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Financial Maths, 2ADV M1 2025 HSC 13

The numbers, 75, \(p\), \(q\), 2025, form a geometric sequence.

Find the values of \(p\) and \(q\).   (2 marks)

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\(p=225, \ q=675\)

Show Worked Solution

\(a=75, \ 75r=p, \ 75r^2=q, \ 75r^3=2025\)

\(\text{Using}\ \ 75r^3=2025:\)

\(r=\sqrt[3]{\dfrac{2025}{75}}=3\)

\(p=75 \times 3 = 225\)

\(q=75 \times 3^{2}=675\)

Financial Maths, 2ADV M1 2023 HSC 21

The fourth term of a geometric sequence is 48 .

The eighth term of the same sequence is `3/16`.

Find the possible value(s) of the common ratio and the corresponding first term(s).  (3 marks)

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`a=3072,\ r=1/4, or`

`a=-3072,\ r=-1/4`

Show Worked Solution

`T_4=ar^3=48\ …\ (1)`

`T_8=ar^7=3/16\ …\ (2)`

`(ar^7)/(ar^3)` `=(3/16)/48`  
`r^4` `=1/256`  
`r` `=+-1/4`  

 
`text{If}\ \ r=1/4`

`a(1/4)^3` `=48`  
`a/64` `=48`  
`a` `=3072`  

 
`text{If}\ \ r=-1/4,\ \ a=-3072`

Financial Maths, 2ADV M1 SM-Bank 4 MC

The first four terms of a geometric sequence are 

`4, – 8, 16, – 32`

The sum of the first ten terms of this sequence is

  1. `– 2048`
  2. `– 1364`
  3. `684`
  4. `1367`
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`B`

Show Worked Solution

`4, –8, 16, –32,\ …`

`text(GP where)\ \ \ a` `= 4`
`r` `=t_(2)/t_(1)= (–8)/4=–2`
`S_n` `=(a(r^n – 1))/(r – 1)`
`:.S_10`  `=[4[(–2)^10 – 1]]/(–2 – 1)`
  `= – 1364`

`=> B`

Financial Maths, 2ADV M1 SM-Bank 2 MC

The first four terms of a geometric sequence are  6400, `t_2` , 8100, – 9112.5

The value of  `t_2` is

  1. `– 7250` 
  2. `– 7200`
  3. `– 1700`
  4. `7200`
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`B`

Show Worked Solution

`text(GP is)\ \ 6400,  t_2,  8100,  –9112.5`

`r` `=t_2/t_1 = t_3/t_2`
`:. t_2 / 6400` `= (–9112.5) / 8100`
`t_2` `= (–9112.5 × 6400) / 8100`
  `= – 7200`

`=>  B`

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`

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`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`

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 2014 HSC 8 MC

Which expression is a term of the geometric series  `3x − 6x^2 + 12x^3 − ...` ?

  1. `3072 x^10`
  2. `–3072 x^10 `
  3. `3072 x^11`
  4. `–3072 x^11 `
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`C`

Show Worked Solution

`3x\ – 6x^2 + 12x^3\ – …`

`a` `= 3x`
`r` `= (T_2)/(T_1) = (-6x^2)/(3x) = -2x`

 

`:.\ T_n = ar^n` `= 3x (-2x)^n` 
  `= 3(-2)^n x^(n + 1)`

`text(If)\ n = 9`

`T_9 = 3(–2)^9 x^(9 + 1) = -1536 x^10`

`text(If)\ n = 10`

`T_10 = 3(–2)^10 x^(10 + 1) = 3072 x^11`

`=>  C`

Financial Maths, 2ADV M1 2011 HSC 9d

 

  1. Rationalise the denominator in the expression  `1/(sqrtn + sqrt(n+1))`  where  `n`  is an integer and  `n >= 1`.   (1 mark)

  2. Using your result from part (i), or otherwise, find the value of the sum
     
         `1/(sqrt1+ sqrt2) + 1/(sqrt2 + sqrt3) + 1/(sqrt3 + sqrt4) + ... + 1/(sqrt99 + sqrt100)`   (2 marks)

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  1. `sqrt(n+1)\ – sqrtn`
  2. `9`
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i.  `1/(sqrtn + sqrt(n+1)) xx (sqrtn\ – sqrt(n+1))/(sqrtn\ – sqrt(n+1))`

MARKER’S COMMENT: Students connecting the information between parts (i) and (ii) were easily the most successful. AGAIN, the information from earlier parts of multi-part questions is gold plated information for directing your answer.

`= (sqrtn\ – sqrt(n+1))/((sqrtn)^2\ – (sqrt(n+1))^2)`

`= (sqrtn\ – sqrt(n+1))/(n\ – (n + 1))`

`= (sqrtn\ – sqrt(n+1))/-1`

`= sqrt(n+1)\ – sqrtn`

 

ii.  `1/(sqrt1 + sqrt2) + 1/(sqrt2 + sqrt3) + 1/(sqrt3 + sqrt4) + … + 1/(sqrt99 + sqrt100)`

`= (sqrt2\ – sqrt1) + (sqrt3\ – sqrt2) + (sqrt4\ – sqrt3) + … + (sqrt100\ – sqrt99)`

`= – sqrt1 + sqrt 100`

♦♦♦ Mean mark 15%.

`= -1 + 10`

`= 9`

 

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)

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  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`