = 62

Worked exampLe 6

eBook plus

Find the 16th and nth term in an arithmetic sequence with the
4th term 15 and 8th term 37.
ThInk

int-0875
Worked example 6

WrITe

1

Write the formula for the nth term of the
arithmetic sequence.

tn = a + (n - 1) × d

2

Substitute n = 4 and t4 = 15 into the formula
and label it equation [1].

t4: a + 3d = 15

[1]

3

Substitute n = 8 and t8 = 37 into the formula
and label it equation [2].

t8: a + 7d = 37

[2]

4

Solve the simultaneous equations.
On a Calculator page, press:
•  MENU b
•  3: Algebra 3
•  1: Solve 1
Complete the entry line as:
solve(a + 3d = 15 and a + 7d = 37, a)
Then press ENTER ·.

5

Write the solution.

Solving a + 3d = 15 and a + 7d = 37 for a and d
gives
−3
11
a=
,d =
2
2

6

To find the nth term of the arithmetic
sequence, substitute the values of a and d
into the general formula and simplify.

tn =

To find the 16th term, substitute n = 16 into
the formula, established in the previous step
and evaluate. Alternatively generate the
sequence using a calculation.

If n = 16, t16 =

7

222

Tutorial

  =

−3

11
+ (n - 1) ×  2
2
− 3 + 11n − 11

2
11n − 14
tn =
, n ∈ {1, 2, 3, . . .}
2

maths Quest 11 Standard General mathematics for the TI-nspire

11 × 16 − 14
2
= 81

rememBer

1. An arithmetic sequence is one where successive terms have a common difference. This
common difference is given the symbol d. Thus tn + 1 - tn = d for all values of n. The
first term in the sequence is given the symbol a.
2. If x, y, z are successive terms in an arithmetic sequence then y is called an arithmetic
x+z
mean and is given by y =
.
2
That is, the middle term is the average of the outer two terms.
3. An arithmetic sequence can be written as a, a + d, a + 2d, … and so the nth term tn is:
tn = a + (n - 1)d using the function notation, or
tn + 1 = tn + d, t1 = a using the iterative notation.
exerCISe

5B

arithmetic sequences
1 We3
a

Show that the following sequences are arithmetic.

{-12, -7, -2,

b {-0.12, 3.48, 7.08, …}

…}

−3 3 9

d {2.3, -1.7, -5.7, …}

c { 8 , 8 , 8 , …}
5 −1 − 7
, ,
9 9

…}

f {18, -18, -54, …}

g {5 3 , 715 , 815, …}

13

h {x + 9, 2x + 7, 3x + 5, …}

i {3x2 - 4x, 5x2 - 2x, 7x2, …}

j {3(2 - x), 2(2 - x), 2 - x, …}

e {9,

2

4

2 We4
State which of the following are arithmetic sequences by finding the difference
between successive terms. For those which are arithmetic, find the next term in the sequence,
t4, and consequently find the functional definition for the nth term for the sequence, tn.
a tn: {3, 5, 7, …}
b tn: {4, 7, 11, …}
c tn: {3, 6, 12, …}
d tn: {-3, 0, 3, …}
2 11 9

e tn: {-2, -6, -10, …}

f tn: { 7 , 14 , 7 , …}

3 3 3

3 3 9

g tn: { 4 , 2 , 1, …}
1 − 3 − 13
, 4,
2

i tn: { 4 ,

h tn: { 4 , 2 , 4 , …}
j tn: {2p + 3, 4p + 1, 6p - 1, …}

…}

3 Find the term given in brackets for each of the following arithmetic sequences.
a {4, 9, 14, …}, (t21)
b {-2, 10, 22, …}, (t58)
c { 27, 12, 3, …} (t100)
d {2, -11, -24, …} (t2025)
4

Find the functional definition for the nth term of the following arithmetic sequences:
a where the first term is 5 and the common difference is -3

eBook plus

b where the first term is 2.5 and the common difference is

1
2

-3

c where the first term is and the common difference is 3
d where the first term is 2x and the common difference is 5x.

Digital doc

Spreadsheet 117
Arithmetic sequences

5 Find the nth term in the arithmetic sequence where the first term is 6 and the
third term is 10.
6 Find the nth term in the arithmetic sequence where the first term is 3 and the third term
is 13.

Chapter 5

Sequences and series

223

7 mC The monthly enrolment of students in social dance class follows an arithmetic sequence.
If there were 4 students enrolled in the first month of operation and 16 students in the fourth
month, how many students will be enrolled in the fifth month, assuming that the pattern will
continue?
A 18
B 20
C 22
D 24
E 32
8 We5 Find the missing terms in this arithmetic sequence: {16, m, 27, n}
9 Find the missing terms in the arithmetic sequence below.
x - 3y,
, -3x + 5y,
,…
10 We6 Find the 4th term and nth term in the arithmetic
sequence whose first term is 6 and whose 7th term is -10.

eBook plus
Digital doc

SkillSHEET 5.1
Using elimination to solve
simultaneous equations

11 If t10 = 100 and t15 = 175, find the first term, the common difference and hence the nth term for
the arithmetic sequence.
−1

3

12 If t10 =  2 and t13 = 4 , find the first term, the common difference and hence the nth term for the
arithmetic sequence.
13 Insert four evenly spaced numbers between 8 and 36.
14 For the arithmetic sequence {22, m, n, 37, …}, find the values for m and n.
15 For the following arithmetic sequences, find the iterative definition and use it in a spreadsheet
to generate the first 50 numbers in the sequence using the Fill Down option.
a tn: {3, 7, 11, …}
b tn: {-3, 0, 3, …}
c tn: {-2, -6, -10…}
3 3 9

e tn: { 4 , 2 , 4 , …}

2

11 9

d tn: { 7 , 14 , 7 , …}
1 − 3 − 13
, 4,
2

f tn: { 4 ,

…}

g tn: {2p + 3, 4p + 1, 6p - 1, …}
16 The first three terms in an arithmetic sequence are 37, 32, 27 and another term is -3. Which
term is -3?
17 Find the value of x such that the following forms an arithmetic progression:
… x, 3x + 4, 10x - 7 …
18 mC For the following sequence t : {4, 11, 18, …}, the difference between the 4th and the
10th term is:
A 35
B 42
C 49
D 56
E 63
19 mC The tenth term in an arithmetic sequence is 12 and the third term is -2. The first term in
the sequence is:
A -7
B -3
C -5
D -8
E -6
20 The ratio (division) between the first term and the second term in an arithmetic sequence is 3 .
4
The ratio between the second term and the third term is 45 .
a Calculate the ratio of the third term to the fourth term.
b Find the ratio of the nth and the (n + 1)th term in the sequence.

5C

arithmetic series
Often we have a sequence of numbers and we wish to know their sum. For an example, we
return to the oil drops on the racetrack from the start of the previous section on arithmetic
sequences. The distance covered by the car each second illustrated the concept of an arithmetic
sequence.
The total distance covered by the car is the sum of the individual distances covered
each second. So after one second the car has travelled 10 m, after two seconds the car has

224

maths Quest 11 Standard General mathematics for the TI-nspire

travelled 10 + 18 m = 28 m, after three seconds the car has travelled a total distance of
10 + 18 + 26 m = 54 m, and so on.
10 metres

18 metres

28 m

26 metres

54 m

34 metres

42 metres

88 m

130 m

A series, Sn, is the sum of a sequence of n terms t1 + t2 + t3 + … + tn.
Thus:


S1 = t1
S

2 = t1 + t2
S

3 = t1 + t2 + t3
Sn = t1 + t2 + t3 + … + tn - 2 + tn - 1 + tn.
For an arithmetic sequence, the sum of the first n terms, Sn, can be written in two ways:
1. The first term in the arithmetic sequence is a, the common differ­ence is d, and the last term —
that is, the nth term — in the sequence is l.
Sn = a + (a + d ) + (a + 2d ) + … (a + (n - 3)d ) + (a + (n - 2)d + a + (n - 1)d

= a + (a + d ) + (a + 2d ) + … (l - 3d ) + (l - 2d ) + (l - d ) + l
[1]
2. We can write the sum Sn in reverse order starting with the nth term and summing back to the
first term a:

Sn = l + (l - d ) + (l - 2d ) + … + (a + 2d ) + (a + d ) + a.
[2]
If we add equation [1] and equation [2] together and recognise that there are n terms each of
which equal (a + l) we get:
2Sn = (a + l) + (a + l) + … n times

= n(a + l)
and so:
n

Sn = (a + l )
2
n
or since l is the nth term, l = a + (n - 1)d, so Sn = [a + a + (n - 1)d ]
2
n

Sn = [2a + (n - 1)d ]
2
The sum of the first n terms in the arithmetic sequence is given by
n
Sn = (a + l)
2
where a is the first term and l is the last term; or alternatively, since l = a + (n - 1)d, by
n
Sn = (2a + (n - 1)d )
2
where a is the first term and d is the common difference.
If we know the first term, a, the common difference, d, and the number of terms, n, that
we wish to add together we can calculate the sum directly without having to add up all the
individual terms.
It is worthwhile also to note that Sn + 1 = Sn + tn + 1. This tells us that the next term in the
series Sn + 1 is the present sum Sn plus the next term in the sequence tn + 1. This result is useful
in spreadsheets where one column gives the sequence and an adjacent column is used to give
the series.

Chapter 5  Sequences and series

225

Worked exampLe 7

eBook plus

a Find the sum of the first 20 terms in the sequence tn: {12, 25, 38, …}.
b For the sequence tn = 2n + 7, n ∈ {1, 2, 3, …}, find i S5 and i i Sn.
ThInk
a

b

WrITe

1

Write the formula for the sum of
the first n terms in the arithmetic
sequence.

2

Identify the variables.

3

Substitute values of a, d and n into
the formula and evaluate.

1

First, define the sequence as follows.
On a Calculator page, press:
•  MENU b
•  1: Actions 1
•  1: Defi ne 1
Complete the entry line as:
Defi ne t(n) = 2 × n + 7
Then press ENTER ·.

2

i To find S5, choose the Sum template

from the Maths Expression template 
and complete the entry line as:

a Sn =

Tutorial

int-0876
Worked example 7

n
(2a + (n - 1)d )
2

a = 12, d = 25 - 12 = 13, n = 20
20
S20 =
(2 × 12 + 19 × 13)
2
S20 = 2710
b

i The sum of the first 5 terms of the sequence,

S5 = 65.

5

∑ (t (n))
n=1

Then press ENTER ·.
ii Repeat for Sn, by completing the entry
line as:
n

i i The sum of n terms of the sequence,

Sn = n2 + 8n.

∑ (t ( x))
x =1

Then press ENTER ·.
Note: To find the sum from 1 to n, use
t(x) instead of t(n).

rememBer

n
1. The sum of the first n terms of an arithmetic sequence is Sn =  (2a + (n - 1)d)
dd) and so
2
to find Sn the values for a and d need to be found for the sequence whose series is
required.
2. In general, tn + 1 = Sn + 1 - Sn.

226

maths Quest 11 Standard General mathematics for the TI-nspire

exerCISe

5C

arithmetic series
1 We7 Consider the following sequences and find the sums of the terms as indicated.
a tn: {1, 2, 3, …}. Find S10, S50, S100.
eBook plus
b tn: {1, 3, 5, …}. Find S5, S10, S20.
c tn = 3n + 7, n ∈ {1, 2, 3, …}. Find S5, S10, Sn.
Digital doc
d tn = -4n + 15, n ∈ {1, 2, 3, …}. Find S5, S10, Sn.
Spreadsheet 117
e tn + 1 = tn + 5.5, t1 = 2.5. Find S5, S10, S20.
Arithmetic series
f tn + 1 = tn + p, t1 = 2p. Find S5, S10, S20.
g The first term is 4 and the common difference is 3. Find S4, S16, S64.
1
h The first term is 14 and the common difference is -3 2 . Find S4, S9, S14.
i The first term is 50 and the 10th term is -40. Find S10.
j The 5th term is 10 and the 8th term is 16. Find S5, S50, S500.

2 a Find the sum of the first 50 positive integers.
b Find the sum of the first 100 positive integers.
3 a Find the sum of all the half-integers between 0 and 100.
1
1
1
1
Note: The sequence of half-integers is { 2 , 1 2 , 2 2 , 3 2 , …}
b Compare your answer with that for question 2b.
4 Find the sum of the first 12 terms of an arithmetic sequence in which the second term is 8 and
thirteenth term is 41.
5 A sequence of numbers is defined by tn: {15, 9, 3, -3, …}.
a Find the sum of the first 13, 16 and 19 terms in the sequence.
b Find the sum of all the terms between and including t10 and t15.
6 A sequence of numbers is defined by tn = 2n - 7, n ∈ {1, 2, 3, …}. Find
a the sum of the first 20 terms
b the sum of all the terms between and including t21 and t40
c the average of the first 40 terms.
Hint: You need to find the sum first.
7 mC Nathan decided to build up his muscles for summer by doing push-ups every day. He
started with 10 push-ups a day. Every second day Nathan increased the number of pushups by 2. (That is, he did 10 push-ups on day 1 and day 2; 12 push-ups on day 3 and day 4;
14 push-ups on day 5 and day 6 and so on. The total number of push-ups that Nathan did after
6 weeks is:
A 682
B 720
C 840
D 1260
E 2142
8 mC Rachel has a box of 100 lego blocks, which she intends to use for building towers.
She uses 4 blocks to build the first tower, 5 blocks to build the second tower, 6 blocks to
build the third tower etc. What is the greatest number of towers that Rachel can build from
the available lego blocks?
A 10
B 11
C 12
D 13
E 14
9 Find the equation that gives the sum of the first n positive integers.
10 a Show that the sum of the first n odd integers is equal to the perfect square n2.
b Show that the sum of the first n even integers is equal to n2 + n.
11 A sequence is 5, 7, 9, 11, … How many consecutive terms need to be added to obtain 357?
12 Consider the sum of the first n integers. For what value of n will the sum first exceed 1000?
13 a Find the sum of all integers divisible by 3 which lie between 200 and 400.
b Find the sum of all integers divisible by 6 which lie between 200 and 400.

Chapter 5

Sequences and series

227

14 The first term in an arithmetic sequence is 5 and the sum of the first 20 terms is 1240. Find the
common difference, d.
15 The sum of the first four terms of an arithmetic sequence is 58, and the sum of the next four
terms is twice that number. Find the sum of the following four terms.
16 The sum of a series is given by Sn = 4n2 + 3n. Use the result that
tn + 1 = Sn + 1 - Sn to prove that the sequence of numbers, tn, whose series is
Sn = 4n2 + 3n is arithmetic. Find both the functional and iterative
equations for the sequence, tn.

5d

eBook plus
Digital doc

WorkSHEET 5.1

Geometric sequences
A farmer is breeding worms which he
hopes to sell to local shire councils for use
in the decomposition of waste at rubbish
dumps. Worms reproduce readily and the
farmer expects a 10% increase per week in
the mass of worms that he is farming. A
10% increase per week would mean that
the mass of worms would increase by a
10

constant factor of (1 + 100 ) or 1.1.
He starts off with 10 kg of worms.
By the beginning of the second week he
will expect 10 × 1.1 = 11 kg of worms,
by the start of the third week he would
expect 11 × 1.1 = 10 × (1.1)2 = 12.1 kg of
worms, and so on. This is an example of a
geometric sequence.
A geometric sequence is the sequence
where each term is obtained by
multiplying the preceding term by a
certain constant factor.
The first term is 10 and the common factor here is 1.10, which represents a 10% increase on
the previous term. We can put the results of the above example into a table:
n

tn

tn

1

10 × (1.1)0

10

2

10 × (1.1)1

11

3

10 × (1.1)2

12.1

4

10 × (1.1)3

13.31

n

10 × (1.1)n - 1

10 × (1.1)n - 1

From this table we can see that
t2 = 1.1 × t1, t3 = 1.1 × t2
and so on. In general:
tn + 1 = 1.1 × tn
The common factor or common ratio
whose value is 1.1 for this example can
be found by dividing any two
tn +1
successive terms:
.
tn

A geometric sequence, t, can be written in terms of the first term, a, and the common ratio, r.
Thus:
t : {a, ar, ar2, ar 3, … , ar n - 1, …}
The first term t1 = a, the second term t2 = ar, the third term t3 = ar2 and consequently the nth
term, tn is ar n - 1.
228

maths Quest 11 Standard General mathematics for the TI-nspire

For a geometric sequence:

tn = ar n - 1
where a is the first term and r the common ratio, given by
t
r = n+1
tn
If we consider three consecutive terms in a geometric sequence, x, y and z, then
y
z
=r=
x
y
where r is the common factor.
Thus the middle term, y, called the geometric mean, can be calculated in terms of the outer
two terms, x and z.
For a geometric sequence …, x, y, z, … :
y2 = xz
Worked Example 8

State whether the sequence is geometric by finding the ratio of successive terms:
t: {2, 6, 18, …}. If it is geometric, find the next term in the sequence, t4, and the nth term for the
sequence, tn.
Think

Write

1

Find the ratio

t2
.
t1

t2 6
=
t1 2
=3

2

Find the ratio

t3
.
t2

t3 18
=
6
t2
=3

3

Compare the ratios and make your conclusion.

Since

t 2 t3
= = 3, the sequence is geometric
t1 t2

with the common ratio r = 3.
4

Since the sequence is geometric, to find the fourth
term, multiply the preceding (third) term by the
common ratio.

t4 = t3 × r
= 18 × 3
= 54

5

Write the general formula for the nth term of the
geometric sequence.

tn = ar n - 1

6

Identify the values of a and r.

a = 2; r = 3

7

Substitute the values of a and r into the general formula. tn = 2 × 3n - 1

Worked Example 9

Find the nth term and the 10th term in the geometric sequence, where the first term is 3 and the
third term is 12.
Think

Write

1

Write the general formula for the nth term in the
geometric sequence.

tn = ar n - 1

2

State the value of a (the first term in the sequence)
and the value of the third term.

a = 3; t3 = 12

Chapter 5  Sequences and series

229

3

Substitute all known values into the general formula.

12 = 3 × r 3 - 1
= 3 × r2

4

Solve for r (note that there are 2 possible solutions).

r2 =

12
3

=4
r=± 4
= ±2
5

Substitute the values of a and r into the general equation. So tn = 3 × 2n - 1, or tn = 3 × (-2)n - 1
Since there are 2 possible values for r, you must show
both expressions for the nth term of the sequence.

6

Find the 10th term by substituting n = 10 into each of
the two expressions for the nth term.

When n = 10, t10 = 3 × 210 - 1 (using r = 2)
= 3 × 29
= 1536
or t10 = 3 × (-2)10 - 1 (using r = -2)
= 3 × (-2)9
= -1536

Worked exampLe 10

eBook plus

The fifth term in a geometric sequence is 14 and the seventh term is 0.56.
Find the common ratio, r, the first term, a, and the nth term for the sequence.
ThInk

WrITe

Tutorial

int-0877
Worked example 10

Method 1: Technology-free
1

Write the general rule for the nth term of the
geometric sequence.

tn = ar n - 1

2

Use the information about the 5th term to
form an equation. Label it [1].

When n = 5, tn = 14
14 = a × r 5 - 1
14 = a × r4

[1]

When n = 7, tn = 0.56
0.56 = a × r7 - 1
0.56 = a × r 6

[2]

3

230

Similarly, use information about the 7th term
to form an equation. Label it [2].

4

Solve equations simultaneously: Divide 
equation [2] by equation [1] to eliminate a.

[2]
ar 6 0.56
gives
=
[1]
14
ar 4

5

Solve for r.

r2 = 0.04

6

Since there are two solutions, we have to
perform two sets of computations. Consider
the positive value of r first. Substitute the
value of r into either of the two equations,
say equation [1], and solve for a.

r = ± 0.04
= ± 0.2
If r = 0.2
Substitute r into [1]:
a × (0.2)4 = 14
0.0016a = 14
a = 14 ÷ 0.0016
= 8750

7

Substitute the values of r and a into the general
equation to find the expression for the nth term.

The nth term is:
tn = 8750 × (0.2)n - 1

8

Now consider the negative value of r.

If r = -0.2

maths Quest 11 Standard General mathematics for the TI-nspire

9

Substitute the value of r into either of the two
equations, say equation [1], and solve for a.
(Note that the value of a is the same for both
values of r.)

10

Substitute the values of r and a into the
general formula to find the second expression
for the nth term of the sequence.

Substitute r into [1]
a × (− 0.2)4 = 14
0.0016a = 14
a = 14 ÷ 0.00016
= 8750
The nth term is:
tn = 8750 × (-0.2)n - 1

or
tn = 8750 × (0.2)n - 1

Method 2: Technology-enabled
1

Repeat steps 1–3 from Method 1 to obtain
equations [1] and [2].

14 = a × r 4
0.56 =

[1]

a × r6

[2]

2

Solve equations [1] and [2] simultaneously.
On a Calculator page, press:
• MENU b
• 3: Algebra 3
• 1: Solve 1
Complete the entry line as:
solve(14 = a × r 4 and 0.56 = a × r6, a)
Then press ENTER ·.

3

Write the solution.

Solving 14 = a × r and 0.56 = a × r6 for a and r
gives
a = 8750 and r = -0.2, or a = 8750 and r = 0.2

4

Substitute the values of r and a into the
general formula to find the expression of the
nth term of the sequence.
Note: Since there are two possible values of r,
there will be two different expressions for tn.

The nth term is:
tn = 8750 × (−0.2)n - 1,
or
tn = 8750 × (0.2)n - 1

REMEMBER

1. A geometric sequence is one where each successive term is obtained by multiplying
the preceding term by the constant number. This number is called the common ratio
t
and is given the symbol r. Thus n + 1 = r for all values of n. The first term in the
t
sequence is given the symbol a. n
2. If x, y, z are successive terms in the geometric sequence then y is called a geometric
mean and is given by y2 = xz.
3. A geometric sequence can be written as a, ar, ar 2, … and so the nth term tn is:
tn = ar n - 1 using the function notation, or
tn + 1 = rtn, t1 = a using the iterative notation.

Chapter 5  Sequences and series

231

exerCISe

5d

Geometric sequences

eBook plus

Digital doc
1 We8 State which of the following are geometric sequences by finding
Spreadsheet 117
the ratio of successive terms. For those which are geometric, find the Geometric sequences
next term in the sequence, t4 and the nth term for the sequence, tn.
a tn: {3, 6, 9, …}
b tn: {4, 12, 36, …}
c tn: {3, 6, 12, …}

d tn: {4, 6, 9, …}
e tn: {-3, 1, 13, …}
f tn: {2, -6, 18, …}

g tn: { 2 ,
7

j tn: { 1 ,

6
, 9,
14 14
-3

4

2

3 3 3

3 3 9

h tn: { 4 , 2 , 1 , …}

…}

i tn: { 4 , 2 , 4 , …}

k tn: {2p, 4p 2, 8p 3, …}

, 9, …}

2 For each of the following:
i show that the sequence is geometric
ii find the nth term and consequently the 6th and the 10th terms.
a t : {5, 10, 20, …}
b t : {2, 5, 12.5, …}
c t : {1, -3, 9, …}
1
d t : {2, 4, 8, …}
e t : {2.3, 3.45, 5.175, …}
f t : { 2 , 1, 2, …}
−1 1 1
3 -1 1
g t : { , , , …}
h t : { 5 , , , …}
i t : {x, 3x4, 9x7, …}
3 12

j t : {1,
x

48

2, 4,
x2 x3

5 15

…}

3 We9 Find the nth term and the 10th term in the geometric sequence where:
a the first term is 2 and the third term is 18 (Why are there two possible answers?)
b the first term is 1 and the third term is 4 (Why are there two possible answers?)
c the first term is 5 and the fourth term is 40
d the first term is -1 and the second term is 2
1
e the first term is 9 and the third term is 81
. (Why are there two possible answers?)

3
4 Find the 4th term in the geometric sequence where the first term is 6 and the 7th term is 32
.

5 mC A geometric sequence is such that a > 0 and r < 0.
Which of the graphs below could represent the first seven terms of this sequence?
A tn
B tn
C tn
7
6
5
4
3
2
1

7
6
5
4
3
2
1

0

D

232

tn
5
4
3
2
1
0
1
2
3
4
5

1 2 3 4 5 6 7

n

0

E

1 2 3 4 5 6 7

n

tn
6
5
4
3
2
1
0
1
2
3
4
5

1 2 3 4 5 6 7

n

1 2 3 4 5 6 7

maths Quest 11 Standard General mathematics for the TI-nspire

5
4
3
2
1
0
1
2
3
4
5

n

1 2 3 4 5 6 7

n

  6 A crystal measured 12.0 cm in length at the beginning of a chemistry experiment. Each day it
increased in length by 3%. The length of the crystal after 14 days growth is closest to:
A 12.4 cm
B 16.7 cm
Exam tip   This question required students to determine the
C 17.0 cm
length of a crystal after 14 days growth. Only 25% of students
D 17.6 cm
gave the correct response. Thirty-nine per cent of students chose
E 18.2 cm
wrong option that corresponded to the length of the crystal at the
start of the 14th day, which only allowed for 13 days’ growth.
[Assessment report 2006]

[©VCAA 2006]

  7 Find the nth term in the geometric sequence where the first term is 3 and the fourth term is
6 2.
  8 For the geometric sequence 3, m, n, 192, … , find the values for m and n.
  9 Consider the geometric sequence t : {16, m, 81, n, …}. Find the values of m and n, if it is
known that both are positive numbers.
10 For the geometric sequence a, 15, b, 0.0375, … , find the values of a and b, given that they are
positive numbers.
11   WE10  The third term in a geometric sequence is 100 and the fifth term is 400. Find the
common ratio, r, the first term, a, and the nth term for the sequence.
27
12 If t2 = 12 and t5 = 16
, find the first term, a, the common factor, r, and hence the nth term for the
geometric sequence.

13 Find the value of x such that the following sequence forms a geometric progression:
x - 1, 3x + 4, 6x + 8.
14 Insert three terms in between 8, _, _, _,

1
32

such that the sequence of numbers is geometric­.

15 The difference between the first term and the second term in a geometric sequence is 6. The
difference between the second term and the third term is 3.
a Calculate the difference between the third term and the fourth term.
b Find the nth term in the sequence.
16 The first two terms in a geometric sequence are 120, 24, and the kth term is 0.0384. Find the
value for k.

5E

Geometric series
When we add up or sum the terms in a sequence we get the series for that sequence. If we look
at the geometric sequence {2, 6, 18, 54, …} where the first term t1 = a = 2 and the common ratio
is 3 we can quickly calculate the first few terms in the series of this sequence.

S1 = t1 = 2

S2 = t1 + t2 = 2 + 6 = 8

S3 = t1 + t2 + t3 = 2 + 6 + 18 = 26

S4 = t1 + t2 + t3 + t4 = 2 + 6 + 18 + 54 = 80
In general the sum of the first n terms is:
Sn = t1 + t2 + t3 + … + tn - 2 + tn - 1 + tn.
For a geometric sequence the first term is a, the second term is ar, the third term is ar 2 and so
on up to the nth term which is ar n − 1. Thus:

Sn = a + ar + ar 2 + … + ar n - 3 + ar n − 2 + ar n − 1
[1]
If we multiply equation [1] by r we get:

rSn = ar + ar 2 + ar 3 + … ar n - 2 + ar n - 1 + ar n
[2]
Chapter 5  Sequences and series

233

Note that on the right-hand side of equations [1] and [2] all but two terms are common,
namely the first term in equation [1], a, and the last term in equation [2], ar n. If we take the
difference between equa­tion [2] and equation [1] we get:

rSn - Sn = ar n - a
[2] - [1]
∴ (r - 1)Sn = a(r n - 1)
a(r n − 1)

; r ≠ 1
(r cannot equal 1)
Sn =
r −1
We now have an equation which allows us to calculate the sum of the first n terms of a
geometric sequence.
The sum of the first n terms of a geometric sequence is given by:
a( r n − 1)
;r≠1
Sn =
r−1
where a is the first term of the sequence and r is the common ratio.
Worked Example 11

Find the sum of the first 5 terms (S5) of these geometric sequences.
−1
1
a tn: {1, 4, 16, .  .  .}
b tn = 2 (2)n - 1, n ∈ {1, 2, 3, .  .  .}
c tn + 1 = 4 tn, t1 = 2
Think
a

1

2
3

4

b

1
2

3

Write

a(r n − 1)
Write the general formula for the sum
a Sn =
r −1
of the first n terms of the geometric
sequence.
Write the sequence.
tn: {1, 4, 16. ...}
Identify the variables: a is the first term; r
a = 1; r = 4 = 4; n = 5
1
can be established by finding the ratio; n
is known from the question.
1(4 5 − 1)
Substitute the values of a, r and n into the
S5 =
4 −1
formula and evaluate.
1024 − 1
=
3
= 341
b tn = 2 (2)n −1, n ∈ {1, 2, 3, ...}
Write the sequence.
Define the sequence as follows.
On a Calculator page, press:
• MENU b
• 1:Actions 1
• 1:Define 1
Complete the entry line as:
Define t(n) = 2 × 2n −1
Then Press ENTER ·
To find S5, choose the Sum template
from the Expression template and
complete the entry line as:
5

∑ (t (n))
n=1

Press ENTER · after each entry.
Write the solution.

234

The sum of the first 5 terms of the sequence is
S5 = 62.

Maths Quest 11 Standard General Mathematics for the TI-Nspire

c

1
2

3

1

c tn + 1 = tn, t1 =
4

Write the sequence.

1
;
4

This is an iterative formula, so the
coefficient of tn is our r; a = t1; n is
known from the question.

r=

Substitute values of a, r and n into
the general formula for the sum and
evaluate.

S5 =
=
=

a=

−1
2

−1
2

;n=5

−1 1 5
[( ) − 1]
2 4
1
−1
4
−1
1
× ( 1024
− 1)
2
−3
4
− 341
512

The infinite sum of a geometric sequence where r < 1
When the constant ratio, r, is less than 1 but greater than -1, that is, {r: -1 < r < 1}, each
successive term in the sequence gets closer to zero. This can readily be shown with the
following two examples.
g: {2, -1, 12 ,
1

−1
4

, …} where a = 2 and r =

1

−1
2
1

h: {40, 2 , 160 , …} where a = 40 and r = 80
In both the examples, successive terms approach zero as n increases. In the second case
the approach is more rapid than in the first and the first sequence alternates positive and
negative. A simple investigation with a spreadsheet will quickly reveal that for geometric
sequences with the size or magnitude of r < 1 the series eventually settles down to a near
constant value. We say that the series converges to a value S∞ which is the sum to infinity of
all terms in the geometric sequence. We can find the value S∞ by recognising that as n → ∞
the term r n → 0, provided r is between -1 and 1. We write this technically as -1 < r < 1 or
|r| < 1. The symbol |r| means the magnitude or size of r. Using our equation for the sum of
the first n terms:
Sn =

a(r n − 1)
;r≠1
r −1

Taking -1 as a common factor from the numerator and denominator:
Sn =

a(1 − r n )
1− r

As n → ∞, r n → 0 and hence 1 - r n → 1. Thus the top line or numer­ator will equal a when
n → ∞:
a
S∞ =
; |r | <1
1− r
We now have an equation which allows us to calculate the sum to infinity, S∞ of a ­geometric
sequence.
The sum to infinity S∞ of the geometric sequence is given by:
S∞ =

a
; |r| <1
1− r

where a is the first term of the sequence and r is the common ratio whose magnitude is less
than one.

Chapter 5  Sequences and series

235

Worked exampLe 12
a Find the sum to infinity for the sequence tn: {10, 1, 0.1, . . .}.
b Find the fourth term in the geometric sequence whose first term is 6 and whose sum to infinity

is 10.

ThInk
a

b

WrITe
a

tn = ar n - 1

1

Write the formula for the nth term of the
geometric sequence.

2

From the question we know that the first
term, a, is 10 and r = 0.1.

3

Write the formula for the sum to infinity.

S∞ =

a
; |r | <1
1− r

4

Substitute a = 10 and r = 0.1 into the
formula and evaluate.

S∞ =

10
1 − 0.1

S∞ =

10
0.9

a = 10, r = 0.1

= 100 = 11 1
9

a
; |r | <1
b S∞ =
1− r
a = 6; S∞ = 10

1

Write the formula for the sum to infinity.

2

From the question it is known that the
infinite sum is equal to 10 and that
the first term a is 6. Write down this
information.

3

Substitute known values into the formula.

10 =

4

Solve for r.

10(1 - r) = 6
10 - 10r = 6
10r = 4
r = 0.4

5

Write the general formula for the nth term
of the geometric sequence.

tn = ar n - 1

6

To find the 4th term substitute a = 6, n = 4
and r = 0.4 into the formula and evaluate.

t4 = 6 × (0.4)3
= 0.384

9

6
1− r

rememBer

1. The sum of the first n terms in a geometric sequence is:
a(1 − r n )
with r ≠ 1
Sn =
1− r
a(r n − 1)
with r ≠ 1
r −1
2. When the magnitude of r is less than one, that is, -1 < r < 1, the sum of a geometric
sequence to infinity, S∞ is given by:
a
S∞ =
1− r
or

236

Sn =

maths Quest 11 Standard General mathematics for the TI-nspire

exerCISe

5e

Geometric series
1 We 11 Consider the following sequences and find the terms indicated.
a tn: {1, 2, 4, …}. Find S5, S10, S20.
b tn: {1, 3, 9, …}. Find S5, S10, S20.
c tn = 3(-2)n - 1, n ∈ {1, 2, 3, …}. Find S5, S10, S20.
d tn = -4(1.2)n - 1, n ∈ {1, 2, 3, …}. Find S1, S10, S20.
e tn + 1 = 2tn, t1 = 3 . Find S1, S5, S10.
f tn + 1 =

−1

2

2

−2

tn, t1 =  3 . Find S1, S5, S10.

g The first term is 3000 and the common ratio is 1.05. Find S4, S16, S64.
h The first term is 1400 and the common ratio is -1.1. Find S4, S9, S14.
i The first term is 20; every other term is obtained by multiplying the preceding term by 5.
Find S5, S10.
j The first term is -2; every other term is obtained by multiplying
−1
eBook plus
the preceding term by 2 . Find S5, S10.
2 Consider the following geometric sequences and find the terms indicated.
a The first term is 440 and the 12th term is 880. Find S6.
b The 5th term is 1 and the 8th term is 8. Find S1, S10, S20.

Digital doc

Spreadsheet 117
Geometric series

3 Find the sum of the first 12 terms of a geometric sequence in which
the second term is 83 and the fifth term is 9.

4 What minimum number of terms of the series 2 + 3 + 4 12 + … must be taken to give a sum in
excess of 100?
5 The sum of the first four terms of a geometric sequence is 312, and the sum of the next four
terms is 625 times that number. Find the sum of the following four terms.
6 Find the sum of all powers of 2 between 500 and 50 000.
7 Find the sum of all powers of 4 between 500 and 50 000.
8 mC In preparation for an upcoming school production Lena spent 2 hours reading and
memorising the script on the first day after receiving it. The next day she spent another
1 hour and 48 minutes on the same task and the day after that she spent 1 hour and
37.2 minutes. If the time that Lena spent daily working on memorising the script followed
the same pattern for two weeks, the total time that she thus spent was closest to:
A 14 hours 15 minutes
B 14 hours 25 minutes
C 15 hours 25 minutes
D 16 hours
E 16 hours 15 minutes
9 We 12a
a tn: {1,
c tn: {1,
e tn: {1,

Find the sum to infinity for the following geometric sequences.
1 1
, , …}
2 4
1 1
, , …}
3 9
−2 4 −8
, , , …}
3 9 9

b tn: {1,
d tn: {1,

−1 1 −1
, , ,
2 4 8
2 4
, , …}
3 9

…}

1 1

10 For the infinite geometric sequence { 12 , 4 , 8 , …}, find the sum to infinity. Consequently, find
what proportion each of the first three terms contributes to this sum as a percentage.
1

1

11 For the infinite geometric sequence {1, 4 , 16, …}, find the sum to infinity. Consequently, find
what proportion each of the first three terms contributes to this sum as a percentage.
3

12 For the infinite geometric sequence {1, 4 , 169 , …}, find the sum to infinity. Consequently, find
what proportion each of the first three terms contributes to this sum as a percentage.

Chapter 5

Sequences and series

237



n −1

13 A sequence of numbers is defined by tn = 3 12  , n ∈ {1, 2, 3, …}.
a Find the sum of the first 20 terms.
b Find the sum of all the terms between and including t21 and t40.
c Find the sum to infinity, S∞.
14 A sequence of numbers is defined by tn: {9, -3, 1, …}.
a Find the sum of the first 9 terms.
b Find the sum of all the terms between and including t10 and t15.
c Find the sum to infinity, S∞.
15 The first term of the geometric sequence is 5 and the fourth term is 0.078 125. Find the sum to
infinity.
16 The sum of the first four terms of a geometric sequence is 30 and the sum to infinity is 32. Find
the first three terms of the sequence.
17 For the geometric sequence 5 + 3 , 5 − 3 , …, find the common factor,
r, and the sum of the infinite series, S∞.

eBook plus

18 If 1 + 3x + 9x2 + … = 23 , find the value of x.

WorkSHEET 5.2

Digital doc

19 We 12b The first term in a geometric sequence is 4 and S∞ = 6.
Find the common factor, r.
20 If the common ratio for a geometric sequence is 0.99 and
the sum to infinity is 100, what is the value of the first and
second terms in the sequence?
21 A student stands at one side of a road 10 metres wide, and
walks half-way across. The student then walks half of the
remaining distance across the road, then half the remaining
distance again and so on.
a Will the student ever make it past the other side of the
road?
b Does the width of the road affect your answer?

5F

applications of sequences
and series

eBook plus
Interactivity

int-0806

Applications of
This section consists of a mixture of problems where the work covered in
sequences and series
the first five exercises is applied to a variety of situations.
The following general guidelines can assist you in solving the problems.
1. Read the question carefully.
2.  Decide whether the information suggests an arithmetic or geometric sequence. Check to 
see if there is a constant difference between successive terms or a constant ratio. If there is
neither, look for a simple number pattern such as the difference between successive terms
changing in a regular way.
3. Write the information from the problem using appropriate notation. For example, if you are told
that the 5th term is 12, write t5 = 12. If the sequence is arithmetic, you then have an equation to
work with, namely: a + 4d = 12. If you know the sequence is geometric, then ar 4 = 12.
4. Defi ne what you have to calculate and write an appropriate formula or formulas. For example, 
if you have to find the 10th number in a sequence which you know is geometric, you have an
equation: t10 = ar9. This can be calculated if a and r are known or can be established.
5. Use algebra to find what is required in the problem.

238

maths Quest 11 Standard General mathematics for the TI-nspire

Worked exampLe 13

eBook plus

In 1970 the cost of 1 megabyte of computer memory was $2025. In 1980 the cost for
Tutorial
the same amount of memory had reduced to $45 and by 1990 the cost had dropped
int-0878
to $1.
Worked example 13
a What was the cost of 1 megabyte of memory in the year 2000?
b How much memory, in megabytes, could you buy for $10 in the year 2010 based on the current
trend?
ThInk
a

1

Present the given information in a table.

a

Year

1970

1980

1990

2000

2010

Cost ($)

2025

45

1

?

?

1

1

Study the table. The information
suggests a geometric sequence for the
cost at each ten-year interval. Verify
this by checking for a constant ratio
between successive terms.

45 ÷ 2025 = 45 and 1 ÷ 45 = 45 so the three terms
form a geometric sequence with common ratio

To find the cost in the year 2000, find
the fourth term in the sequence by
multiplying the preceding (third) term
by the common ratio.

t4 = t3 × r
1
t4 = 1 × 45
1
  = 45

4

Interpret the result and clearly answer
the question.

In the year 2000 one would have paid about
2 cents for a megabyte of memory.

1

If the cost of 1 megabyte can be found
in the year 2010 then the amount of
memory purchased for $10 can be
determined. To find the predicted cost
in the year 2010 the fifth term in the
sequence needs to be determined.

2

3

b

WrITe

1

r = 45.

= 0.022 …

b t5 = t4 × r

=
  =

1
1
45 × 45
1
of a
2025

dollar per megabyte

2

Take the reciprocal of t5 to get the
amount of memory per dollar.

The amount of memory per dollar is
2025 megabytes.

3

Find the amount of memory that can be
purchased for $10.

So $10 would buy 10 × 2025
= 20 250 megabytes.

Worked exampLe 14

Express the recurring decimal 0.131 313 13 … as a proper fraction.
ThInk

WrITe

1

Express the given number as a geometric
series.

0.131 313 … = 0.13 + 0.001 3 + 0.000 013 …

2

State the values of a and r.

a = 0.13 and r = 0.00013
= 0.01
.13

3

Find the sum to infinity, S∞:
Write the formula for the sum to infinity.

S∞ =

a
1− r

Chapter 5

Sequences and series

239

0.13
1 − 0.01
0.13
S∞ =
0.99

4

Substitute values of a and r into the formula
and simplify.

S∞ =

5

Multiply both numerator and denominator by
100 to get rid of the decimal point.

S∞ =

13
99

REMEMBER

To solve problems, use the following guidelines.
1. Identify the type of the sequence by checking whether there is a common difference, or
a common ratio.
2. Translate given information into mathematical statements, using appropriate notation.
3. Define what you have to find and write appropriate formula(s).
4. Use algebra to find what is required.
Exercise

5F

Applications of sequences and series
  1   WE 13  In 1970 the Smith family purchased a small house for $60  000. Over the following
years, the value of their property rose steadily. In 1975 the value of the house was $69  000 and
in 1980 it reached $79  350.
a Assuming that the pattern continues through the years, find (to the nearest dollar) the
value of the Smiths’ house in i 1985, ii 1995.
b By what factor will the value of the house have increased by the year 2010, compared to
the original value?
  2 An accountant has been working with the same company for 15 years. She commenced on a
salary of $28  000 dollars and has received a $2500 increase each year.
a What type of sequence of numbers does her annual income follow?
b How much did she earn in her 15th year of employment?
c How much has she earned from the company altogether?
d What was her percentage increase at the end of i her first and ii her fourteenth year of
employment?
  3 A chemist has been working with the same company for 15 years. He commenced on a salary
of $28  000 dollars and has received a 4% increase each year.
a What type of sequence of numbers
does his annual income follow?
b How much did he earn in his
15th year of employment?
c How much has he earned from the
company altogether?
d What was his increase in salary
at the end of i his first and ii his
­fourteenth year of employment?
  4 A biologist is growing a tissue culture
in a Petri dish. The initial mass of the
culture was 20 milligrams. By the end
of the first day the culture was a mass of
28 milligrams.

240

Maths Quest 11 Standard General Mathematics for the TI-Nspire

a Assuming that the daily growth is arithmetic, find the mass of the ­culture after the
second, third, tenth and nth day.
b On what day will the culture mass first exceed 200 milligrams?
c Assuming that the daily growth is geometric, find the mass of the culture after the second,
third, tenth and nth day.
d On what day will the culture mass first exceed 200 milligrams?
  5 Logs of wood can be stacked so that there is one more log on each descending layer than on
the previous layer. The top row has 6 logs and there are 20 rows.
a How many logs are in the stack altogether?
b The logs are to be separated into two equal piles. They are sepa­rated by removing logs
from the top of the pile. How many rows down will workers take away before they
remove half the stack?
  6 As I was going to St Ives
I met a man with seven wives.
Every wife had seven sacks,
Every sack had seven cats,
Every cat had seven kits.
Kits, cats, sacks and wives,
How many were coming from St Ives?

Note: This is a variation on the original
riddle, which asks ‘How many were
going to St Ives’.

  7 Thoughtful Frank has 100 movie tickets to give away to people at a local shopping centre. He
gives the first person one ticket, the next person two tickets, the third person three tickets and
so on until he can no longer give the nth person n tickets. How many tickets did the last lucky
person receive? How many tickets did Frank have left?
  8 Kind-hearted Kate has 200 movie tickets to give away to people at the ­shopping centre. She
gives the first person one ticket, the next person two tickets, the third person four tickets and
so on following a geometric progression until she can no longer give the nth person 2(n - 1)
tickets. How many tickets did the last lucky person receive? How many tickets did Kate have
left?
  9 The King of Persia, so the story goes, offered Xanadu any reward to secure the safety of his
kingdom. As his reward, Xanadu requested a chessboard with one grain of rice on the first
square, two grains on the second, four on the third and so on until the 64th square had its share
of rice deposited.
a Find the total number of grains of rice that the king needed to supply.
b If each grain of rice weighs 0.10 grams, how many kilograms of rice does this represent?
(Note: There are 103 grams in 1 kilo­gram.)
10 As legend has it, the King of Constantinople offered Xanadu’s cousin Yittrius any reward to
secure the safety of his city. This Yittrius accepted: she requested a chessboard with one grain
of rice on the first square, three grains of rice on the second square, five grains of rice on the
third square and so on until the 64th square had its share of rice deposited.
a Find the total number of grains of rice that the king needed to supply.
b If each grain of rice weighs 0.10 gram, how many kilograms of rice does this represent?
(Note: There are 103 grams in 1 kilo­gram.)
11 A student is 3.0 m from the door to a classroom and decides that he will take a 1.0 m step
followed by a step of half that distance, and half again and so on until he gets to the classroom
door. Show that he will never get any closer than one metre from the door.
12 A hiker walks 36 km on the first day and 23 that distance on the second. Every day thereafter
she walks 23 of the distance she walked on the day before. Will the hiker cover the distance of
100 km to complete the walk and on what day will she complete the task?

Chapter 5  Sequences and series

241

13 We 14
Recurring decimals can be expressed as rational numbers. Find the fraction
equivalent of the following recurring decimal numbers by writing the decimal number as a sum
of infinite terms.
a 0.1111 … = 0.1 + 0.01 + 0.001 + …
b 0.333 333 333 … 
c 0.5757 …
d 2.343 434 … 
e 3.142 142 142 …
f 21.2121 …
g 16.666 …
14 In 1990, 100 students enrolled for a hypocorisma subject at a local university. Each subsequent
year for the next decade the enrolment increased by 20%.
a Find the number of students enrolled in hypocorisma in 1995.
b Over the course of the decade find the total number of students who had enrolled in
hypocorisma.
15 For tax purposes, the value of a computer used for a business depreciates by 8.5% of the initial
cost each year. For economic reasons the business sells its computers when they first depreciate
to less than half their initial value. After how many years will a computer used by this business
be sold?
16 The side lengths of a right-angled triangle form the successive terms of an arithmetic sequence.
The perimeter of the triangle is 72 m. What are the side lengths of the triangle?
17 A circular board is divided into a series of concentric
circles of radius 1 cm, 2 cm, 3 cm and 4 cm as shown
at right.
a Find the areas of each of the successive shaded
regions and show that they form an arithmetic
progression.
b A dart is fired at the board at random and hits
the board. What is the probability of striking each
of the four regions of the board?
(Note: The probability of striking a region = area of
region ÷ total area.)

4 cm
3 cm
1 cm

2 cm

18 A bullet is fired vertically up into the air. In the first second it has an average speed of 180 m/s;
that is, it travels 180 m up into the air during the first second. Each second its speed diminishes
by 12 m/s. Thus during the 2nd second the bullet has an average speed only 168 m/s and
accordingly travels 168 m further up into the air.
a Find an equation for the average speed of the bullet for the nth second that it is in the air.
b Find the time when the average speed of the bullet is equal to zero.
c Find the maximum height of the bullet above where it was fired.
19 Coffee cools according to Newton’s Law of
Cooling in which the temperature of the coffee
above room temperature drops by a constant
fraction each unit of time. The table below shows
the temperature of a cup of coffee in a room at
20 °C each minute after it was made.
Remember to subtract the room temperature
from the temperature of the coffee before you
do your calculations.
Time (min)
1
2
3

242

Temperature (°C)
80.0
74.0
68.6

maths Quest 11 Standard General mathematics for the TI-nspire

The person who made the coffee will drink it only if it has a temperature in excess of 50 °C.
What is the minimum time after the cup of coffee has been made before it will not be
drinkable?
20 Two arithmetic sequences, tn and un, are multiplied together. That is, each term is multiplied by
the other to form a new term.
tn = 2n - 3, n ∈ {1, 2, 3, …} and
un = 3n, n ∈ (1, 2, 3, …}
Show that the new sequence of numbers t1 × u1, t2 × u2, t3 × u3, … is an arithmetic series and
hence find the arithmetic sequence for that new series.
(Hint: For a sequence, an, with a series An, an = An - An - 1.)

Chapter 5  Sequences and series

243

SummarY
Describing sequences

•  A sequence is a string of numbers or expressions. It may contain a finite or infinite number of terms and may
or may not follow a recognisable pattern.
•  A sequence can be described in a number of ways.
1. As a list tn: {1, 7, 2, 6, …} (note that t3 = 2)
2. As a function: tn = 2n - n2, n ∈ {1, 2, 3, …} (note that t5 = 2 × 5 - 52 = -15)
3. As a recursive or iterative formula: tn + 1 = 2tn - 3, t1 = 6
(note that t2 = 2 × 6 - 3 = 9)
•  The logistic equation is a model of population growth of the general form:
tn + 1 = atn(1 – tn),
where 0 < t0 < 1 and a is a constant.
•  A convergent sequence is a sequence whose terms settle at a certain fixed number, x, called the limit of the
sequence. This can be written as tn → x. A sequence whose terms grow further and further apart is called
divergent. That is, a sequence is divergent if tn → ∞, or tn → -∞ as n → ∞. A sequence whose terms fluctuate
between two (or more) values is called oscillating.
Arithmetic sequences

•  An arithmetic sequence is one whose successive terms have a common difference. This common difference
is given the symbol d. Thus tn + 1 - tn = d for all values of n. The first term in the sequence is given the
symbol a.
•  If x, y, z are successive terms in an arithmetic sequence then the middle term (y) is called an arithmetic mean
and is equal to the average of the two outer terms (x and z):
y=

x+z
2

•  An arithmetic sequence can be written as a, a + d, a + 2d, … and so the nth term, tn, is:
tn = a + (n - 1)d using the function notation, or
tn + 1 = tn + d, t1 = a using the iterative notation.
Arithmetic series

•  The sum of the first n terms of the arithmetic sequence is given by
Sn =
•  In general, tn + 1 = Sn + 1 - Sn.

n
(2a + ( n − 1)d )
2

Geometric sequences

•  A geometric sequence is one in which each successive term is obtained by multiplying the preceding term by
a constant number. This number is called the common ratio and is given
tn +1
the symbol r. Thus
= r for all values of n. The first term in the sequence is given the symbol a.
tn
•  If x, y, z are successive terms in an arithmetic sequence then y is called a geometric mean and is given by
y2 = xz.
•  A geometric sequence can be written as a, ar, ar 2, … and so the nth term, tn, is
tn = ar n - 1 using the function notation, or
tn + 1 = rtn, t1 = a using the iterative notation.

244

maths Quest 11 Standard General mathematics for the TI-nspire

Geometric series

The sum of the first n terms in a geometric sequence is given by
a(1 − r n )
with r ≠ 1
Sn =
1− r
a(r n − 1)
or  Sn =
with r ≠ 1
r −1
• When the magnitude of r is less than one, that is, -1 < r < 1, the sum of a geometric sequence
a
to infinity S∞ is given by S∞ =
.
1− r
Applications of arithmetic and geometric sequences and series

• To solve problems, use the following guidelines.
1. Identify the type of sequence by checking whether there is a common difference, or a common ratio.
2. Translate given information into mathematical statements, using appropriate notation.
3. Define what you have to find and write appropriate formula(s).
4. Use algebra to find what is required.

Chapter 5  Sequences and series

245

chapter review
Multiple choice

1 Consider the sequence tn + 1 = 2tn + 4; t3 = 12. The
second term in the sequence is:
A 10
B 6
C 28
D 4
E 8
2 A series is listed as 3, 10, 21, 36, … The next term
in the series is:
A 51
B 52
C 53
D 54
E 55
3 Paula started a stamp collection. She decided to buy
a number of new stamps every week.
The number of stamps bought in the nth week, tn,
is defined by the equation
tn = tn − 1 + tn − 2, where t1 = 1 and t2 = 2.
The total number of stamps in her collection after
five weeks is:
A 8
B 12
C 15
D 19
E 24

[©VCAA 2006]
Exam tip   This question was an application of a
Fibonacci-related sequence, which required students
to determine and then sum the first five terms in the
sequence. Option A, which was chosen by 29% of
students, corresponded to the value of the fifth term in
the sequence.
[Assessment report 2006]

4 The 23rd term in the sequence of numbers
{7, 3, -1, …} is:
A -88 B -81 C -74 D -83 E 90
5 Consider the arithmetic sequence 52, a, 41, b. The
numerical value of the expression a - 3b is:
1
1
1
A -60 B -64 2 C -67 2 D -71 E 72 2
6 Consider the arithmetic sequence x - 2y, 3x - 4y,
4x - 7y, … An expression for y in terms of x is:
A y = x
B y = -x
C y = -2x
D y = 2x
E y = 3x
7 A car is accelerating such that in the 1st second
it travels 2.0 metres, in the 2nd second it travels
3.5 metres, in the 3rd second it travels 5.0 metres,
and so on for a total of 15 seconds. The total
distance travelled by the car is:
A 630 m
B 93.75 m
C 187.5 m
D 375 m
E 315 m
8 The sum of the first four terms in an arithmetic
sequence is 70. The sum of the first six terms is 63.
The sixth term of the sequence is equal to:
A -14
B -7
C 0
D 7
E 14
246

9 For a geometric sequence, the 4th term is 5 and the
7th term is -625. The second term in the sequence
is:
A -2.5
B -1.25
C 0.25
D 0.25
E 0.20
10 t
n

n

The graph above shows the first six terms of a
sequence.
This sequence is best described as:
A geometric with -1 < r < 0
B geometric with 0 < r < 1
C geometric with r > 1
D arithmetic with d < 0
E arithmetic with 0 < d < 1
[©VCAA 2005]
11 The sum of an infinite geometric sequence is 5.6
with the common ratio equal to 0.20. The sum of
the first four terms of the geometric sequence is
closest to:
A 5.0
B 5.2
C 5.4
D 5.6
E 5.8
12 The sum of the first 10 terms of a geometric
sequence is 400. The next term in the sequence
is 3 times the previous term. The first term in the
sequence is:
A

17

731

D

200
781



B

400

1473

E

10
387

C

100
7381

13 A healthy eating and gym program is designed to
help football recruits build body weight over an
extended period of time.
Roh, a new recruit who initially weighs 73.4 kg,
decides to follow the program.
In the first week he gains 400 g in body weight.
In the second week he gains 380 g in body weight.
In the third week he gains 361 g in body weight.
If Roh continues to follow this program
indefinitely, and this pattern of weight gain

Maths Quest 11 Standard General Mathematics for the TI-Nspire

remains the same, his eventual body weight will
be closest to:
A 74.5 kg
B 77.1 kg
C 77.3 kg
D 80.0 kg
E 81.4 kg

[©VCAA 2006]
Exam tip   This question required students to

model an athlete’s weight gain over an extended
period of time using an infinite geometric sequence.
Only 33% of students gave the correct response. The
relatively even distribution responses across the
incorrect options suggests that the majority of
students guessed their answer to this question.
[Assessment report 2006]

14 The sequence tn = 81 ×  1  n - 1 and the sequence
3
t
sn = 1 - 0.1n are combined to form the ratio n .
sn
When n = 9 the value of the ratio is:
1
1
1
A 27
B 270
C 243
D

1
81



E

10
81

Short answer

1 Write the iterative definition for each of the
following sequences:
a {7, 11, 19, 35, 67, …} b {-2, 5, 26, 677, …}
2 For each of the following values of a and t0:
a a = 0.3, t0 = 0.4
b a = 2.8, t0 = 0.6
c a = 3.1, t0 = 0.8
  i use the logistic equation tn + 1 = atn(1 – tn),
to generate a sequence of 5 terms
ii state whether the sequence is convergent,
divergent, or oscillating (you may need to
generate more terms first)
iii if the sequence is convergent, state its limit.
3 For the arithmetic sequence where t3 = 10 and
t6 = 478, find
a the functional rule for the nth term in the
sequence
b the iterative rule for the sequence.
4 A car at a racetrack starts from rest and travels
0.5 m in the 1st second and 1.0 m in the 2nd second
following an arithmetic progression in the distances
covered each subsequent second.
a How far will it travel during the 10th second?
b After 10 seconds of motion, how far will it
have travelled in total?
c To the nearest whole second, how long will it
take to travel 1000 m (1 km)?
5 Noel is an enthusiastic runner. He decides to
follow a training program over several weeks. The
distance (in kilometres) that Noel runs each day
forms an arithmetic sequence. The distances for the

first three days of this training program are shown
below.
Day
Distance (km)

1
  4.2

2
  4.5

3
  4.8

a What distance (in kilometres) will Noel run on
Day 5?
b On which day will Noel first run more than
7 km?
c Determine the total distance that Noel runs
from Day 3 to Day 12 inclusive.
Write your answer in kilometres, correct to
1 decimal place.
Exam tip   Some students listed, then added, all the
terms from t3 to t12, and received full marks. Others
stopped at t11 and did not gain full marks. A number
of students understood that S12 − S2 was required, but
many incorrectly said this was equal to S9 or S10.

[Assessment report 2005]

d An expression for the nth term of this sequence
can be written as tn = 0.3n + b
Determine the value of b.

[©VCAA 2005]
Exam tip   A common incorrect answer was 4.2
(i.e. the value of a).
[Assessment report 2005]

6 Consider the geometric sequence gn = 1.4 × 1.5n - 1;
n = 1, 2, 3, …
a Find the first and 5th terms of the sequence.
b Find the sum of the first 10 terms.
c Find the percentage contribution that the 10th
term makes to this sum.
7 At Bugas Heights a radiation leak in a waste
disposal tank potentially exposes staff to a
1000 milli-rem h dose on the first day of the
accident, a 800 milli-rem h dose on the second day
after the accident and a 640 milli-rem h dose on the
third day following the accident.
a Assuming a geometric sequence, find the amount
of potential exposure dose by the 10th day.
b Find the total potential exposure dose in the
first 5 days.
8 Australian Heating is a company that produces
heating systems. The number of heating systems
produced annually is modelled by an increasing
geometric sequence. The number of heating
systems produced in each of the first three years is
shown in the table below.
Year
Number of heating
systems produced

1
2000

2
2200

3
2420

Chapter 5  Sequences and series

247

a Show that the common ratio, r, of this
geometric sequence is 1.1.
Exam tip   Most students were able to show some
appropriate division or multiplication that showed
the common ratio was 1.1.

Exam tip   A number of students did not round this
answer to a whole number of heating systems. Several
tabulated the sequence and added the terms rather
than use the appropriate series formula. Some students
found only the number of heating systems produced
in the tenth year.

[Assessment report 2004]

[Assessment report 2004]

b What is the annual percentage increase in the
number of heating systems produced each
year?
Exam tip   Many students got this wrong. A
common incorrect answer was 110%.
[Assessment report 2004]

c How many heating systems will be produced
in year 5? Write your answer correct to the
nearest whole number.
Exam tip   Some students did not round this
answer to a whole number of heating systems.
[Assessment report 2004]

d The number of heating systems produced
annually continues to follow this pattern. In
total, how many heating systems will they
produce in the first 10 years of operation?

[©VCAA 2004]

9 The infinite sum of a geometric sequence is 99 and
the first term is 10. Find the common ratio for the
sequence.
10 Find the sum of the following expressions:
a 1 +

1
4

1
1
+ 16
+ 64
… b   1 -

2
3

+

4
9

-

8
27



11 a For the sequence defined by tn + 1 = 1 (tn + 2 );
2

t

n

t1 = 1, find the values of t2, t3, … t10, using a
spreadsheet or calculator.
b The terms in this sequence give successive
approximations of a real number. What is this
number?
12 Find the fraction equivalent of the following
recurring decimals:
a 0.222  222
b 2.454  545  454

Extended Response

1 Consider a square of side length 2 units.
a What is the perimeter of the square?
b Each of the four midpoints form the vertices of a new square inscribed within the original square. Find
the perimeter of this new square.
c Repeat the process to find the perimeter of a third square inscribed within the second.
d Give an expression for the perimeter of the nth square.
2 Consider the following iterative definitions:
1

a tn + 1 = tn – 3 , t1 = 8
4

b tn + 1 = atn, t1 = b2
c tn + 1 = 3tn2 – 1.5, t1 = 0.5
If each of these definitions is used to generate a sequence of numbers:
i decide whether the sequence is arithmetic, geometric or neither, and
ii find its fourth term.
3 In January 2004, Rachel and Nathan inherited a small trout farm from their Uncle Michael. They were
told that in any given year the trout population, pn + 1, could be easily calculated using the formula
pn + 1 = 0.5pn(1 – pn), where pn is the number of trout (in thousands) in the preceding year. They were also told
that on the day of their inheritance the farm housed 800 fish.
a Use the above formula to predict (to the nearest whole number) the size of the fish population on Rachel
and Nathan’s farm for the next three years; that is, for
i January 2005
ii January 2006
iii January 2007.

248

Maths Quest 11 Standard General Mathematics for the TI-Nspire

b What will happen to the size of the fish population if it continues to change according to the formula?
How long will it take?
After extensive research, Rachel and Nathan decided to modernise their newly acquired farm. A new
feeding system and other improvements were installed, and were completed within the first two years (that
is, by January 2006). As a result, in any given year the trout population, pn + 1, could now be calculated
using the formula pn + 1 = 1.6pn(1 – pn), where pn is the number of trout (in thousands) in the preceding year.
c Use this new formula, and the figure obtained in part a ii (that is, the trout population in January 2006)
as your starting point to predict the size of the trout population for January 2007 and January 2008.
d Do Rachel and Nathan still run the risk of losing all of their trout stock? Explain your answer.
e Will the size of the trout population ever reach and increase beyond the initial number of 800 fish? Give
reasons for your answer.
4 On an island in the Pacific Ocean the population of a species of insect (species A) is increasing geometrically
with a population of 10  000 in 1990 and an annual growth rate of 12.0%. Another species of insect (species B)
is also increasing its population, but arithmetically with numbers 15  000 in 1990 and an annual increment of
1000 per annum.
a Using a spreadsheet or other method, determine the difference in the numbers of the two species during
the last decade of the twentieth century (that is, up to 1999).
b In what year will the first species be greater in number than the second species, assuming that growth
rates remain fixed?
A scientist has a mathematical model where the species can cohabit provided that they have equal numbers
in the year 2000.
c If the growth rate in species A is to remain unchanged, what would the annual increment in species B
need to be to achieve this?
d If the annual increment in species B is to remain unchanged, what would the growth rate in species A
need to be to achieve this?
5 a A series is given by the equation
Sn = 2n2 + 3n.
Show that the sequence is arithmetic and give the expression for the nth term in the sequence, tn.
b A series is given by the equation Sn = an2 + bn. Show that the sequence is arithmetic and give the
expression for the nth term in the sequence, tn, in terms of a and b.
6 Maria intends to follow a healthy eating plan. She will reduce her daily kilojoules intake by a constant amount
each day over a period of 14 days. The graph below shows Maria’s kilojoules intake for the first five days.

8700

Kilojoule intake

8600
8500
8400
8300
8200
8100
8000
7900
0

a What was Maria’s kilojoules intake on day 1?
b Determine Maria’s kilojoules intake on day 6.

1

2

3 4
Day

5

6

Exam tip   A common incorrect answer was 7945.
[Assessment report 2007]

Chapter 5  Sequences and series

249

c Maria’s kilojoules intake on the nth day is given by the
equation Kn = a − 150 × n. Determine the value of a.
d On which day will Maria’s daily kilojoules intake 
be 6750?

exam TIp

A common incorrect answer was 8700.
[Assessment report 2007]

exam TIp

A common incorrect answer was

Day 13.
[Assessment report 2007]

Maria’s brother, Rupert, believes he will benefi t by reducing his daily kilojoules intake. His kilojoules 
intake over 14 days will follow a geometric sequence with a common ratio of 0.95.
On day 1, Rupert’s kilojoules intake was 12 000.
e By what percentage is Rupert’s kilojoules intake
exam TIp
A common incorrect answer was 95%
reduced each day?
or 0.95.

[Assessment report 2007]

f Determine Rupert’s kilojoules intake on day 3.

exam TIp
A frequent error was to use 1200
instead of 12 000.
[Assessment report 2007]

g Write an equation that gives Rupert’s kilojoules intake
Rn on the nth day.

exam TIp

A formula in terms of n was required.
[Assessment report 2007]

h Find the difference between Rupert’s kilojoules intake on day 9 and day 10.
Write your answer correct to the nearest kilojoule.
i Determine Rupert’s total kilojoules intake from day 8 
exam TIp   Many students calculated and listed S ,
8
to day 14 inclusive.
S9, S10 etc. and then added the result rather than
subtracting the results of two geometric series.
Others tried S14 − S8 rather than S14 − S7.

[Assessment report 2007]

Write your answer correct to the nearest kilojoule.
Maria decides to improve her fi tness level by cycling each day. The time in minutes Mn that Maria cycles on 
the nth day is modelled by the equation
exam TIp
For 1 mark, students had to show two
Mn + 1 = 0.75Mn + 8 where M2 = 20
subtractions
not having the same result and two
j For how many minutes will Maria cycle on day 4?
quotients not having the same result. Some only
k Show that the time Maria cycles each day does not 
showed the calculations and did not state that the
follow an arithmetic or a geometric sequence.
results within each pair were not equal. Others
merely showed the results of the calculations without
indicating where these results came from, or their
significance.
[Assessment report 2007]

l For how many minutes will Maria cycle on day 1?
Rupert decides to include both swimming and running in his exercise plan. On day 1, Rupert swims 100 m
and runs 500 m. Each day he will increase the distance he swims and the distance he runs. His swimming
distance will increase by 50 m each day. His running distance will increase by 2% of the distance he ran on
the previous day.
m On which day will the distance Rupert swims first be greater than the distance he runs?
[©VCAA 2007]

eBook plus
Digital doc

Test Yourself
Chapter 5

250

maths Quest 11 Standard General mathematics for the TI-nspire

exam TIp
A method mark was available for
setting up an equation to find where the two
distances were equal. Many students reached the 
wrong answer with working out that could not be
clearly followed and thus earned no marks.
[Assessment report 2007]

eBook plus

aCTIvITIeS

Chapter opener
Digital doc

•  10 Quick Questions: Warm up with ten quick
questions on sequences and series. (page 212)
5A

Describing sequences

Tutorial

•  We1 int-0874: Watch how to find terms in three
different sequences, by recognising the pattern and
by using an expression. (page 214)
5B

Arithmetic sequences

Tutorial

•  We6 int-0875: Watch how to find specific terms and
a general expression for a term in an arithmetic
sequence. (page 222)
Digital docs

•  Spreadsheet 117: Investigate arithmetic sequences.
(page 223)
•  SkillSHEET 5.1: Practise using elimination to solve
simultaneous equations. (page 224)
5C

Arithmetic series

Tutorial

•  We7 int-0876: Watch how to find the sum of the first
20 terms in an arithmetic sequence. (page 226)
Digital docs

•  Spreadsheet 117: Investigate arithmetic series.
(page 227)
•  WorkSHEET 5.1: Determine terms, differences 
between terms, and sums of terms in arithmetic
sequences. (page 228)
5D

Geometric sequences

Tutorial

•  We10 int-0877: Watch how to find a general
expression for a term in a geometric sequence.
(page 230)
Digital doc

5E

Geometric series

Digital docs

•  Spreadsheet 117: Investigate geometric series.
(page 237)
•  WorkSHEET 5.2: Determine terms, differences 
between terms, and sums of terms in geometric
sequences. (page 238)
5F

Applications of sequences and series

Interactivity

•  Sequences and series int-0806: Consolidate your
understanding of sequences and series. (page 238)
Tutorial

•  We13 int-0878: Watch how to apply the model
of a geometric sequence to a real-life problem.
(page 239)
Chapter review
Digital doc

•  Test Yourself: Take the end-of-chapter test to test
your progress. (page 250)
To access eBookPLUS activities, log on to
www.jacplus.com.au

•  Spreadsheet 117: Investigate geometric sequences.
(page 232)

Chapter 5

Sequences and series

251

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