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Financial Maths, 2ADV M1 2017 HSC 15b

Anita opens a savings account. At the start of each month she deposits `$X` into the savings account. At the end of each month, after interest is added into the savings account, the bank withdraws $2500 from the savings account as a loan repayment. Let `M_n` be the amount in the savings account after the `n`th  withdrawal.

The savings account pays interest of 4.2% per annum compounded monthly.

  1. Show that after the second withdrawal the amount in the savings account is given by
    `qquad qquad M_2 = X(1.0035^2 + 1.0035) - 2500 (1.0035 + 1)`.  (2 marks)

  2. Find the value of  `X`  so that the amount in the savings account is $80 000 after the last withdrawal of the fourth year.  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `$ 4019.42\ text{(nearest cent)}`
Show Worked Solution
i.   `M_1` `= X (1 + 4.2/(12 xx 100)) – 2500`
    `= X (1.0035) – 2500`
  `M_2` `= X (1.0035) + M_1 (1.0035) – 2500`
    `=X(1.0035)+(1.0035)[X(1.0035)-2500]-2500`
    `= X (1.0035) + X (1.0035^2) – 2500 (1.0035) – 2500`
    `= X (1.0035^2 + 1.0035) – 2500 (1.0035 + 1)`
♦ Mean mark part (ii) 48%.

 

ii.  

`M_3 = X (1.0035^3 + … + 1.0035) – 2500 (1.0035^2 + 1.0035 + 1)`

`vdots`

`M_n = X (1.0035^n + … + 1.0035) – 2500 (1.0035^(n – 1) + … + 1)`

`text(Find)\ X\ text(such that)\ \ M_n = 80\ 000\ \ text(when)\ n = 48`

`80\ 000 = X (1.0035^48 + …\ 1.0035) – 2500 (1.0035^47 + … + 1)`

`text(Using)\ S_n = (a(r^n – 1))/(r – 1)`

`80\ 000` `= X [(1.0035(1.0035^48 – 1))/(1.0035 – 1)] – 2500 [(1(1.0035^48 – 1))/(1.0035 – 1)]`
`80\ 000` `= X (52.351…) – 130\ 421.20…`
   
`:. X` `= (80\ 000 + 130\ 421.20…)/(52.351…)`
  `= $4019.42\ text{(nearest cent)}`