Arithmetic and Geometric Series

132 Pages·2017·0.75 MB·English

ArithmeticandGeometricSeries
HSC Revision Day
Arithmetic and Geometric Series
WoononaHighSchool
2017
Teacher: Paul Hancock
School: Woonona High School
E-mail: [email protected]
PaulHancock (WoononaHighSchool) Series T12017 1/23
Year Marks
2012 14
2013 12
2014 11
2015 8
2016 11
The average number of marks per exam is 11.2.
ArithmeticandGeometricSeries Review
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.
PaulHancock (WoononaHighSchool) Series T12017 2/23
The average number of marks per exam is 11.2.
ArithmeticandGeometricSeries Review
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 Marks
2012 14
2013 12
2014 11
2015 8
2016 11
PaulHancock (WoononaHighSchool) Series T12017 2/23
ArithmeticandGeometricSeries Review
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 Marks
2012 14
2013 12
2014 11
2015 8
2016 11
The average number of marks per exam is 11.2.
PaulHancock (WoononaHighSchool) Series T12017 2/23
Well, the first term is a=3 and d =7−3=11−7=4.
So the nth term is given by T =3+(n−1)×4 =4n−1.
n
Hence the 15th term is T =4(15)−1=59
15
(A)
ArithmeticandGeometricSeries MultipleChoice
2015 - Question 3
Example 1 (Q3)
Answer:
PaulHancock (WoononaHighSchool) Series T12017 3/23
and d =7−3=11−7=4.
So the nth term is given by T =3+(n−1)×4 =4n−1.
n
Hence the 15th term is T =4(15)−1=59
15
(A)
ArithmeticandGeometricSeries MultipleChoice
2015 - Question 3
Example 1 (Q3)
Well, the first term is a=3
Answer:
PaulHancock (WoononaHighSchool) Series T12017 3/23
So the nth term is given by T =3+(n−1)×4 =4n−1.
n
Hence the 15th term is T =4(15)−1=59
15
(A)
ArithmeticandGeometricSeries MultipleChoice
2015 - Question 3
Example 1 (Q3)
Well, the first term is a=3 and d =7−3=11−7=4.
Answer:
PaulHancock (WoononaHighSchool) Series T12017 3/23
=4n−1.
Hence the 15th term is T =4(15)−1=59
15
(A)
ArithmeticandGeometricSeries MultipleChoice
2015 - Question 3
Example 1 (Q3)
Well, the first term is a=3 and d =7−3=11−7=4.
So the nth term is given by T =3+(n−1)×4
n
Answer:
PaulHancock (WoononaHighSchool) Series T12017 3/23
Hence the 15th term is T =4(15)−1=59
15
(A)
ArithmeticandGeometricSeries MultipleChoice
2015 - Question 3
Example 1 (Q3)
Well, the first term is a=3 and d =7−3=11−7=4.
So the nth term is given by T =3+(n−1)×4 =4n−1.
n
Answer:
PaulHancock (WoononaHighSchool) Series T12017 3/23
(A)
ArithmeticandGeometricSeries MultipleChoice
2015 - Question 3
Example 1 (Q3)
Well, the first term is a=3 and d =7−3=11−7=4.
So the nth term is given by T =3+(n−1)×4 =4n−1.
n
Hence the 15th term is T =4(15)−1=59
15
Answer:
PaulHancock (WoononaHighSchool) Series T12017 3/23