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Financial Maths, 2ADV M1 SM-Bank 7

Joe buys a tractor under a buy-back scheme. This scheme gives Joe the right to sell the tractor back to the dealer.

The recurrence relation below can be used to calculate the price Joe sells the tractor back to the dealer `(P_n)`, after `n` years
 

`qquad\ \ \ P_0 = 56\ 000,qquadP_n = P_(n - 1) - 7000`
  

  1. Write the general rule to find the value of  `P_n`  in terms of  `n`.  (1 mark)

  2. After how many years will the dealer offer to buy back Joe's tractor at half of its original value.  (1 mark)

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i.  `P_n = 56\ 000 – 7000n`

ii.  `4\ text(years)`

Show Worked Solution
i.    `P_1` `= P_0 – 7000`
  `P_2` `= P_0 – 7000 – 7000`
    `= 56\ 000 – 7000 xx 2`
  `vdots`  
  `P_n`  `= 56\ 000 – 7000n` 

 

ii.    `text(Half original value)` `= 56\ 000 ÷ 2=$28\ 000`

 

`text(Find)\ \ n\ \ text(such that:)`

`28\ 000` `= 56\ 000 – 7000n`
`7000n` `= 28\ 000`
`:. n` `= 4\ text(years)`

Financial Maths, 2ADV M1 SM-Bank 1 MC

On day 1, Vikki spends 90 minutes on a training program.

On each following day, she spends 10 minutes less on the training program than she did the day before.

Let  `t_n`  be the number of minutes that Vikki spends on the training program on day  `n`.

A recursive equation that can be used to model this situation for  `1 ≤ n ≤ 10`  is

A.   `t_(n + 1) = 0.90t_n` `t_1 = 90`
B.   `t_(n + 1) = 1.10 t_n` `t_1 = 90`
C.   `t_(n + 1) = 1 - 10 t_n` `t_1 = 90`
D.   `t_(n + 1) = t_n - 10` `t_1 = 90`

 

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`D`

Show Worked Solution

`text(Difference equation where each term is 10 minutes)`

`text(less than the preceding term.)`

`∴\ text(Equation)\ \ \t_(n+1) = t_n-10, \ \ t_1 = 90`

`=>  D`