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Financial Maths, 2ADV M1 2024 HSC 30

Suppose the geometric series \(x+x^2+x^3+\ \cdots\) has a limiting sum.

By considering the graph  \(y=-1-\dfrac{1}{x-1}\), or otherwise, find the range of possible values of \(S\).   (3 marks)

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\(S \in \Big( -\dfrac{1}{2}, \infty\Big ) \)

Show Worked Solution

\(x + x^2+x^3\ \cdots \Rightarrow a=x, r=x\)

\(S=\dfrac{a}{1-r} = \dfrac{x}{1-x}=\dfrac{-(1-x)+1}{1-x} = -1+\dfrac{1}{1-x}=-1-\dfrac{1}{x-1}\)
 

\(\text{If limiting sum}\ \Rightarrow \abs{r} \lt 1\ \Rightarrow \ \abs{x} \lt 1\)

\(\Rightarrow \text{Possible values of}\ S =\ \text{range of}\ \ y=-1-\dfrac{1}{x-1}\ \text{in domain}\ \abs{x} \lt 1\)

\(\text{Consider}\ \ y=-1-\dfrac{1}{x-1}:\)

\(\text{Asymptote at}\ \ x=1.\)

\begin{array} {|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & 0 & 1 & 2 \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & -\dfrac{2}{3} &  -\dfrac{1}{2} & \ \ 0\ \  & \infty & -2 \\
\hline
\end{array}

\(\text{As}\ x \rightarrow \infty, \ y \rightarrow -1\)

\(\text{As}\ x \rightarrow -\infty, \ y \rightarrow -1\)

\(\abs{x} \lt 1\ \Rightarrow \  y \in \Big( -\dfrac{1}{2}, \infty\Big ) \)

\(\therefore\ \text{Possible values of}\ S \in \Big( -\dfrac{1}{2}, \infty\Big ) \)

Financial Maths, 2ADV M1 2017 HSC 16b

A geometric series has first term  `a`  and limiting sum 2.

Find all possible values for  `a`.  (3 marks)

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`0 < a < 4`

Show Worked Solution

`S_oo = a/(1 – r)`

♦ Mean mark 39%.

`text(If)\ \ S_oo = 2:`

`2` `= a/(1 – r)`
`2(1 – r)` `= a`
`1 – r` `= a/2`
`r` `= 1 – a/2`

 

`text(S)text(ince)\ \ |r| < 1,`

`|1 – a/2| < 1`

`1 – a/2` `< 1` `or qquad -(1 – a/2)` `< 1`
`a/2` `> 0` `a/2` `< 2`
`a` `> 0` `a` `< 4`

 

`:. 0 < a < 4`

Financial Maths, 2ADV M1 2016 HSC 14d

By summing the geometric series  `1 + x + x^2 + x^3 + x^4`, or otherwise,

find  `lim_(x -> 1) (x^5 - 1)/(x - 1).`  (2 marks)

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`5`

Show Worked Solution

`1 + x + x^2 + x^3 + x^4`

♦ Mean mark 45%.

`=> text(GP where)\ \ a = 1,\ \ r = x,\ \ n = 5`

`S_n` `= (a(r^n – 1))/(r – 1)`
 `S_5` `= ((x^5 – 1))/(x – 1)`

 
`:. lim_(x -> 1) (x^5 – 1)/(x – 1)`

`= lim_(x -> 1) (1 + x + x^2 + x^3 + x^4)`

`= 5`

Financial Maths, 2ADV M1 SM-Bank 7 MC

A dragster is travelling at a speed of 100 km/h.

It increases its speed by

  • 50 km/h in the 1st second
  • 30 km/h in the 2nd second
  • 18 km/h in the 3rd second

and so on in this pattern.

Correct to the nearest whole number, the greatest speed, in km/h, that the dragster will reach is

  1. `125`
  2. `200`
  3. `220`
  4. `225`
Show Answers Only

`D`

Show Worked Solution

`text (Sequence of speed increases is)`

`text (50, 30, 18, …)`

`text (GP where)\ \ \ a`  `= 50, and` 
`r` `= t_2/t_1 = 30/50 = 0.6`

 

`text (S) text (ince)\ \ |\ r\ | < |,`

`S_oo` `= a / (1-r)`
  `= 50/ (1 – 0.6)`
  `= 125`

 
`:.\ text (Max speed is 100 + 125 = 225 km/h)`

`rArr D`

Trigonometry, 2ADV T2 2004 HSC 9a

Consider the geometric series  `1 − tan^2 theta + tan^4 theta − …`

  1. When the limiting sum exists, find its value in simplest form.  (2 marks)

  2. For what values of  `theta`  in the interval
     
        `−pi/2 < theta < pi/2`  does the limiting sum of the series exist?  (2 marks)

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  1. `cos^2 theta`
  2. `− pi/4 < theta < pi/4`
Show Worked Solution

i.   `1 − tan^2 theta + tan^4 theta − …`

`=>\ text(GP where)\ \ a=1,\ \ r=T_2/T_1= − tan^2 theta`

`:. S_∞` `= 1/(1 − (−tan^2 theta))`
  `= 1/(1 + tan^2 theta)`
  `= 1/(sec^2 theta)`
  `= cos^2 theta`

 

ii.   `text(Find)\ \ theta\ \ text(such that)\ \ \ |\ r\ |` `< 1`
  `|−tan^2 theta\ |` `< 1`
  ` tan^2 theta` `< 1`
  `−1 < tan theta` `< 1`
  `:. − pi/4 < theta` `< pi/4`

Financial Maths, 2ADV M1 2015 HSC 11d

Find the limiting sum of the geometric series  `1 - 1/4 + 1/16 - 1/64 + …`.  (2 marks)

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`4/5`

Show Worked Solution

`1 – 1/4 + 1/16 – 1/64 + …`

`r = -1/4,\ \ a=1`

`text(S)text(ince)\ |\ r\ | = 1/4 < 1`

`S_oo` `= a/(1 – r)`
  `= 1/(1 – (-1/4))`
  `= 1/(5/4)`
  `= 4/5`

Trigonometry, 2ADV T1 2005 HSC 9b

Trig Ratios, 2UA 2005 HSC 9b
 

The triangle  `ABC`  has a right angle at `B, \ ∠BAC = theta` and  `AB = 6`. The line `BD` is drawn perpendicular to `AC`. The line `DE` is then drawn perpendicular to `BC`. This process continues indefinitely as shown in the diagram.

  1. Find the length of the interval `BD`, and hence show that the length of the interval  `EF`  is  `6 sin^3\ theta`.  (2 marks)

  2. Show that the limiting sum 
     
    `qquad BD + EF + GH + ···`
     
    is given by  `6 sec\ theta tan\ theta`.  (3 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
Show Worked Solution
i.   Trig Ratios, 2UA 2005 HSC 9b Answer

`text(Show)\ EF = 6\ sin^3\ theta`

`text(In)\ ΔADB`

`sin\ theta` `= (DB)/6`
`DB` `= 6\ sin\ theta`
`∠ABD` `= 90 − theta\ \ \ text{(angle sum of}\ ΔADB)`
`:.∠DBE` `= theta\ \ \ (∠ABE\ text{is a right angle)}`

 

`text(In)\ ΔBDE:`

`sin\ theta` `= (DE)/(DB)`
  `= (DE)/(6\ sin\ theta)`
`DE` `= 6\ sin^2\ theta`
`∠BDE` `= 90 − theta\ \ \ text{(angle sum of}\ ΔDBE)`
`∠EDF` `= theta\ \ \ (∠FDB\ text{is a right angle)}`

 

`text(In)\ ΔDEF:`

`sin\ theta` `= (EF)/(DE)`
  `= (EF)/(6\ sin^2\ theta)`
`:.EF` `= 6\ sin^3\ theta\ \ …text(as required)`

 

ii. `text(Show)\ \ BD + EF + GH\ …`

`text(has limiting sum)\ =6 sec theta tan theta`

`underbrace{6\ sin\ theta + 6\ sin^3\ theta +\ …}_{text(GP where)\ \ a = 6 sin theta, \ \ r = sin^2 theta}`
 

`text(S)text(ince)\ \ 0 < theta < 90^@`

`−1` `< sin\ theta` `< 1`
`0` `< sin^2\ theta` `< 1`

 
`:. |\ r\ | < 1`
 

`:.S_∞` `= a/(1 − r)`
  `= (6\ sin\ theta)/(1 − sin^2\ theta)`
  `= (6\ sin\ theta)/(cos^2\ theta)`
  `= 6 xx 1/(cos\ theta) xx (sin\ theta)/(cos\ theta)`
  `= 6 sec\ theta\ tan\ theta\ \ …text(as required.)`

Financial Maths, 2ADV M1 2006 HSC 1f

Find the limiting sum of the geometric series  `13/5 + 13/25 + 13/125 + …`  (2 marks)

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`13/4`

Show Worked Solution

`13/5 + 13/25 + 13/125`

`=>\ text(GP where)\ \ a=13/5,\ text(and)`

`r = T_2/T_1 = 13/25 ÷ 13/5 = 1/5`

`text(S)text(ince)\ |\ r\ | < 1`

`S_oo` `= a/(1-r)`
  `= (13/5)/(1 – 1/5)`
  `= 13/5 xx 5/4`
  `= 13/4`

Financial Maths, 2ADV M1 SM-Bank 11

Express the recurring decimal  `0.323232...`  as a fraction.  (2 marks)

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`32/99`

Show Worked Solution
`0.3232…` `=32/100+32/10^4+32/10^6+…`
  `=32/100(1+1/10^2+1/10^4+…)`
`=>\ text(GP where)\ a=1,\ \ r=1/10^2`
  `=32/100(a/(1-r))`
  `=32/100(1/(1-1/100))`
  `=32/100(1/(99/100))`
  `=32/99`

Financial Maths, 2ADV M1 2014 HSC 14d

At the beginning of every 8-hour period, a patient is given 10 mL of a particular drug.

During each of these 8-hour periods, the patient’s body partially breaks down the drug. Only  `1/3`  of the total amount of the drug present in the patient’s body at the beginning of each 8-hour period remains at the end of that period.  

  1. How much of the drug is in the patient’s body immediately after the second dose is given?    (1 mark)

  2. Show that the total amount of the drug in the patient’s body never exceeds 15 mL.     (2 marks)

Show Answers Only
  1. `13.33\ text{mL  (2 d.p.)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.   `text(Let)\ \ A =\ text(Amount of drug in body)`

`text(Initially)\ A = 10`

`text(After 8 hours)\ \ \ A` `=1/3 xx 10`
`text(After 2nd dose)\ \ A` `= 10 + 1/3 xx 10\ text(mL)`
  `=13.33\ text{mL  (2 d.p.)}`

 

ii.   `text(After the 3rd dose)`

`A_3` `= 10 + 1/3 (10 + 1/3 xx 10)`
  `= 10 + 1/3 xx 10 + (1/3)^2 xx 10`

 
`  =>\ text(GP where)\ a = 10,\ r = 1/3`

`text(S)text(ince)\ \ |\ r\ | < 1:`

`S_oo` `= a/(1\ – r)`
  `= 10/(1\ – 1/3)`
  `= 10/(2/3)`
  `= 15`

 

 
`:.\ text(The amount of the drug will never exceed 15 mL.)`

Financial Maths, 2ADV M1 2008 HSC 5b

Consider the geometric series

`5+10x+20x^2+40x^3+\ ...`

  1. For what values of `x` does this series have a limiting sum?    (2 marks)

  2. The limiting sum of this series is `100`.

     

    Find the value of `x`.     (2 marks)

Show Answers Only
  1. `-1/2<x<1/2`
  2. `19/40`
Show Worked Solutions

i.   `text(Limiting sum when)\ |\ r\ |<1`

`r=T_2/T_1=(10x)/5=2x`

`:.\ |\ 2x\ |<1`

`text(If)\ \ 2x` `>0` `text(If)\ \ 2x` `<0`
`2x` `<1` `-(2x)` `<1`
`x` `<1/2` `2x` `> -1`
    `x` `> -1/2`

 

`:. text(Limiting sum when)\ \ -1/2<x<1/2`

 

ii.  `text(Given)\  S_oo=100,  text(find) \ x`

`=> S_oo=a/(1-r)=100`

` 5/(1-2x)` `=100`
`100(1-2x)` `=5`
`200x` `=95`
`:.\ x` `=95/200=19/40`

Financial Maths, 2ADV M1 2009 HSC 4a

A tree grows from ground level to a height of 1.2 metres in one year. In each subsequent year, it grows `9/10` as much as it did in the previous year.

Find the limiting height of the tree.     (2 marks)

 

Show Answer Only

`12\ text(m)`

Show Worked Solution

`a=1.2, \ \ \ r=9/10`

`text(S)text(ince)\ \ |\ r\ |<1,`

`S_oo` `=a/(1-r)`
  `=1.2/(1-(9/10))`
  `=12\ text(m)`

 

`:.\ text(Limiting height of tree is 12 m.)`

Financial Maths, 2ADV M1 2012 HSC 15a

Rectangles of the same height are cut from a strip and arranged in a row. The first rectangle has width 10cm. The width of each subsequent rectangle is 96% of the width of the previous rectangle.
 

2012 15a
 

  1. Find the length of the strip required to make the first ten rectangles.     (2 marks)

  2. Explain why a strip of 3m is sufficient to make any number of rectangles.     (1 mark)

Show Answers Only
  1. `83.8\ text{cm  (1 d.p.)}`
  2. `S_oo=2.5\ text(m)\ \ =>\ \ text(sufficient.)`
Show Worked Solutions
i.    `T_1` `=a=10`
`T_2` `=ar=10xx0.96=9.6`
`T_3` `=ar^2=10xx0.96^2=9.216`

 
`=>\ text(GP where)\ \ a=10\ \ text(and)\ \ r=0.96`

MARKER’S COMMENT: A common error was to find `T_10` instead of `S_10`
`S_10` `=\ text(Length of strip for 10 rectangles)`
  `=(a(1-r^n))/(1-r)`
  `=10((1-0.96^10)/(1-0.96))`
  `=83.8\ text{cm   (to 1 d.p.)}`

 

ii.   `text(S)text(ince)\ |\ r\ |<\ 1`

`S_oo` `=a/(1-r)`
  `=10/(1-0.96)`
  `=250\ text(cm)`

 

`:.\ text(S)text(ince  3 m > 2.5 m, it is sufficient.)`

Probability, 2ADV S1 2013 HSC 15d

Pat and Chandra are playing a game. They take turns throwing two dice. The game is won by the first player to throw a double six. Pat starts the game.

  1. Find the probability that Pat wins the game on the first throw.     (1 mark)

  2. What is the probability that Pat wins the game on the first or on the second throw?     (2 marks)

  3. Find the probability that Pat eventually wins the game.     (2 marks)

Show Answers Only
  1. `1/36`
  2. `(2521)/(46\ 656)\ \ text(or)\ \ 0.054`
  3. `36/71`
Show Worked Solutions

i.   `P\ text{(Pat wins on 1st throw)}=P(W)`

`P(W)` `=P\ text{(Pat throws 2 sixes)}`
  `=1/6 xx 1/6`
  `=1/36`

 

ii.  `text(Let)\ P(L)=P text{(loss for either player on a throw)}=35/36`

`P text{(Pat wins on 1st or 2nd throw)}` 

♦♦ Mean mark 33%
MARKER’S COMMENT: Many students did not account for Chandra having to lose when Pat wins on the 2nd attempt.

`=P(W) + P(LL W)`

`=1/36\ + \ (35/36)xx(35/36)xx(1/36)`

`=(2521)/(46\ 656)`

`=0.054\ \ \ text{(to 3 d.p.)}`

 

iii.  `P\ text{(Pat wins eventually)}`

`=P(W) + P(LL\ W)+P(LL\ LL\ W)+ … `

`=1/36\ +\ (35/36)^2 (1/36)\ +\ (35/36)^2 (35/36)^2 (1/36)\ +…`

 
`=>\ text(GP where)\ \ a=1/36,\ \ r=(35/36)^2=(1225)/(1296)`

♦♦♦ Mean mark 8%!
 COMMENT: Be aware that diminishing probabilities and `S_oo` within the Series and Applications are a natural cross-topic combination.

 
`text(S)text(ince)\ |\ r\ |<\ 1:`

`S_oo` `=a/(1-r)`
  `=(1/36)/(1-(1225/1296))`
  `=1/36 xx 1296/71`
  `=36/71`

 

`:.\ text(Pat’s chances to win eventually are)\  36/71`.