hsc revision day arithmetic and geometric seriesweb/@eis/...answer: (a) paul hancock (woonona high...

Arithmetic and Geometric Series

HSC Revision DayArithmetic and Geometric Series

Woonona High School

2017

Teacher: Paul HancockSchool: Woonona High SchoolE-mail: [email protected]

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.


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)



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)



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)



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)



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)



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)



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)



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)



2014 - Question 8

Example 2 (Q8)

Well, the first term is a = 3x

and r = −6x2

3x = 12x3



Answer:

(C)



2014 - Question 8

Example 2 (Q8)


3x = 12x3

−6x2 = −2x .

So the nth term is given by Tn = 3x × (−2x)n−1 = 3x × (−2)n−1 × xn−1


Answer:

(C)



2014 - Question 8

Example 2 (Q8)


3x = 12x3

−6x2 = −2x .So the nth term is given by Tn = 3x × (−2x)n−1

= 3x × (−2)n−1 × xn−1


Answer:

(C)



2014 - Question 8

Example 2 (Q8)


3x = 12x3



Answer:

(C)



2014 - Question 8

Example 2 (Q8)


3x = 12x3


= 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)



2014 - Question 8

Example 2 (Q8)


3x = 12x3


= 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)



2014 - Question 8

Example 2 (Q8)


3x = 12x3


= 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)



2014 - Question 8

Example 2 (Q8)


3x = 12x3


= 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)


Arithmetic and Geometric Series Standard Questions

2012 - Question 12c



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



2012 - Question 12c


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



2012 - Question 12c


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



2012 - Question 12c



T20 = 3+ (20− 1)× 2= 3+ 19× 2= 3+ 38= 41



2012 - Question 12c



T20 = 3+ (20− 1)× 2

= 3+ 19× 2= 3+ 38= 41



2012 - Question 12c



T20 = 3+ (20− 1)× 2= 3+ 19× 2

= 3+ 38= 41



2012 - Question 12c



T20 = 3+ (20− 1)× 2= 3+ 19× 2= 3+ 38

= 41



2012 - Question 12c



T20 = 3+ (20− 1)× 2= 3+ 19× 2= 3+ 38= 41



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



2012 - Question 12c



S20 =202[2× 3+ (20− 1)× 2]

= 10[6+ 19× 2]

= 10[44]= 440



2012 - Question 12c



S20 =202[2× 3+ (20− 1)× 2]

= 10[6+ 19× 2]= 10[44]

= 440



2012 - Question 12c



S20 =202[2× 3+ (20− 1)× 2]

= 10[6+ 19× 2]= 10[44]= 440



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.



2012 - Question 12c



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)




2012 - Question 12c



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)




2013 - Question 13d



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



2013 - Question 13d


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



2013 - Question 13d


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



2013 - Question 13d


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



2013 - Question 13d


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



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



2013 - Question 13d


A20 ≈ 500000(1.005)240 − 2998[(1.005240 − 1)

1.005 − 1

]

≈ 269903

≈ 270000



2013 - Question 13d


A20 ≈ 500000(1.005)240 − 2998[(1.005240 − 1)

1.005 − 1

]≈ 269903

≈ 270000



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.




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






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






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.





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


2014 - Question 14d



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.



2014 - Question 14d


The initial amount, which we will call A1 was 10 mL.

The amount immediatelyafter the second dose would be 10mL plus 1


So A2 = 10+ 10( 1

3

)= 40

3 = 13 13




2014 - Question 14d


The initial amount, which we will call A1 was 10 mL. The amount immediatelyafter the second dose would be 10mL plus 1


So A2 = 10+ 10( 1

3

)= 40

3 = 13 13




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




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




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




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




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




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




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


Arithmetic and Geometric Series Harder Questions

2012 - Question 15a



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



2012 - Question 15a


L = 10+ 10× 0.96+ 10× 0.962 + · · ·+ 10× 0.969

= 10(1+ 0.96+ 0.962 + · · ·+ 0.969)


= 10(1(1− 0.9610)

1− 0.96

)

≈ 84



2012 - Question 15a


L = 10+ 10× 0.96+ 10× 0.962 + · · ·+ 10× 0.969

= 10(1+ 0.96+ 0.962 + · · ·+ 0.969)


= 10(1(1− 0.9610)

1− 0.96

)≈ 84



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.



2012 - Question 15a



S∞ =1

1− 0.96

S∞ = 25





2012 - Question 15a



S∞ =1

1− 0.96S∞ = 25





2012 - Question 15a



S∞ =1

1− 0.96S∞ = 25

Hence we can now determine the limiting value of L.

L∞ = 10(25) = 250




2012 - Question 15a



S∞ =1

1− 0.96S∞ = 25





2014 - Question 16b



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).



2014 - Question 16b


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)




2014 - Question 16b


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)




2014 - Question 16b


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)




2014 - Question 16b


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)




2014 - Question 16b


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)




2014 - Question 16b


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)




2014 - Question 16b


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)




2014 - Question 16b


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)




2014 - Question 16b


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)




2014 - Question 16b


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)




2014 - Question 16b


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)




2014 - Question 16b


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)




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)





3rd month (beginning)

500(1.003)2 + 500(1.01)(1.003) + 500(1.01)2



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)








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)






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)


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)






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)


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)






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)


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)






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)


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)








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)



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 $)



2014 - Question 16b



S = 500(1.003)60

(( 1.011.003

)60 − 1( 1.011.003 − 1

) )

≈ 500(1.003)60 (74.29) (nearest cent)≈ 44404 (nearest $)



2014 - Question 16b



S = 500(1.003)60

(( 1.011.003

)60 − 1( 1.011.003 − 1

) )≈ 500(1.003)60 (74.29) (nearest cent)

≈ 44404 (nearest $)



2014 - Question 16b



S = 500(1.003)60

(( 1.011.003

)60 − 1( 1.011.003 − 1

) )≈ 500(1.003)60 (74.29) (nearest cent)≈ 44404 (nearest $)


Arithmetic and Geometric Series Finale

Good Luck

And in the words of Douglas Adams:



Good Luck

And in the words of Douglas Adams: "So long and thanks for all the fish."



Good Luck

And in the words of Douglas Adams: "So long and thanks for all the fish."

I mean: "Don’t panic."


hsc revision day arithmetic and geometric seriesweb/@eis/...answer: (a) paul hancock (woonona high...

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