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Arithmetic and Geometric Series
HSC Revision DayArithmetic and Geometric Series
Woonona High School
2017
Teacher: Paul HancockSchool: Woonona High SchoolE-mail: paul.hancock5@det.nsw.edu.au
Paul Hancock (Woonona High School) Series T1 2017 1 / 23
Arithmetic and Geometric Series Review
HSC Question Analysis
Let’s have a quick look at the number of marks allocated to the topic for HSCexams in the current format.
Year Marks2012 142013 122014 112015 82016 11
The average number of marks per exam is 11.2.
Paul Hancock (Woonona High School) Series T1 2017 2 / 23
Arithmetic and Geometric Series Review
HSC Question Analysis
Let’s have a quick look at the number of marks allocated to the topic for HSCexams in the current format.
Year Marks2012 142013 122014 112015 82016 11
The average number of marks per exam is 11.2.
Paul Hancock (Woonona High School) Series T1 2017 2 / 23
Arithmetic and Geometric Series Review
HSC Question Analysis
Let’s have a quick look at the number of marks allocated to the topic for HSCexams in the current format.
Year Marks2012 142013 122014 112015 82016 11
The average number of marks per exam is 11.2.
Paul Hancock (Woonona High School) Series T1 2017 2 / 23
Arithmetic and Geometric Series Multiple Choice
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 Tn = 3+ (n − 1)× 4 = 4n − 1.Hence the 15th term is T15 = 4(15)− 1 = 59
Answer:
(A)
Paul Hancock (Woonona High School) Series T1 2017 3 / 23
Arithmetic and Geometric Series Multiple Choice
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 Tn = 3+ (n − 1)× 4 = 4n − 1.Hence the 15th term is T15 = 4(15)− 1 = 59
Answer:
(A)
Paul Hancock (Woonona High School) Series T1 2017 3 / 23
Arithmetic and Geometric Series Multiple Choice
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 Tn = 3+ (n − 1)× 4 = 4n − 1.Hence the 15th term is T15 = 4(15)− 1 = 59
Answer:
(A)
Paul Hancock (Woonona High School) Series T1 2017 3 / 23
Arithmetic and Geometric Series Multiple Choice
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 Tn = 3+ (n − 1)× 4
= 4n − 1.Hence the 15th term is T15 = 4(15)− 1 = 59
Answer:
(A)
Paul Hancock (Woonona High School) Series T1 2017 3 / 23
Arithmetic and Geometric Series Multiple Choice
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 Tn = 3+ (n − 1)× 4 = 4n − 1.
Hence the 15th term is T15 = 4(15)− 1 = 59
Answer:
(A)
Paul Hancock (Woonona High School) Series T1 2017 3 / 23
Arithmetic and Geometric Series Multiple Choice
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 Tn = 3+ (n − 1)× 4 = 4n − 1.Hence the 15th term is T15 = 4(15)− 1 = 59Answer:
(A)
Paul Hancock (Woonona High School) Series T1 2017 3 / 23
Arithmetic and Geometric Series Multiple Choice
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 Tn = 3+ (n − 1)× 4 = 4n − 1.Hence the 15th term is T15 = 4(15)− 1 = 59Answer: (A)
Paul Hancock (Woonona High School) Series T1 2017 3 / 23
Arithmetic and Geometric Series Multiple Choice
2014 - Question 8
Example 2 (Q8)
Well, the first term is a = 3x and r = −6x2
3x = 12x3
−6x2 = −2x .So the nth term is given by Tn = 3x × (−2x)n−1 = 3x × (−2)n−1 × xn−1
= 3(−2)n−1xn.Now 3072 = 3× 1024 and 1024 = 210 , so n − 1 = 10 and n = 11 .This means the solution is either (c) or (d) , but since n − 1 is even we have:
Answer:
(C)
Paul Hancock (Woonona High School) Series T1 2017 4 / 23
Arithmetic and Geometric Series Multiple Choice
2014 - Question 8
Example 2 (Q8)
Well, the first term is a = 3x
and r = −6x2
3x = 12x3
−6x2 = −2x .So the nth term is given by Tn = 3x × (−2x)n−1 = 3x × (−2)n−1 × xn−1
= 3(−2)n−1xn.Now 3072 = 3× 1024 and 1024 = 210 , so n − 1 = 10 and n = 11 .This means the solution is either (c) or (d) , but since n − 1 is even we have:
Answer:
(C)
Paul Hancock (Woonona High School) Series T1 2017 4 / 23
Arithmetic and Geometric Series Multiple Choice
2014 - Question 8
Example 2 (Q8)
Well, the first term is a = 3x and r = −6x2
3x = 12x3
−6x2 = −2x .
So the nth term is given by Tn = 3x × (−2x)n−1 = 3x × (−2)n−1 × xn−1
= 3(−2)n−1xn.Now 3072 = 3× 1024 and 1024 = 210 , so n − 1 = 10 and n = 11 .This means the solution is either (c) or (d) , but since n − 1 is even we have:
Answer:
(C)
Paul Hancock (Woonona High School) Series T1 2017 4 / 23
Arithmetic and Geometric Series Multiple Choice
2014 - Question 8
Example 2 (Q8)
Well, the first term is a = 3x and r = −6x2
3x = 12x3
−6x2 = −2x .So the nth term is given by Tn = 3x × (−2x)n−1
= 3x × (−2)n−1 × xn−1
= 3(−2)n−1xn.Now 3072 = 3× 1024 and 1024 = 210 , so n − 1 = 10 and n = 11 .This means the solution is either (c) or (d) , but since n − 1 is even we have:
Answer:
(C)
Paul Hancock (Woonona High School) Series T1 2017 4 / 23
Arithmetic and Geometric Series Multiple Choice
2014 - Question 8
Example 2 (Q8)
Well, the first term is a = 3x and r = −6x2
3x = 12x3
−6x2 = −2x .So the nth term is given by Tn = 3x × (−2x)n−1 = 3x × (−2)n−1 × xn−1
= 3(−2)n−1xn.Now 3072 = 3× 1024 and 1024 = 210 , so n − 1 = 10 and n = 11 .This means the solution is either (c) or (d) , but since n − 1 is even we have:
Answer:
(C)
Paul Hancock (Woonona High School) Series T1 2017 4 / 23
Arithmetic and Geometric Series Multiple Choice
2014 - Question 8
Example 2 (Q8)
Well, the first term is a = 3x and r = −6x2
3x = 12x3
−6x2 = −2x .So the nth term is given by Tn = 3x × (−2x)n−1 = 3x × (−2)n−1 × xn−1
= 3(−2)n−1xn.
Now 3072 = 3× 1024 and 1024 = 210 , so n − 1 = 10 and n = 11 .This means the solution is either (c) or (d) , but since n − 1 is even we have:
Answer:
(C)
Paul Hancock (Woonona High School) Series T1 2017 4 / 23
Arithmetic and Geometric Series Multiple Choice
2014 - Question 8
Example 2 (Q8)
Well, the first term is a = 3x and r = −6x2
3x = 12x3
−6x2 = −2x .So the nth term is given by Tn = 3x × (−2x)n−1 = 3x × (−2)n−1 × xn−1
= 3(−2)n−1xn.Now 3072 = 3× 1024 and 1024 = 210 , so n − 1 = 10 and n = 11 .
This means the solution is either (c) or (d) , but since n − 1 is even we have:
Answer:
(C)
Paul Hancock (Woonona High School) Series T1 2017 4 / 23
Arithmetic and Geometric Series Multiple Choice
2014 - Question 8
Example 2 (Q8)
Well, the first term is a = 3x and r = −6x2
3x = 12x3
−6x2 = −2x .So the nth term is given by Tn = 3x × (−2x)n−1 = 3x × (−2)n−1 × xn−1
= 3(−2)n−1xn.Now 3072 = 3× 1024 and 1024 = 210 , so n − 1 = 10 and n = 11 .This means the solution is either (c) or (d) , but since n − 1 is even we have:Answer:
(C)
Paul Hancock (Woonona High School) Series T1 2017 4 / 23
Arithmetic and Geometric Series Multiple Choice
2014 - Question 8
Example 2 (Q8)
Well, the first term is a = 3x and r = −6x2
3x = 12x3
−6x2 = −2x .So the nth term is given by Tn = 3x × (−2x)n−1 = 3x × (−2)n−1 × xn−1
= 3(−2)n−1xn.Now 3072 = 3× 1024 and 1024 = 210 , so n − 1 = 10 and n = 11 .This means the solution is either (c) or (d) , but since n − 1 is even we have:Answer: (C)
Paul Hancock (Woonona High School) Series T1 2017 4 / 23
Arithmetic and Geometric Series Standard Questions
2012 - Question 12c
Paul Hancock (Woonona High School) Series T1 2017 5 / 23
Arithmetic and Geometric Series Standard Questions
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 tilesadded to each future row.This forms an arithmetic series, with first term a = 3 and common differenced = 2.We can use the nth term formula to find out how many tiles Jay will use inthe 20th row.
T20 = 3+ (20− 1)× 2= 3+ 19× 2= 3+ 38= 41
Paul Hancock (Woonona High School) Series T1 2017 6 / 23
Arithmetic and Geometric Series Standard Questions
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 tilesadded to each future row.
This forms an arithmetic series, with first term a = 3 and common differenced = 2.We can use the nth term formula to find out how many tiles Jay will use inthe 20th row.
T20 = 3+ (20− 1)× 2= 3+ 19× 2= 3+ 38= 41
Paul Hancock (Woonona High School) Series T1 2017 6 / 23
Arithmetic and Geometric Series Standard Questions
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 tilesadded to each future row.This forms an arithmetic series, with first term a = 3 and common differenced = 2.
We can use the nth term formula to find out how many tiles Jay will use inthe 20th row.
T20 = 3+ (20− 1)× 2= 3+ 19× 2= 3+ 38= 41
Paul Hancock (Woonona High School) Series T1 2017 6 / 23
Arithmetic and Geometric Series Standard Questions
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 tilesadded to each future row.This forms an arithmetic series, with first term a = 3 and common differenced = 2.We can use the nth term formula to find out how many tiles Jay will use inthe 20th row.
T20 = 3+ (20− 1)× 2= 3+ 19× 2= 3+ 38= 41
Paul Hancock (Woonona High School) Series T1 2017 6 / 23
Arithmetic and Geometric Series Standard Questions
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 tilesadded to each future row.This forms an arithmetic series, with first term a = 3 and common differenced = 2.We can use the nth term formula to find out how many tiles Jay will use inthe 20th row.
T20 = 3+ (20− 1)× 2
= 3+ 19× 2= 3+ 38= 41
Paul Hancock (Woonona High School) Series T1 2017 6 / 23
Arithmetic and Geometric Series Standard Questions
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 tilesadded to each future row.This forms an arithmetic series, with first term a = 3 and common differenced = 2.We can use the nth term formula to find out how many tiles Jay will use inthe 20th row.
T20 = 3+ (20− 1)× 2= 3+ 19× 2
= 3+ 38= 41
Paul Hancock (Woonona High School) Series T1 2017 6 / 23
Arithmetic and Geometric Series Standard Questions
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 tilesadded to each future row.This forms an arithmetic series, with first term a = 3 and common differenced = 2.We can use the nth term formula to find out how many tiles Jay will use inthe 20th row.
T20 = 3+ (20− 1)× 2= 3+ 19× 2= 3+ 38
= 41
Paul Hancock (Woonona High School) Series T1 2017 6 / 23
Arithmetic and Geometric Series Standard Questions
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 tilesadded to each future row.This forms an arithmetic series, with first term a = 3 and common differenced = 2.We can use the nth term formula to find out how many tiles Jay will use inthe 20th row.
T20 = 3+ (20− 1)× 2= 3+ 19× 2= 3+ 38= 41
Paul Hancock (Woonona High School) Series T1 2017 6 / 23
Arithmetic and Geometric Series Standard Questions
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 =202[2× 3+ (20− 1)× 2]
= 10[6+ 19× 2]= 10[44]= 440
Paul Hancock (Woonona High School) Series T1 2017 7 / 23
Arithmetic and Geometric Series Standard Questions
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 =202[2× 3+ (20− 1)× 2]
= 10[6+ 19× 2]= 10[44]= 440
Paul Hancock (Woonona High School) Series T1 2017 7 / 23
Arithmetic and Geometric Series Standard Questions
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 =202[2× 3+ (20− 1)× 2]
= 10[6+ 19× 2]= 10[44]= 440
Paul Hancock (Woonona High School) Series T1 2017 7 / 23
Arithmetic and Geometric Series Standard Questions
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 =202[2× 3+ (20− 1)× 2]
= 10[6+ 19× 2]
= 10[44]= 440
Paul Hancock (Woonona High School) Series T1 2017 7 / 23
Arithmetic and Geometric Series Standard Questions
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 =202[2× 3+ (20− 1)× 2]
= 10[6+ 19× 2]= 10[44]
= 440
Paul Hancock (Woonona High School) Series T1 2017 7 / 23
Arithmetic and Geometric Series Standard Questions
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 =202[2× 3+ (20− 1)× 2]
= 10[6+ 19× 2]= 10[44]= 440
Paul Hancock (Woonona High School) Series T1 2017 7 / 23
Arithmetic and Geometric Series Standard Questions
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 rowsJay can make with 200 tiles.
Sn = 200 =n
2[2× 3+ (n − 1)× 2]
400 = n[6+ 2n − 2]
400 = 4n + 2n2
2n2 + 4n − 400 = 0
n2 + 2n − 200 = 0n = 13.77 (Taking positive solution)
∴ Jay can complete 13 rows with 200 tiles.
Paul Hancock (Woonona High School) Series T1 2017 8 / 23
Arithmetic and Geometric Series Standard Questions
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 rowsJay can make with 200 tiles.
Sn = 200 =n
2[2× 3+ (n − 1)× 2]
400 = n[6+ 2n − 2]
400 = 4n + 2n2
2n2 + 4n − 400 = 0
n2 + 2n − 200 = 0n = 13.77 (Taking positive solution)
∴ Jay can complete 13 rows with 200 tiles.
Paul Hancock (Woonona High School) Series T1 2017 8 / 23
Arithmetic and Geometric Series Standard Questions
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 rowsJay can make with 200 tiles.
Sn = 200 =n
2[2× 3+ (n − 1)× 2]
400 = n[6+ 2n − 2]
400 = 4n + 2n2
2n2 + 4n − 400 = 0
n2 + 2n − 200 = 0n = 13.77 (Taking positive solution)
∴ Jay can complete 13 rows with 200 tiles.
Paul Hancock (Woonona High School) Series T1 2017 8 / 23
Arithmetic and Geometric Series Standard Questions
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 rowsJay can make with 200 tiles.
Sn = 200 =n
2[2× 3+ (n − 1)× 2]
400 = n[6+ 2n − 2]
400 = 4n + 2n2
2n2 + 4n − 400 = 0
n2 + 2n − 200 = 0n = 13.77 (Taking positive solution)
∴ Jay can complete 13 rows with 200 tiles.
Paul Hancock (Woonona High School) Series T1 2017 8 / 23
Arithmetic and Geometric Series Standard Questions
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 rowsJay can make with 200 tiles.
Sn = 200 =n
2[2× 3+ (n − 1)× 2]
400 = n[6+ 2n − 2]
400 = 4n + 2n2
2n2 + 4n − 400 = 0
n2 + 2n − 200 = 0n = 13.77 (Taking positive solution)
∴ Jay can complete 13 rows with 200 tiles.
Paul Hancock (Woonona High School) Series T1 2017 8 / 23
Arithmetic and Geometric Series Standard Questions
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 rowsJay can make with 200 tiles.
Sn = 200 =n
2[2× 3+ (n − 1)× 2]
400 = n[6+ 2n − 2]
400 = 4n + 2n2
2n2 + 4n − 400 = 0
n2 + 2n − 200 = 0n = 13.77 (Taking positive solution)
∴ Jay can complete 13 rows with 200 tiles.
Paul Hancock (Woonona High School) Series T1 2017 8 / 23
Arithmetic and Geometric Series Standard Questions
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 rowsJay can make with 200 tiles.
Sn = 200 =n
2[2× 3+ (n − 1)× 2]
400 = n[6+ 2n − 2]
400 = 4n + 2n2
2n2 + 4n − 400 = 0
n2 + 2n − 200 = 0
n = 13.77 (Taking positive solution)
∴ Jay can complete 13 rows with 200 tiles.
Paul Hancock (Woonona High School) Series T1 2017 8 / 23
Arithmetic and Geometric Series Standard Questions
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 rowsJay can make with 200 tiles.
Sn = 200 =n
2[2× 3+ (n − 1)× 2]
400 = n[6+ 2n − 2]
400 = 4n + 2n2
2n2 + 4n − 400 = 0
n2 + 2n − 200 = 0n = 13.77 (Taking positive solution)
∴ Jay can complete 13 rows with 200 tiles.
Paul Hancock (Woonona High School) Series T1 2017 8 / 23
Arithmetic and Geometric Series Standard Questions
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 rowsJay can make with 200 tiles.
Sn = 200 =n
2[2× 3+ (n − 1)× 2]
400 = n[6+ 2n − 2]
400 = 4n + 2n2
2n2 + 4n − 400 = 0
n2 + 2n − 200 = 0n = 13.77 (Taking positive solution)
∴ Jay can complete 13 rows with 200 tiles.
Paul Hancock (Woonona High School) Series T1 2017 8 / 23
Arithmetic and Geometric Series Standard Questions
2013 - Question 13d
Paul Hancock (Woonona High School) Series T1 2017 9 / 23
Arithmetic and Geometric Series Standard Questions
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 = $500000 and r = 112 × 6
100 = 1200 = 0.005. So,
A30 = 500000(1.005)360 −M(1 + 1.005 + 1.0052 + · · ·+ 1.005360−1) = 0
500000(1.005)360 = M(1 + 1.005 + 1.0052 + · · ·+ 1.005359)
500000(1.005)360 = M
[(1.005360 − 1)
1.005 − 1
]3011000.287606 ≈ 1004.515M
M ≈ 2998
Paul Hancock (Woonona High School) Series T1 2017 10 / 23
Arithmetic and Geometric Series Standard Questions
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 = $500000 and r = 112 × 6
100 = 1200 = 0.005.
So,
A30 = 500000(1.005)360 −M(1 + 1.005 + 1.0052 + · · ·+ 1.005360−1) = 0
500000(1.005)360 = M(1 + 1.005 + 1.0052 + · · ·+ 1.005359)
500000(1.005)360 = M
[(1.005360 − 1)
1.005 − 1
]3011000.287606 ≈ 1004.515M
M ≈ 2998
Paul Hancock (Woonona High School) Series T1 2017 10 / 23
Arithmetic and Geometric Series Standard Questions
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 = $500000 and r = 112 × 6
100 = 1200 = 0.005. So,
A30 = 500000(1.005)360 −M(1 + 1.005 + 1.0052 + · · ·+ 1.005360−1) = 0
500000(1.005)360 = M(1 + 1.005 + 1.0052 + · · ·+ 1.005359)
500000(1.005)360 = M
[(1.005360 − 1)
1.005 − 1
]3011000.287606 ≈ 1004.515M
M ≈ 2998
Paul Hancock (Woonona High School) Series T1 2017 10 / 23
Arithmetic and Geometric Series Standard Questions
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 = $500000 and r = 112 × 6
100 = 1200 = 0.005. So,
A30 = 500000(1.005)360 −M(1 + 1.005 + 1.0052 + · · ·+ 1.005360−1) = 0
500000(1.005)360 = M(1 + 1.005 + 1.0052 + · · ·+ 1.005359)
500000(1.005)360 = M
[(1.005360 − 1)
1.005 − 1
]3011000.287606 ≈ 1004.515M
M ≈ 2998
Paul Hancock (Woonona High School) Series T1 2017 10 / 23
Arithmetic and Geometric Series Standard Questions
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 = $500000 and r = 112 × 6
100 = 1200 = 0.005. So,
A30 = 500000(1.005)360 −M(1 + 1.005 + 1.0052 + · · ·+ 1.005360−1) = 0
500000(1.005)360 = M(1 + 1.005 + 1.0052 + · · ·+ 1.005359)
500000(1.005)360 = M
[(1.005360 − 1)
1.005 − 1
]
3011000.287606 ≈ 1004.515M
M ≈ 2998
Paul Hancock (Woonona High School) Series T1 2017 10 / 23
Arithmetic and Geometric Series Standard Questions
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 = $500000 and r = 112 × 6
100 = 1200 = 0.005. So,
A30 = 500000(1.005)360 −M(1 + 1.005 + 1.0052 + · · ·+ 1.005360−1) = 0
500000(1.005)360 = M(1 + 1.005 + 1.0052 + · · ·+ 1.005359)
500000(1.005)360 = M
[(1.005360 − 1)
1.005 − 1
]3011000.287606 ≈ 1004.515M
M ≈ 2998
Paul Hancock (Woonona High School) Series T1 2017 10 / 23
Arithmetic and Geometric Series Standard Questions
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 = $500000 and r = 112 × 6
100 = 1200 = 0.005. So,
A30 = 500000(1.005)360 −M(1 + 1.005 + 1.0052 + · · ·+ 1.005360−1) = 0
500000(1.005)360 = M(1 + 1.005 + 1.0052 + · · ·+ 1.005359)
500000(1.005)360 = M
[(1.005360 − 1)
1.005 − 1
]3011000.287606 ≈ 1004.515M
M ≈ 2998
Paul Hancock (Woonona High School) Series T1 2017 10 / 23
Arithmetic and Geometric Series Standard Questions
2013 - Question 13d
(ii) Show that the balance owing after 20 years is $270000 to the nearestthousand dollars.
A20 ≈ 500000(1.005)240 − 2998[(1.005240 − 1)
1.005 − 1
]≈ 269903
≈ 270000
Paul Hancock (Woonona High School) Series T1 2017 11 / 23
Arithmetic and Geometric Series Standard Questions
2013 - Question 13d
(ii) Show that the balance owing after 20 years is $270000 to the nearestthousand dollars.
A20 ≈ 500000(1.005)240 − 2998[(1.005240 − 1)
1.005 − 1
]
≈ 269903
≈ 270000
Paul Hancock (Woonona High School) Series T1 2017 11 / 23
Arithmetic and Geometric Series Standard Questions
2013 - Question 13d
(ii) Show that the balance owing after 20 years is $270000 to the nearestthousand dollars.
A20 ≈ 500000(1.005)240 − 2998[(1.005240 − 1)
1.005 − 1
]≈ 269903
≈ 270000
Paul Hancock (Woonona High School) Series T1 2017 11 / 23
Arithmetic and Geometric Series Standard Questions
2013 - Question 13d
(ii) Show that the balance owing after 20 years is $270000 to the nearestthousand dollars.
A20 ≈ 500000(1.005)240 − 2998[(1.005240 − 1)
1.005 − 1
]≈ 269903
≈ 270000
Paul Hancock (Woonona High School) Series T1 2017 11 / 23
Arithmetic and Geometric Series Standard Questions
2013 - Question 13d(iii) After 20 years the family borrows an extra amount, so that the family thenowes a total of $370000. The monthly repayment remains $2998 and the interestrate remains the same. How long will it take to repay the $370000?
Now P = $370000 with all other constants unaltered. So,
An = 370000(1.005)n − 2998[(1.005n − 1)1.005 − 1
]= 0
370000(1.005)n − 599600(1.005)n + 599600 = 0
− 299600(1.005)n + 599600 = 0
299600(1.005)n = 599600
(1.005)n = 2.6114981587
ln(1.005)n = ln(2.6114981587)
n ln(1.005) = ln(2.6114981587)
n = 192.46
But this is 193 months! Which is 16 years and 1 month.
Paul Hancock (Woonona High School) Series T1 2017 12 / 23
Arithmetic and Geometric Series Standard Questions
2013 - Question 13d(iii) After 20 years the family borrows an extra amount, so that the family thenowes a total of $370000. The monthly repayment remains $2998 and the interestrate remains the same. How long will it take to repay the $370000?
Now P = $370000 with all other constants unaltered.
So,
An = 370000(1.005)n − 2998[(1.005n − 1)1.005 − 1
]= 0
370000(1.005)n − 599600(1.005)n + 599600 = 0
− 299600(1.005)n + 599600 = 0
299600(1.005)n = 599600
(1.005)n = 2.6114981587
ln(1.005)n = ln(2.6114981587)
n ln(1.005) = ln(2.6114981587)
n = 192.46
But this is 193 months! Which is 16 years and 1 month.
Paul Hancock (Woonona High School) Series T1 2017 12 / 23
Arithmetic and Geometric Series Standard Questions
2013 - Question 13d(iii) After 20 years the family borrows an extra amount, so that the family thenowes a total of $370000. The monthly repayment remains $2998 and the interestrate remains the same. How long will it take to repay the $370000?
Now P = $370000 with all other constants unaltered. So,
An = 370000(1.005)n − 2998[(1.005n − 1)1.005 − 1
]= 0
370000(1.005)n − 599600(1.005)n + 599600 = 0
− 299600(1.005)n + 599600 = 0
299600(1.005)n = 599600
(1.005)n = 2.6114981587
ln(1.005)n = ln(2.6114981587)
n ln(1.005) = ln(2.6114981587)
n = 192.46
But this is 193 months! Which is 16 years and 1 month.
Paul Hancock (Woonona High School) Series T1 2017 12 / 23
Arithmetic and Geometric Series Standard Questions
2013 - Question 13d(iii) After 20 years the family borrows an extra amount, so that the family thenowes a total of $370000. The monthly repayment remains $2998 and the interestrate remains the same. How long will it take to repay the $370000?
Now P = $370000 with all other constants unaltered. So,
An = 370000(1.005)n − 2998[(1.005n − 1)1.005 − 1
]= 0
370000(1.005)n − 599600(1.005)n + 599600 = 0
− 299600(1.005)n + 599600 = 0
299600(1.005)n = 599600
(1.005)n = 2.6114981587
ln(1.005)n = ln(2.6114981587)
n ln(1.005) = ln(2.6114981587)
n = 192.46
But this is 193 months! Which is 16 years and 1 month.
Paul Hancock (Woonona High School) Series T1 2017 12 / 23
Arithmetic and Geometric Series Standard Questions
2013 - Question 13d(iii) After 20 years the family borrows an extra amount, so that the family thenowes a total of $370000. The monthly repayment remains $2998 and the interestrate remains the same. How long will it take to repay the $370000?
Now P = $370000 with all other constants unaltered. So,
An = 370000(1.005)n − 2998[(1.005n − 1)1.005 − 1
]= 0
370000(1.005)n − 599600(1.005)n + 599600 = 0
− 299600(1.005)n + 599600 = 0
299600(1.005)n = 599600
(1.005)n = 2.6114981587
ln(1.005)n = ln(2.6114981587)
n ln(1.005) = ln(2.6114981587)
n = 192.46
But this is 193 months! Which is 16 years and 1 month.
Paul Hancock (Woonona High School) Series T1 2017 12 / 23
Arithmetic and Geometric Series Standard Questions
2013 - Question 13d(iii) After 20 years the family borrows an extra amount, so that the family thenowes a total of $370000. The monthly repayment remains $2998 and the interestrate remains the same. How long will it take to repay the $370000?
Now P = $370000 with all other constants unaltered. So,
An = 370000(1.005)n − 2998[(1.005n − 1)1.005 − 1
]= 0
370000(1.005)n − 599600(1.005)n + 599600 = 0
− 299600(1.005)n + 599600 = 0
299600(1.005)n = 599600
(1.005)n = 2.6114981587
ln(1.005)n = ln(2.6114981587)
n ln(1.005) = ln(2.6114981587)
n = 192.46
But this is 193 months! Which is 16 years and 1 month.
Paul Hancock (Woonona High School) Series T1 2017 12 / 23
Arithmetic and Geometric Series Standard Questions
2013 - Question 13d(iii) After 20 years the family borrows an extra amount, so that the family thenowes a total of $370000. The monthly repayment remains $2998 and the interestrate remains the same. How long will it take to repay the $370000?
Now P = $370000 with all other constants unaltered. So,
An = 370000(1.005)n − 2998[(1.005n − 1)1.005 − 1
]= 0
370000(1.005)n − 599600(1.005)n + 599600 = 0
− 299600(1.005)n + 599600 = 0
299600(1.005)n = 599600
(1.005)n = 2.6114981587
ln(1.005)n = ln(2.6114981587)
n ln(1.005) = ln(2.6114981587)
n = 192.46
But this is 193 months! Which is 16 years and 1 month.
Paul Hancock (Woonona High School) Series T1 2017 12 / 23
Arithmetic and Geometric Series Standard Questions
2013 - Question 13d(iii) After 20 years the family borrows an extra amount, so that the family thenowes a total of $370000. The monthly repayment remains $2998 and the interestrate remains the same. How long will it take to repay the $370000?
Now P = $370000 with all other constants unaltered. So,
An = 370000(1.005)n − 2998[(1.005n − 1)1.005 − 1
]= 0
370000(1.005)n − 599600(1.005)n + 599600 = 0
− 299600(1.005)n + 599600 = 0
299600(1.005)n = 599600
(1.005)n = 2.6114981587
ln(1.005)n = ln(2.6114981587)
n ln(1.005) = ln(2.6114981587)
n = 192.46
But this is 193 months! Which is 16 years and 1 month.
Paul Hancock (Woonona High School) Series T1 2017 12 / 23
Arithmetic and Geometric Series Standard Questions
2013 - Question 13d(iii) After 20 years the family borrows an extra amount, so that the family thenowes a total of $370000. The monthly repayment remains $2998 and the interestrate remains the same. How long will it take to repay the $370000?
Now P = $370000 with all other constants unaltered. So,
An = 370000(1.005)n − 2998[(1.005n − 1)1.005 − 1
]= 0
370000(1.005)n − 599600(1.005)n + 599600 = 0
− 299600(1.005)n + 599600 = 0
299600(1.005)n = 599600
(1.005)n = 2.6114981587
ln(1.005)n = ln(2.6114981587)
n ln(1.005) = ln(2.6114981587)
n = 192.46
But this is 193 months!
Which is 16 years and 1 month.
Paul Hancock (Woonona High School) Series T1 2017 12 / 23
Arithmetic and Geometric Series Standard Questions
2013 - Question 13d(iii) After 20 years the family borrows an extra amount, so that the family thenowes a total of $370000. The monthly repayment remains $2998 and the interestrate remains the same. How long will it take to repay the $370000?
Now P = $370000 with all other constants unaltered. So,
An = 370000(1.005)n − 2998[(1.005n − 1)1.005 − 1
]= 0
370000(1.005)n − 599600(1.005)n + 599600 = 0
− 299600(1.005)n + 599600 = 0
299600(1.005)n = 599600
(1.005)n = 2.6114981587
ln(1.005)n = ln(2.6114981587)
n ln(1.005) = ln(2.6114981587)
n = 192.46
But this is 193 months! Which is 16 years and 1 month.Paul Hancock (Woonona High School) Series T1 2017 12 / 23
Arithmetic and Geometric Series Standard Questions
2014 - Question 14d
Paul Hancock (Woonona High School) Series T1 2017 13 / 23
Arithmetic and Geometric Series Standard Questions
2014 - Question 14d
(i) How much of the drug is in the patient’s body immediately after the seconddose is given?
The initial amount, which we will call A1 was 10 mL. The amount immediatelyafter the second dose would be 10mL plus 1
3 the initial amount.
So A2 = 10+ 10( 1
3
)= 40
3 = 13 13
∴ the amount immediately after the second dose was 13 13 mL.
Paul Hancock (Woonona High School) Series T1 2017 14 / 23
Arithmetic and Geometric Series Standard Questions
2014 - Question 14d
(i) How much of the drug is in the patient’s body immediately after the seconddose is given?
The initial amount, which we will call A1 was 10 mL.
The amount immediatelyafter the second dose would be 10mL plus 1
3 the initial amount.
So A2 = 10+ 10( 1
3
)= 40
3 = 13 13
∴ the amount immediately after the second dose was 13 13 mL.
Paul Hancock (Woonona High School) Series T1 2017 14 / 23
Arithmetic and Geometric Series Standard Questions
2014 - Question 14d
(i) How much of the drug is in the patient’s body immediately after the seconddose is given?
The initial amount, which we will call A1 was 10 mL. The amount immediatelyafter the second dose would be 10mL plus 1
3 the initial amount.
So A2 = 10+ 10( 1
3
)= 40
3 = 13 13
∴ the amount immediately after the second dose was 13 13 mL.
Paul Hancock (Woonona High School) Series T1 2017 14 / 23
Arithmetic and Geometric Series Standard Questions
2014 - Question 14d
(i) How much of the drug is in the patient’s body immediately after the seconddose is given?
The initial amount, which we will call A1 was 10 mL. The amount immediatelyafter the second dose would be 10mL plus 1
3 the initial amount.
So A2 = 10+ 10( 1
3
)= 40
3 = 13 13
∴ the amount immediately after the second dose was 13 13 mL.
Paul Hancock (Woonona High School) Series T1 2017 14 / 23
Arithmetic and Geometric Series Standard Questions
2014 - Question 14d
(i) How much of the drug is in the patient’s body immediately after the seconddose is given?
The initial amount, which we will call A1 was 10 mL. The amount immediatelyafter the second dose would be 10mL plus 1
3 the initial amount.
So A2 = 10+ 10( 1
3
)= 40
3 = 13 13
∴ the amount immediately after the second dose was 13 13 mL.
Paul Hancock (Woonona High School) Series T1 2017 14 / 23
Arithmetic and Geometric Series Standard Questions
2014 - Question 14d(ii) Show that the total amount of the drug in the patient’s body never exceeds 15mL.
A1 = 10
A2 = 10+ 10(13
)A3 = 10+
(10+ 10
(13
))(13
)= 10+ 10
(13
)+ 10
(13
)2
...
An = 10
[1+
13+
(13
)2
+ · · ·+(13
)n]
= 10
[1
1− 13
]= 10
[123
]= 10
[32
]= 15
Paul Hancock (Woonona High School) Series T1 2017 15 / 23
Arithmetic and Geometric Series Standard Questions
2014 - Question 14d(ii) Show that the total amount of the drug in the patient’s body never exceeds 15mL.
A1 = 10
A2 = 10+ 10(13
)A3 = 10+
(10+ 10
(13
))(13
)= 10+ 10
(13
)+ 10
(13
)2
...
An = 10
[1+
13+
(13
)2
+ · · ·+(13
)n]
= 10
[1
1− 13
]= 10
[123
]= 10
[32
]= 15
Paul Hancock (Woonona High School) Series T1 2017 15 / 23
Arithmetic and Geometric Series Standard Questions
2014 - Question 14d(ii) Show that the total amount of the drug in the patient’s body never exceeds 15mL.
A1 = 10
A2 = 10+ 10(13
)
A3 = 10+(10+ 10
(13
))(13
)= 10+ 10
(13
)+ 10
(13
)2
...
An = 10
[1+
13+
(13
)2
+ · · ·+(13
)n]
= 10
[1
1− 13
]= 10
[123
]= 10
[32
]= 15
Paul Hancock (Woonona High School) Series T1 2017 15 / 23
Arithmetic and Geometric Series Standard Questions
2014 - Question 14d(ii) Show that the total amount of the drug in the patient’s body never exceeds 15mL.
A1 = 10
A2 = 10+ 10(13
)A3 = 10+
(10+ 10
(13
))(13
)= 10+ 10
(13
)+ 10
(13
)2
...
An = 10
[1+
13+
(13
)2
+ · · ·+(13
)n]
= 10
[1
1− 13
]= 10
[123
]= 10
[32
]= 15
Paul Hancock (Woonona High School) Series T1 2017 15 / 23
Arithmetic and Geometric Series Standard Questions
2014 - Question 14d(ii) Show that the total amount of the drug in the patient’s body never exceeds 15mL.
A1 = 10
A2 = 10+ 10(13
)A3 = 10+
(10+ 10
(13
))(13
)= 10+ 10
(13
)+ 10
(13
)2
...
An = 10
[1+
13+
(13
)2
+ · · ·+(13
)n]
= 10
[1
1− 13
]= 10
[123
]= 10
[32
]= 15
Paul Hancock (Woonona High School) Series T1 2017 15 / 23
Arithmetic and Geometric Series Standard Questions
2014 - Question 14d(ii) Show that the total amount of the drug in the patient’s body never exceeds 15mL.
A1 = 10
A2 = 10+ 10(13
)A3 = 10+
(10+ 10
(13
))(13
)= 10+ 10
(13
)+ 10
(13
)2
...
An = 10
[1+
13+
(13
)2
+ · · ·+(13
)n]
= 10
[1
1− 13
]= 10
[123
]= 10
[32
]= 15
Paul Hancock (Woonona High School) Series T1 2017 15 / 23
Arithmetic and Geometric Series Standard Questions
2014 - Question 14d(ii) Show that the total amount of the drug in the patient’s body never exceeds 15mL.
A1 = 10
A2 = 10+ 10(13
)A3 = 10+
(10+ 10
(13
))(13
)= 10+ 10
(13
)+ 10
(13
)2
...
An = 10
[1+
13+
(13
)2
+ · · ·+(13
)n]
= 10
[1
1− 13
]
= 10
[123
]= 10
[32
]= 15
Paul Hancock (Woonona High School) Series T1 2017 15 / 23
Arithmetic and Geometric Series Standard Questions
2014 - Question 14d(ii) Show that the total amount of the drug in the patient’s body never exceeds 15mL.
A1 = 10
A2 = 10+ 10(13
)A3 = 10+
(10+ 10
(13
))(13
)= 10+ 10
(13
)+ 10
(13
)2
...
An = 10
[1+
13+
(13
)2
+ · · ·+(13
)n]
= 10
[1
1− 13
]= 10
[123
]
= 10[32
]= 15
Paul Hancock (Woonona High School) Series T1 2017 15 / 23
Arithmetic and Geometric Series Standard Questions
2014 - Question 14d(ii) Show that the total amount of the drug in the patient’s body never exceeds 15mL.
A1 = 10
A2 = 10+ 10(13
)A3 = 10+
(10+ 10
(13
))(13
)= 10+ 10
(13
)+ 10
(13
)2
...
An = 10
[1+
13+
(13
)2
+ · · ·+(13
)n]
= 10
[1
1− 13
]= 10
[123
]= 10
[32
]
= 15
Paul Hancock (Woonona High School) Series T1 2017 15 / 23
Arithmetic and Geometric Series Standard Questions
2014 - Question 14d(ii) Show that the total amount of the drug in the patient’s body never exceeds 15mL.
A1 = 10
A2 = 10+ 10(13
)A3 = 10+
(10+ 10
(13
))(13
)= 10+ 10
(13
)+ 10
(13
)2
...
An = 10
[1+
13+
(13
)2
+ · · ·+(13
)n]
= 10
[1
1− 13
]= 10
[123
]= 10
[32
]= 15
Paul Hancock (Woonona High School) Series T1 2017 15 / 23
Arithmetic and Geometric Series Harder Questions
2012 - Question 15a
Paul Hancock (Woonona High School) Series T1 2017 16 / 23
Arithmetic and Geometric Series Harder Questions
2012 - Question 15a
(i) Find the length of the strip required to make the first ten rectangles.
L = 10+ 10× 0.96+ 10× 0.962 + · · ·+ 10× 0.969
= 10(1+ 0.96+ 0.962 + · · ·+ 0.969)
The bracketed term is a geometric series with 10 terms, a = 1and r = 0.96.
= 10(1(1− 0.9610)
1− 0.96
)≈ 84
Paul Hancock (Woonona High School) Series T1 2017 17 / 23
Arithmetic and Geometric Series Harder Questions
2012 - Question 15a
(i) Find the length of the strip required to make the first ten rectangles.
L = 10+ 10× 0.96+ 10× 0.962 + · · ·+ 10× 0.969
= 10(1+ 0.96+ 0.962 + · · ·+ 0.969)
The bracketed term is a geometric series with 10 terms, a = 1and r = 0.96.
= 10(1(1− 0.9610)
1− 0.96
)≈ 84
Paul Hancock (Woonona High School) Series T1 2017 17 / 23
Arithmetic and Geometric Series Harder Questions
2012 - Question 15a
(i) Find the length of the strip required to make the first ten rectangles.
L = 10+ 10× 0.96+ 10× 0.962 + · · ·+ 10× 0.969
= 10(1+ 0.96+ 0.962 + · · ·+ 0.969)
The bracketed term is a geometric series with 10 terms, a = 1and r = 0.96.
= 10(1(1− 0.9610)
1− 0.96
)≈ 84
Paul Hancock (Woonona High School) Series T1 2017 17 / 23
Arithmetic and Geometric Series Harder Questions
2012 - Question 15a
(i) Find the length of the strip required to make the first ten rectangles.
L = 10+ 10× 0.96+ 10× 0.962 + · · ·+ 10× 0.969
= 10(1+ 0.96+ 0.962 + · · ·+ 0.969)
The bracketed term is a geometric series with 10 terms, a = 1and r = 0.96.
= 10(1(1− 0.9610)
1− 0.96
)≈ 84
Paul Hancock (Woonona High School) Series T1 2017 17 / 23
Arithmetic and Geometric Series Harder Questions
2012 - Question 15a
(i) Find the length of the strip required to make the first ten rectangles.
L = 10+ 10× 0.96+ 10× 0.962 + · · ·+ 10× 0.969
= 10(1+ 0.96+ 0.962 + · · ·+ 0.969)
The bracketed term is a geometric series with 10 terms, a = 1and r = 0.96.
= 10(1(1− 0.9610)
1− 0.96
)
≈ 84
Paul Hancock (Woonona High School) Series T1 2017 17 / 23
Arithmetic and Geometric Series Harder Questions
2012 - Question 15a
(i) Find the length of the strip required to make the first ten rectangles.
L = 10+ 10× 0.96+ 10× 0.962 + · · ·+ 10× 0.969
= 10(1+ 0.96+ 0.962 + · · ·+ 0.969)
The bracketed term is a geometric series with 10 terms, a = 1and r = 0.96.
= 10(1(1− 0.9610)
1− 0.96
)≈ 84
Paul Hancock (Woonona High School) Series T1 2017 17 / 23
Arithmetic and Geometric Series Harder Questions
2012 - Question 15a
(ii) Explain why a strip of length 3 m is sufficient to make any number ofrectangles.
Considering just the geometric series from (i), we note that r = 0.96 < 1. Hencethere is a limiting sum. Let that sum be S∞ .
S∞ =1
1− 0.96S∞ = 25
Hence we can now determine the limiting value of L.L∞ = 10(25) = 250
Now since 250 < 300, a 3 m strip is more than sufficient.
Paul Hancock (Woonona High School) Series T1 2017 18 / 23
Arithmetic and Geometric Series Harder Questions
2012 - Question 15a
(ii) Explain why a strip of length 3 m is sufficient to make any number ofrectangles.
Considering just the geometric series from (i), we note that r = 0.96 < 1. Hencethere is a limiting sum. Let that sum be S∞ .
S∞ =1
1− 0.96S∞ = 25
Hence we can now determine the limiting value of L.L∞ = 10(25) = 250
Now since 250 < 300, a 3 m strip is more than sufficient.
Paul Hancock (Woonona High School) Series T1 2017 18 / 23
Arithmetic and Geometric Series Harder Questions
2012 - Question 15a
(ii) Explain why a strip of length 3 m is sufficient to make any number ofrectangles.
Considering just the geometric series from (i), we note that r = 0.96 < 1. Hencethere is a limiting sum. Let that sum be S∞ .
S∞ =1
1− 0.96
S∞ = 25
Hence we can now determine the limiting value of L.L∞ = 10(25) = 250
Now since 250 < 300, a 3 m strip is more than sufficient.
Paul Hancock (Woonona High School) Series T1 2017 18 / 23
Arithmetic and Geometric Series Harder Questions
2012 - Question 15a
(ii) Explain why a strip of length 3 m is sufficient to make any number ofrectangles.
Considering just the geometric series from (i), we note that r = 0.96 < 1. Hencethere is a limiting sum. Let that sum be S∞ .
S∞ =1
1− 0.96S∞ = 25
Hence we can now determine the limiting value of L.L∞ = 10(25) = 250
Now since 250 < 300, a 3 m strip is more than sufficient.
Paul Hancock (Woonona High School) Series T1 2017 18 / 23
Arithmetic and Geometric Series Harder Questions
2012 - Question 15a
(ii) Explain why a strip of length 3 m is sufficient to make any number ofrectangles.
Considering just the geometric series from (i), we note that r = 0.96 < 1. Hencethere is a limiting sum. Let that sum be S∞ .
S∞ =1
1− 0.96S∞ = 25
Hence we can now determine the limiting value of L.
L∞ = 10(25) = 250
Now since 250 < 300, a 3 m strip is more than sufficient.
Paul Hancock (Woonona High School) Series T1 2017 18 / 23
Arithmetic and Geometric Series Harder Questions
2012 - Question 15a
(ii) Explain why a strip of length 3 m is sufficient to make any number ofrectangles.
Considering just the geometric series from (i), we note that r = 0.96 < 1. Hencethere is a limiting sum. Let that sum be S∞ .
S∞ =1
1− 0.96S∞ = 25
Hence we can now determine the limiting value of L.L∞ = 10(25) = 250
Now since 250 < 300, a 3 m strip is more than sufficient.
Paul Hancock (Woonona High School) Series T1 2017 18 / 23
Arithmetic and Geometric Series Harder Questions
2012 - Question 15a
(ii) Explain why a strip of length 3 m is sufficient to make any number ofrectangles.
Considering just the geometric series from (i), we note that r = 0.96 < 1. Hencethere is a limiting sum. Let that sum be S∞ .
S∞ =1
1− 0.96S∞ = 25
Hence we can now determine the limiting value of L.L∞ = 10(25) = 250
Now since 250 < 300, a 3 m strip is more than sufficient.
Paul Hancock (Woonona High School) Series T1 2017 18 / 23
Arithmetic and Geometric Series Harder Questions
2014 - Question 16b
Paul Hancock (Woonona High School) Series T1 2017 19 / 23
Arithmetic and Geometric Series Harder Questions
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) (deposit added) 5001st month (end) (interest added) 500(1.003)2nd month (beginning) (deposit added) 500(1.003) + 500(1.01)2nd month (end) (interest added) (500(1.003) + 500(1.01)) (1.003)
Now (500(1.003) + 500(1.01)) (1.003) = 500(1.003)2 + 500(1.01)(1.003), so thebalance of the account at the end of the second month was$500(1.003)2 + $500(1.01)(1.003).
Paul Hancock (Woonona High School) Series T1 2017 20 / 23
Arithmetic and Geometric Series Harder Questions
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)
(deposit added) 5001st month (end) (interest added) 500(1.003)2nd month (beginning) (deposit added) 500(1.003) + 500(1.01)2nd month (end) (interest added) (500(1.003) + 500(1.01)) (1.003)
Now (500(1.003) + 500(1.01)) (1.003) = 500(1.003)2 + 500(1.01)(1.003), so thebalance of the account at the end of the second month was$500(1.003)2 + $500(1.01)(1.003).
Paul Hancock (Woonona High School) Series T1 2017 20 / 23
Arithmetic and Geometric Series Harder Questions
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) (deposit added)
5001st month (end) (interest added) 500(1.003)2nd month (beginning) (deposit added) 500(1.003) + 500(1.01)2nd month (end) (interest added) (500(1.003) + 500(1.01)) (1.003)
Now (500(1.003) + 500(1.01)) (1.003) = 500(1.003)2 + 500(1.01)(1.003), so thebalance of the account at the end of the second month was$500(1.003)2 + $500(1.01)(1.003).
Paul Hancock (Woonona High School) Series T1 2017 20 / 23
Arithmetic and Geometric Series Harder Questions
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) (deposit added) 500
1st month (end) (interest added) 500(1.003)2nd month (beginning) (deposit added) 500(1.003) + 500(1.01)2nd month (end) (interest added) (500(1.003) + 500(1.01)) (1.003)
Now (500(1.003) + 500(1.01)) (1.003) = 500(1.003)2 + 500(1.01)(1.003), so thebalance of the account at the end of the second month was$500(1.003)2 + $500(1.01)(1.003).
Paul Hancock (Woonona High School) Series T1 2017 20 / 23
Arithmetic and Geometric Series Harder Questions
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) (deposit added) 5001st month (end)
(interest added) 500(1.003)2nd month (beginning) (deposit added) 500(1.003) + 500(1.01)2nd month (end) (interest added) (500(1.003) + 500(1.01)) (1.003)
Now (500(1.003) + 500(1.01)) (1.003) = 500(1.003)2 + 500(1.01)(1.003), so thebalance of the account at the end of the second month was$500(1.003)2 + $500(1.01)(1.003).
Paul Hancock (Woonona High School) Series T1 2017 20 / 23
Arithmetic and Geometric Series Harder Questions
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) (deposit added) 5001st month (end) (interest added)
500(1.003)2nd month (beginning) (deposit added) 500(1.003) + 500(1.01)2nd month (end) (interest added) (500(1.003) + 500(1.01)) (1.003)
Now (500(1.003) + 500(1.01)) (1.003) = 500(1.003)2 + 500(1.01)(1.003), so thebalance of the account at the end of the second month was$500(1.003)2 + $500(1.01)(1.003).
Paul Hancock (Woonona High School) Series T1 2017 20 / 23
Arithmetic and Geometric Series Harder Questions
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) (deposit added) 5001st month (end) (interest added) 500(1.003)
2nd month (beginning) (deposit added) 500(1.003) + 500(1.01)2nd month (end) (interest added) (500(1.003) + 500(1.01)) (1.003)
Now (500(1.003) + 500(1.01)) (1.003) = 500(1.003)2 + 500(1.01)(1.003), so thebalance of the account at the end of the second month was$500(1.003)2 + $500(1.01)(1.003).
Paul Hancock (Woonona High School) Series T1 2017 20 / 23
Arithmetic and Geometric Series Harder Questions
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) (deposit added) 5001st month (end) (interest added) 500(1.003)2nd month (beginning)
(deposit added) 500(1.003) + 500(1.01)2nd month (end) (interest added) (500(1.003) + 500(1.01)) (1.003)
Now (500(1.003) + 500(1.01)) (1.003) = 500(1.003)2 + 500(1.01)(1.003), so thebalance of the account at the end of the second month was$500(1.003)2 + $500(1.01)(1.003).
Paul Hancock (Woonona High School) Series T1 2017 20 / 23
Arithmetic and Geometric Series Harder Questions
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) (deposit added) 5001st month (end) (interest added) 500(1.003)2nd month (beginning) (deposit added)
500(1.003) + 500(1.01)2nd month (end) (interest added) (500(1.003) + 500(1.01)) (1.003)
Now (500(1.003) + 500(1.01)) (1.003) = 500(1.003)2 + 500(1.01)(1.003), so thebalance of the account at the end of the second month was$500(1.003)2 + $500(1.01)(1.003).
Paul Hancock (Woonona High School) Series T1 2017 20 / 23
Arithmetic and Geometric Series Harder Questions
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) (deposit added) 5001st month (end) (interest added) 500(1.003)2nd month (beginning) (deposit added) 500(1.003) + 500(1.01)
2nd month (end) (interest added) (500(1.003) + 500(1.01)) (1.003)
Now (500(1.003) + 500(1.01)) (1.003) = 500(1.003)2 + 500(1.01)(1.003), so thebalance of the account at the end of the second month was$500(1.003)2 + $500(1.01)(1.003).
Paul Hancock (Woonona High School) Series T1 2017 20 / 23
Arithmetic and Geometric Series Harder Questions
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) (deposit added) 5001st month (end) (interest added) 500(1.003)2nd month (beginning) (deposit added) 500(1.003) + 500(1.01)2nd month (end)
(interest added) (500(1.003) + 500(1.01)) (1.003)
Now (500(1.003) + 500(1.01)) (1.003) = 500(1.003)2 + 500(1.01)(1.003), so thebalance of the account at the end of the second month was$500(1.003)2 + $500(1.01)(1.003).
Paul Hancock (Woonona High School) Series T1 2017 20 / 23
Arithmetic and Geometric Series Harder Questions
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) (deposit added) 5001st month (end) (interest added) 500(1.003)2nd month (beginning) (deposit added) 500(1.003) + 500(1.01)2nd month (end) (interest added)
(500(1.003) + 500(1.01)) (1.003)
Now (500(1.003) + 500(1.01)) (1.003) = 500(1.003)2 + 500(1.01)(1.003), so thebalance of the account at the end of the second month was$500(1.003)2 + $500(1.01)(1.003).
Paul Hancock (Woonona High School) Series T1 2017 20 / 23
Arithmetic and Geometric Series Harder Questions
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) (deposit added) 5001st month (end) (interest added) 500(1.003)2nd month (beginning) (deposit added) 500(1.003) + 500(1.01)2nd month (end) (interest added) (500(1.003) + 500(1.01)) (1.003)
Now (500(1.003) + 500(1.01)) (1.003) = 500(1.003)2 + 500(1.01)(1.003), so thebalance of the account at the end of the second month was$500(1.003)2 + $500(1.01)(1.003).
Paul Hancock (Woonona High School) Series T1 2017 20 / 23
Arithmetic and Geometric Series Harder Questions
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) (deposit added) 5001st month (end) (interest added) 500(1.003)2nd month (beginning) (deposit added) 500(1.003) + 500(1.01)2nd month (end) (interest added) (500(1.003) + 500(1.01)) (1.003)
Now (500(1.003) + 500(1.01)) (1.003) = 500(1.003)2 + 500(1.01)(1.003), so thebalance of the account at the end of the second month was$500(1.003)2 + $500(1.01)(1.003).
Paul Hancock (Woonona High School) Series T1 2017 20 / 23
Arithmetic and Geometric Series Harder Questions
2014 - Question 16b(ii) Find the balance of the account at the end of the 60th month, correct to thenearest dollar.
Continuing on from (i):
3rd month (beginning) 500(1.003)2 + 500(1.01)(1.003) + 500(1.01)2
3rd month (end) 500(1.003)3 + 500(1.01)(1.003)2 + 500(1.01)2(1.003)...60th month (end) 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
Let B be the balance owing at the end of the 60th month. Then:
B = 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
= 500(1.003)60 + 500(1.003)60 1.011.003
+ · · ·+ 500(1.003)60(
1.011.003
)59
= 500(1.003)60
(1 +
1.011.003
+ · · ·+(
1.011.003
)59)
Paul Hancock (Woonona High School) Series T1 2017 21 / 23
Arithmetic and Geometric Series Harder Questions
2014 - Question 16b(ii) Find the balance of the account at the end of the 60th month, correct to thenearest dollar.
Continuing on from (i):
3rd month (beginning) 500(1.003)2 + 500(1.01)(1.003) + 500(1.01)2
3rd month (end) 500(1.003)3 + 500(1.01)(1.003)2 + 500(1.01)2(1.003)...60th month (end) 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
Let B be the balance owing at the end of the 60th month. Then:
B = 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
= 500(1.003)60 + 500(1.003)60 1.011.003
+ · · ·+ 500(1.003)60(
1.011.003
)59
= 500(1.003)60
(1 +
1.011.003
+ · · ·+(
1.011.003
)59)
Paul Hancock (Woonona High School) Series T1 2017 21 / 23
Arithmetic and Geometric Series Harder Questions
2014 - Question 16b(ii) Find the balance of the account at the end of the 60th month, correct to thenearest dollar.
Continuing on from (i):
3rd month (beginning)
500(1.003)2 + 500(1.01)(1.003) + 500(1.01)2
3rd month (end) 500(1.003)3 + 500(1.01)(1.003)2 + 500(1.01)2(1.003)...60th month (end) 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
Let B be the balance owing at the end of the 60th month. Then:
B = 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
= 500(1.003)60 + 500(1.003)60 1.011.003
+ · · ·+ 500(1.003)60(
1.011.003
)59
= 500(1.003)60
(1 +
1.011.003
+ · · ·+(
1.011.003
)59)
Paul Hancock (Woonona High School) Series T1 2017 21 / 23
Arithmetic and Geometric Series Harder Questions
2014 - Question 16b(ii) Find the balance of the account at the end of the 60th month, correct to thenearest dollar.
Continuing on from (i):
3rd month (beginning) 500(1.003)2 + 500(1.01)(1.003) + 500(1.01)2
3rd month (end) 500(1.003)3 + 500(1.01)(1.003)2 + 500(1.01)2(1.003)...60th month (end) 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
Let B be the balance owing at the end of the 60th month. Then:
B = 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
= 500(1.003)60 + 500(1.003)60 1.011.003
+ · · ·+ 500(1.003)60(
1.011.003
)59
= 500(1.003)60
(1 +
1.011.003
+ · · ·+(
1.011.003
)59)
Paul Hancock (Woonona High School) Series T1 2017 21 / 23
Arithmetic and Geometric Series Harder Questions
2014 - Question 16b(ii) Find the balance of the account at the end of the 60th month, correct to thenearest dollar.
Continuing on from (i):
3rd month (beginning) 500(1.003)2 + 500(1.01)(1.003) + 500(1.01)2
3rd month (end)
500(1.003)3 + 500(1.01)(1.003)2 + 500(1.01)2(1.003)...60th month (end) 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
Let B be the balance owing at the end of the 60th month. Then:
B = 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
= 500(1.003)60 + 500(1.003)60 1.011.003
+ · · ·+ 500(1.003)60(
1.011.003
)59
= 500(1.003)60
(1 +
1.011.003
+ · · ·+(
1.011.003
)59)
Paul Hancock (Woonona High School) Series T1 2017 21 / 23
Arithmetic and Geometric Series Harder Questions
2014 - Question 16b(ii) Find the balance of the account at the end of the 60th month, correct to thenearest dollar.
Continuing on from (i):
3rd month (beginning) 500(1.003)2 + 500(1.01)(1.003) + 500(1.01)2
3rd month (end) 500(1.003)3 + 500(1.01)(1.003)2 + 500(1.01)2(1.003)
...60th month (end) 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
Let B be the balance owing at the end of the 60th month. Then:
B = 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
= 500(1.003)60 + 500(1.003)60 1.011.003
+ · · ·+ 500(1.003)60(
1.011.003
)59
= 500(1.003)60
(1 +
1.011.003
+ · · ·+(
1.011.003
)59)
Paul Hancock (Woonona High School) Series T1 2017 21 / 23
Arithmetic and Geometric Series Harder Questions
2014 - Question 16b(ii) Find the balance of the account at the end of the 60th month, correct to thenearest dollar.
Continuing on from (i):
3rd month (beginning) 500(1.003)2 + 500(1.01)(1.003) + 500(1.01)2
3rd month (end) 500(1.003)3 + 500(1.01)(1.003)2 + 500(1.01)2(1.003)...
60th month (end) 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
Let B be the balance owing at the end of the 60th month. Then:
B = 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
= 500(1.003)60 + 500(1.003)60 1.011.003
+ · · ·+ 500(1.003)60(
1.011.003
)59
= 500(1.003)60
(1 +
1.011.003
+ · · ·+(
1.011.003
)59)
Paul Hancock (Woonona High School) Series T1 2017 21 / 23
Arithmetic and Geometric Series Harder Questions
2014 - Question 16b(ii) Find the balance of the account at the end of the 60th month, correct to thenearest dollar.
Continuing on from (i):
3rd month (beginning) 500(1.003)2 + 500(1.01)(1.003) + 500(1.01)2
3rd month (end) 500(1.003)3 + 500(1.01)(1.003)2 + 500(1.01)2(1.003)...60th month (end)
500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
Let B be the balance owing at the end of the 60th month. Then:
B = 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
= 500(1.003)60 + 500(1.003)60 1.011.003
+ · · ·+ 500(1.003)60(
1.011.003
)59
= 500(1.003)60
(1 +
1.011.003
+ · · ·+(
1.011.003
)59)
Paul Hancock (Woonona High School) Series T1 2017 21 / 23
Arithmetic and Geometric Series Harder Questions
2014 - Question 16b(ii) Find the balance of the account at the end of the 60th month, correct to thenearest dollar.
Continuing on from (i):
3rd month (beginning) 500(1.003)2 + 500(1.01)(1.003) + 500(1.01)2
3rd month (end) 500(1.003)3 + 500(1.01)(1.003)2 + 500(1.01)2(1.003)...60th month (end) 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
Let B be the balance owing at the end of the 60th month. Then:
B = 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
= 500(1.003)60 + 500(1.003)60 1.011.003
+ · · ·+ 500(1.003)60(
1.011.003
)59
= 500(1.003)60
(1 +
1.011.003
+ · · ·+(
1.011.003
)59)
Paul Hancock (Woonona High School) Series T1 2017 21 / 23
Arithmetic and Geometric Series Harder Questions
2014 - Question 16b(ii) Find the balance of the account at the end of the 60th month, correct to thenearest dollar.
Continuing on from (i):
3rd month (beginning) 500(1.003)2 + 500(1.01)(1.003) + 500(1.01)2
3rd month (end) 500(1.003)3 + 500(1.01)(1.003)2 + 500(1.01)2(1.003)...60th month (end) 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
Let B be the balance owing at the end of the 60th month. Then:
B = 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
= 500(1.003)60 + 500(1.003)60 1.011.003
+ · · ·+ 500(1.003)60(
1.011.003
)59
= 500(1.003)60
(1 +
1.011.003
+ · · ·+(
1.011.003
)59)
Paul Hancock (Woonona High School) Series T1 2017 21 / 23
Arithmetic and Geometric Series Harder Questions
2014 - Question 16b(ii) Find the balance of the account at the end of the 60th month, correct to thenearest dollar.
Continuing on from (i):
3rd month (beginning) 500(1.003)2 + 500(1.01)(1.003) + 500(1.01)2
3rd month (end) 500(1.003)3 + 500(1.01)(1.003)2 + 500(1.01)2(1.003)...60th month (end) 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
Let B be the balance owing at the end of the 60th month. Then:
B = 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
= 500(1.003)60 + 500(1.003)60 1.011.003
+ · · ·+ 500(1.003)60(
1.011.003
)59
= 500(1.003)60
(1 +
1.011.003
+ · · ·+(
1.011.003
)59)
Paul Hancock (Woonona High School) Series T1 2017 21 / 23
Arithmetic and Geometric Series Harder Questions
2014 - Question 16b(ii) Find the balance of the account at the end of the 60th month, correct to thenearest dollar.
Continuing on from (i):
3rd month (beginning) 500(1.003)2 + 500(1.01)(1.003) + 500(1.01)2
3rd month (end) 500(1.003)3 + 500(1.01)(1.003)2 + 500(1.01)2(1.003)...60th month (end) 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
Let B be the balance owing at the end of the 60th month. Then:
B = 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
= 500(1.003)60 + 500(1.003)60 1.011.003
+ · · ·+ 500(1.003)60(
1.011.003
)59
= 500(1.003)60
(1 +
1.011.003
+ · · ·+(
1.011.003
)59)
Paul Hancock (Woonona High School) Series T1 2017 21 / 23
Arithmetic and Geometric Series Harder Questions
2014 - Question 16b(ii) Find the balance of the account at the end of the 60th month, correct to thenearest dollar.
Continuing on from (i):
3rd month (beginning) 500(1.003)2 + 500(1.01)(1.003) + 500(1.01)2
3rd month (end) 500(1.003)3 + 500(1.01)(1.003)2 + 500(1.01)2(1.003)...60th month (end) 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
Let B be the balance owing at the end of the 60th month. Then:
B = 500(1.003)60 + 500(1.01)(1.003)59 + · · ·+ 500(1.01)59(1.003)
= 500(1.003)60 + 500(1.003)60 1.011.003
+ · · ·+ 500(1.003)60(
1.011.003
)59
= 500(1.003)60
(1 +
1.011.003
+ · · ·+(
1.011.003
)59)
Paul Hancock (Woonona High School) Series T1 2017 21 / 23
Arithmetic and Geometric Series Harder Questions
2014 - Question 16b
(ii) Find the balance of the account at the end of the 60th month, correct to thenearest dollar.
The bracketed term is a geometric series with 60 terms, a = 1 and r = 1.011.003 .
S = 500(1.003)60
(( 1.011.003
)60 − 1( 1.011.003 − 1
) )≈ 500(1.003)60 (74.29) (nearest cent)≈ 44404 (nearest $)
Paul Hancock (Woonona High School) Series T1 2017 22 / 23
Arithmetic and Geometric Series Harder Questions
2014 - Question 16b
(ii) Find the balance of the account at the end of the 60th month, correct to thenearest dollar.
The bracketed term is a geometric series with 60 terms, a = 1 and r = 1.011.003 .
S = 500(1.003)60
(( 1.011.003
)60 − 1( 1.011.003 − 1
) )≈ 500(1.003)60 (74.29) (nearest cent)≈ 44404 (nearest $)
Paul Hancock (Woonona High School) Series T1 2017 22 / 23
Arithmetic and Geometric Series Harder Questions
2014 - Question 16b
(ii) Find the balance of the account at the end of the 60th month, correct to thenearest dollar.
The bracketed term is a geometric series with 60 terms, a = 1 and r = 1.011.003 .
S = 500(1.003)60
(( 1.011.003
)60 − 1( 1.011.003 − 1
) )
≈ 500(1.003)60 (74.29) (nearest cent)≈ 44404 (nearest $)
Paul Hancock (Woonona High School) Series T1 2017 22 / 23
Arithmetic and Geometric Series Harder Questions
2014 - Question 16b
(ii) Find the balance of the account at the end of the 60th month, correct to thenearest dollar.
The bracketed term is a geometric series with 60 terms, a = 1 and r = 1.011.003 .
S = 500(1.003)60
(( 1.011.003
)60 − 1( 1.011.003 − 1
) )≈ 500(1.003)60 (74.29) (nearest cent)
≈ 44404 (nearest $)
Paul Hancock (Woonona High School) Series T1 2017 22 / 23
Arithmetic and Geometric Series Harder Questions
2014 - Question 16b
(ii) Find the balance of the account at the end of the 60th month, correct to thenearest dollar.
The bracketed term is a geometric series with 60 terms, a = 1 and r = 1.011.003 .
S = 500(1.003)60
(( 1.011.003
)60 − 1( 1.011.003 − 1
) )≈ 500(1.003)60 (74.29) (nearest cent)≈ 44404 (nearest $)
Paul Hancock (Woonona High School) Series T1 2017 22 / 23
Arithmetic and Geometric Series Finale
Good Luck
And in the words of Douglas Adams:
Paul Hancock (Woonona High School) Series T1 2017 23 / 23
Arithmetic and Geometric Series Finale
Good Luck
And in the words of Douglas Adams: "So long and thanks for all the fish."
Paul Hancock (Woonona High School) Series T1 2017 23 / 23
Arithmetic and Geometric Series Finale
Good Luck
And in the words of Douglas Adams: "So long and thanks for all the fish."
I mean: "Don’t panic."
Paul Hancock (Woonona High School) Series T1 2017 23 / 23
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