1. Arithemtic and Geometric Series
Paul Hancock
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HSC Question Analysis
Let’s have a quick look at the number of marks allocated to the topic for
HSC exams in the current format.
Year
2012
2013
2014
2015
Marks
14
12
11
8
Arithmetic and Geometric Series
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The Syllabus
The Mathematics Syllabus can be found at:
http://www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/
maths23u_syl.pdf
The pages relevant to us today are 44-49. Let’s take a look at some
important pieces of information from the syllabus.
”The definitions of series, term, nth term and sum to n terms should
be understood.”
”The definition of an arithmetic series and its common difference
should be understood. The formulae for the nth term and the sum to
n terms should be derived.”
”The definitions of a geometric series and its common ratio should be
understood, and the formulae for the nth term and sum to n terms
derived.”
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Arithmetic Series
a is called the first term.
d is called the common difference.
Tn is the nth term of the series.
d = T2 − T1 = T3 − T2 = · · · = Tn − Tn−1 .
The nth term of the series is given by Tn = a + (n − 1)d
The sum of the first n terms of the series is given by
Sn = n2 [a + l] = n2 [2a + (n − 1)d].
Where l is the last term.
Arithmetic and Geometric Series
Review
Geometric Series
a is called the first term.
r is called the common ratio.
Tn is the nth term of the series.
d=
The
T3
T2
Tn
T1 = T2 = · · · = Tn−1 .
nth term of the series is
given by Tn = ar n−1
a(1−r n )
1−r .
a
S∞ = 1−r
.
The sum of the first n terms of the series is given by Sn =
The series has a limiting sum, if |r | < 1. It is given by
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Multiple Choice
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2015 - Question 3
Example (Q3)
Well, the first term is
and
th
So the n term is given by Tn =
Hence the 15th term is T15 =
Answer:
Arithmetic and Geometric Series
2014 - Question 8
Example (Q8)
Well, the first term is
and
So the nth term is given by Tn =
Now
and
This means the solution is either
have:
Answer:
Arithmetic and Geometric Series
, so
, but since n − 1 is
.
we
Multiple Choice
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Standard Questions
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2012 - Question 12c
Arithmetic and Geometric Series
2012 - Question 12c
(i) How many tiles would Jay use in row 20?
The number of tiles in each row is given by 3, 5 and 7, with 2 more
tiles added to each future row.
This forms an arithmetic series, with first term a = and common
difference d = .
We can use the nth term formula to find out how many tiles Jay will
use in the 20th row.
T20 =
=
=
=
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2012 - Question 12c
(ii) How many tiles would Jay use altogether to make the first 20 rows?
We can use the the formula for the sum of 20 terms.
S20 =
=
=
=
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2012 - Question 12c
(iii) Jay has only 200 tiles. How many complete rows of the pattern can
Jay make?
We can use the sum of n terms formula to find out how many
complete rows Jay can make with 200 tiles.
Sn =
=
=
=
=0
=0
n=
∴ Jay can complete
Arithmetic and Geometric Series
(Taking positive solution)
rows with 200 tiles.
Standard Questions
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Standard Questions
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2013 - Question 12c
Arithmetic and Geometric Series
2013 - Question 12c
(i) Show that in the 10th year Kim’s annual salary is higher than Alex’s
annual salary.
Kim
Alex
A0 =
A0 =
A1 =
A1 =
=
An =
An =
A10 =
A10 =
Hence after the 10th year Kim is earning more than Alex.
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2013 - Question 12c
(ii) In the first 10 years how much, in total, does Kim earn?
Sn =
=
Hence after 10 years Kim has earned $
Arithmetic and Geometric Series
Standard Questions
in total.
2013 - Question 12c
(iii) Every year, Alex saves
take her to save $87500?
1
3
of her annual salary. How many years does it
We can use the sum of n terms formula along with our working in (i) to
solve the problem.
Sn =
87500 =
350 =
=0
n=
∴ It takes 7 years for Alex to earns $87500.
Arithmetic and Geometric Series
Standard Questions
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Standard Questions
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2013 - Question 13d
Arithmetic and Geometric Series
2013 - Question 13d
(i) The loan is to be repaid over 30 years. Show that the monthly
repayment is $2998 to the nearest dollar.
Here P = $
and r =
So,
A30 =
=
=
≈
M≈
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2013 - Question 13d
(ii) Show that the balance owing after 20 years is $270000 to the nearest
thousand dollars.
A20 ≈
≈
≈
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2013 - Question 13d
(iii) After 20 years the family borrows an extra amount, so that the family
then owes a total of $370000. The monthly repayment remains $2998 and
the interest rate remains the same. How long will it take to repay the
$370000?
Now P = $
with all other constants unaltered. So,
An =
=
=
=
=
=
=
=
n=
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2013 - Question 13d
(iii) After 20 years the family borrows an extra amount, so that the family
then owes a total of $370000. The monthly repayment remains $2998 and
the interest rate remains the same. How long will it take to repay the
$370000?
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Standard Questions
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2014 - Question 14d
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2014 - Question 14d
(i) How much of the drug is in the patient’s body immediately after the
second dose is given?
The initial amount, which we will call A1 was 10 mL. The amount
immediately after the second dose would be 10mL plus 31 the initial
amount.
∴ the amount immediately after the second dose was
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2014 - Question 14d
(ii) Show that the total amount of the drug in the patient’s body never
exceeds 15 mL.
A1 = 10
A2 =
A3 =
..
.
An =
=
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Standard Questions
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2015 - Question 14c
Arithmetic and Geometric Series
2015 - Question 14c
(i) Show that A2 = 100000(1.006)2 − M(1 + 1.006).
A0 =
A1 =
A1 =
=
A2 =
=
=
=
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2015 - Question 14c
(ii) Show that An =
100000(1.006)n
−M
(1.006)n −1
0.006
.
From (i) we have a formula for A2 and will continue to determine the
pattern for An .
A2 =
A3 =
=
=
=
..
.
An =
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2015 - Question 14c
(ii) Show that An =
100000(1.006)n
−M
(1.006)n −1
0.006
.
An =
An =
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2015 - Question 14c
(iii) Sam makes monthly repayments of $780. Show that after making 120
monthly repayments the amount owing is $68500 to the nearest $100.
A120 =
≈
(Nearest $)
≈
≈
≈
Arithmetic and Geometric Series
(Nearest $100)
Standard Questions
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2015 - Question 14c
(iv) Immediately after making the 120th repayment, Sam makes a one-off
payment, reducing the amount owing to $48500. The interest rate and
monthly repayment remain unchanged. After how many more months will
the amount owing be completely repaid?
Well this changes the formula for An in only one way, the intial amount is
no longer $100000 it is now $48500. An is otherwise unchanged. We wish
to find a value of n for which An =0.
An =
=0
=0
=0
=
=
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2015 - Question 14c
(iv) Immediately after making the 120th repayment, Sam makes a one-off
payment, reducing the amount owing to $48500. The interest rate and
monthly repayment remain unchanged. After how many more months will
the amount owing be completely repaid?
=
=
n=
n≈
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2012 - Question 15a
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Harder Questions
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2012 - Question 15a
(i) Find the length of the strip required to make the first ten rectangles.
L=
=
=
approx
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2012 - Question 15a
(ii) Explain why a strip of length 3 m is sufficient to make any number of
rectangles.
Considering just the geometric series from (i), we note that
.
.
=
=
L∞ =
Now since
, a 3 m strip is more than sufficient.
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Harder Questions
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2014 - Question 16b
Arithmetic and Geometric Series
2014 - Question 16b
(i) Explain why the balance of the account at the end of the second month
is $500(1.003)2 + $500(1.01)(1.003).
1st month (beginning)
1st month (end)
2nd month (beginning)
2nd month (end)
(deposit added)
(interest added)
(deposit added)
(interest added)
Now
so the balance of the account at the end of the second month was
$500(1.003)2 + $500(1.01)(1.003).
Arithmetic and Geometric Series
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2014 - Question 16b
(ii) Find the balance of the account at the end of the 60th month, correct
to the nearest dollar.
Continuing on from (i):
3rd month (beginning)
3rd month (end)
..
.
60th month (end)
Let B be the balance owing at the end of the 60th month. Then:
B=
=
=
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2014 - Question 16b
(ii) Find the balance of the account at the end of the 60th month, correct
to the nearest dollar.
The bracketed term is a geometric series with 60 terms,
. Let the sum of the series be S, then:
and
S=
≈
≈
Arithmetic and Geometric Series
(nearest cent)
(nearest $)
Harder Questions
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