smc-2512-70-Recurrence relations

Financial Maths, GEN1 2024 VCAA 24 MC

André invested $18 000 in an account for five years, with interest compounding monthly.

He adds an extra payment into the account each month immediately after the interest is calculated.

For the first two years, the balance of the account, in dollars, after $n$ months, $A_n$, can be modelled by the recurrence relation

$A_0=18\,000, \quad A_{n+1}=1.002 A_n+100$

After two years, André decides he would like the account to reach a balance of $30 000 at the end of the five years.

He must increase the value of the monthly extra payment to achieve this.

The minimum value of the new payment for the last three years is closest to

$189.55
$195.45
$202.35
$246.55

Show Answers Only

$A$

Show Worked Solution

$\text{Step 1: Using CAS}$

$\text{Annual interest rate}\ = (1.002-1) \times 12 = 0.024 = 2.4\% $

$\text{Compounding periods (2 years)}\ = 2 \times 12=24$

$\text{Value of investment after 2 years}\ =$21\,340.18$

♦ Mean mark 46%.

$\text{Step 2: Using CAS}$

$\text{Next 3 years:}\ N=3 \times 12 = 36$

$\text{Minimum value of new payment}\ =$189.55$

$\Rightarrow A$

Financial Maths, GEN2 2023 VCAA 6

Arthur invests $600 000 in an annuity that provides him with a monthly payment of $3973.00.

Interest is calculated monthly.

Three lines of the amortisation table for this annuity are shown below.

\begin{array} {|c|c|}
\hline
\textbf{Payment} & \textbf{Payment} & \textbf{Interest} & \textbf{Principal reduction} & \textbf{Balance} \\
\textbf{number} & \textbf{(\$) } & \textbf{(\$) } & \textbf{(\$) } & \textbf{(\$) }\\
\hline
\rule{0pt}{2.5ex} 0 \rule[-1ex]{0pt}{0pt} & 0.00 & 0.00 & 0.00 & 600\ 000.00 \\
\hline
\rule{0pt}{2.5ex} 1 \rule[-1ex]{0pt}{0pt} & 3973.00 & 2520.00 & 1453.00& 598\ 547.00\\
\hline
\rule{0pt}{2.5ex} 2 \rule[-1ex]{0pt}{0pt} & 3973.00 & 2513.90 & 1459.10 & 597\ 087.90 \\
\hline
\end{array}

The interest rate for the annuity is 0.42% per month.
Determine the interest rate per annum. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Using the values in the table, complete the next line of the amortisation table.
Write your answers in the spaces provided in the table below.
Round all values to the nearest cent. (1 mark)

--- 0 WORK AREA LINES (style=lined) ---

Let $V_n$ be the balance of Arthur's annuity, in dollars, after $n$ months.
Write a recurrence relation in terms of $V_0, V_{n+1}$ and $V_n$ that can model the value of the annuity from month to month. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
The amortisation tables above show that the balance of the annuity reduces each month.
If the balance of an annuity remained constant from month to month, what name would be given to this type of annuity? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $I\% (\text{annual}) = 12 \times 0.42 = 5.04\% $

b. $\text{Row 3 calculations are as follows:}$

$\text{Payment}\ = \$3973.00\ \text{(remains constant)}$

$\text{Interest}\ = 597\ 087.90 \times 0.0042 = \$2507.77 $

$\text{Principal reduction}\ = 3973.00-2507.77 = \$1465.23 $

$\text{Balance}\ = 597\ 087.90-1465.23 = \$595\ 622.67$

c. $V_0 = 600\ 000$

$V_{n+1} = 1.0042 \times V_n-3973$

d. $\text{Perpetuity}$

Show Worked Solution

a. $I\% (\text{annual}) = 12 \times 0.42 = 5.04\% $

b. $\text{Row 3 calculations are as follows:}$

$\text{Payment}\ = \$3973.00\ \text{(remains constant)}$

$\text{Interest}\ = 597\ 087.90 \times 0.0042 = \$2507.77 $

$\text{Principal reduction}\ = 3973.00+2507.77 = \$1465.23 $

$\text{Balance}\ = 597\ 087.90-1465.23 = \$595\ 622.67$

♦♦ Mean mark (b) 40%.

c. $V_0 = 600\ 000$

$V_{n+1} = 1.0042 \times V_n-3973$

♦ Mean mark (c) 41%.

d. $\text{Perpetuity}$

CORE, FUR2 2021 VCAA 9

Sienna invests $152 431 into an annuity from which she will receive a regular monthly payment of $900 for 25 years. The interest rate for this annuity is 5.1% per annum, compounding monthly.

Let `V` be the balance of the annuity after `n` monthly payments. A recurrence relation written in terms of `V_0 , V_{n + 1}` and `V_n` can model the value of this annuity from month to month.
Showing recursive calculations, determine the value of the annuity after two months.
Round your answer to the nearest cent. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
After two years, the interest rate for this annuity will fail to 4.6%.
To ensure that she will still receive the same number of $900 monthly payments, Sienna will add an extra one-pff amount into the annuity at this time.
Determine the value of this extra amount that Sienna will add.
Round your answer to the nearest cent. (1 mark)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`$151 \ 925.59`
`$7039.20`

Show Worked Solution

a. `text{Payments are monthly} \ => \ r = 5.1/12 = 0.425text(%) \ text{per month}`

`R`	`= 1 + r/100 = 1.00425`
`V_1`	`= RV_0-text{payment}`
	`= 1.00425 xx 152 \ 431-900`
	`= $152\ 178.83`
`:.V_2`	`= 1.00425 xx 152\ 178.83-900`
	`= $ 151 \ 925.59`

b. `text{Find} \ V_24 \ text{(annuity value after 2 years) by TVM Solver:}`

`N`	`= 24`
`I text{(%)}`	`= 5.1`
`PV`	`= -152 \ 431`
`PMT`	`= 900`
`FV`	`= ?`
`text(P/Y)`	`= text(C/Y) = 12`

`=> FV = 146 \ 073.7405`

`text{Find}\ PV\ text{of annuity needed (by TVM solver):}`

`N`	`= 243 xx 12 = 276`
`Itext{(%)}`	`= 4.6`
`PV`	`= ?`
`PMT`	`= 900`
`FV`	`= 0`
`text(P/Y)`	`= text(C/Y) = 12`

`=> PV = -153\ 112.9399`

`:. \ text{Amount to add}`	`= 153 \ 112.94-146 \ 073.74`
	`= $ 7039.20`

CORE, FUR2 2021 VCAA 6

Sienna invests $420 000 in a perpetuity from which she will receive a regular monthly payment of $1890.

The perpetuity earns interest at the rate of 5.4% per annum.

Determine the total amount, in dollars, that Sienna will receive after one year of monthly payments. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Write down the value of the perpetuity after Sienna has received one year of monthly payments. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
Let `S_n` be the value of Sienna's perpetuity after `n` months.
Complete the recurrence relation, in terms of `S_0`, `S_{n + 1}` and `S_n`, that would model the value of this perpetuity over time. Write your answers in the boxes provided. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

`S_n =`

, `S_{n+1} =`

`xx S_n-1890`

Show Answers Only

`$ 22 \ 680`
`$ 420 \ 000`
`S_n = 420 \ 000, \ \ S_{n+1} = 1.0045 xx S_n-1890`

Show Worked Solution

a.	`text{Total amount}`	`= 12 xx 1890`
		`= $ 22 \ 680`

b. `text{Value} = $ 420 \ 00 \ text{(balance remains the same after each payment).}`

c. `S_0 = 420 \ 000 \ => \ S_n = 420 \ 000`

`S_{n+1} = RS_n-1890 \ \ text{where}\ \ R = 1 + r/100`

`r = 5.4/12 = 0.45`

`:. \ S_{n+1} = 1.0045 xx S_n-1890`

CORE, FUR2 2020 VCAA 10

Samuel now invests $500 000 in an annuity from which he receives a regular monthly payment.

The balance of the annuity, in dollars, after `n` months, `A_n` , can be modelled by a recurrence relation of the form

`A_0 = 500\ 000, qquad A_(n+1) = kA_n - 2000`

Calculate the balance of this annuity after two months if `k = 1.0024`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Calculate the annual compound interest rate percentage for this annuity if `k = 1.0024`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
For what value of `k` would this investment act as a simple perpetuity? (1 mark)

--- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

`$498\ 398.08`
`2.88 text(%)`
`1.004`

Show Worked Solution

a.	`A_1 = 1.0024 xx 500\ 000-2000 = $499\ 200`
	`A_2 = 1.0024 xx 499\ 200-2000 = $498\ 398.08`

♦ Mean mark 48%.

b.	`text(Monthly interest rate)`	`= (1.0024-1) xx 100 = 0.24text(%)`
	`text(Annual interest rate)`	`= 12 xx 0.24 = 2.88text(%)`

♦ Mean mark 36%.

c.	`text(Perpetuity would occur when)`
	`k xx 500\ 000-2000`	`= 500\ 000`
	`k`	`= (502\ 000)/(500\ 000)`
		`= 1.004`

Financial Maths, GEN2 2019 NHT 7

Tisha plays drums in the same band as Marlon.

She would like to buy a new drum kit and has saved $2500.

Tisha could invest this money in an account that pays interest compounding monthly.

The balance of this investment after `n` months, `T_n` could be determined using the recurrence relation below

`T_0 = 2500, \ \ \ \ T_(n+1) = 1.0036 xx T_n`

Calculate the total interest that would be earned by Tisha's investment in the first five months.

Round your answer to the nearest cent. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Tisha could invest the $2500 in a different account that pays interest at the rate of 4.08% per annum, compounding monthly. She would make a payment of $150 into this account every month.

Let `V_n` be the value of Tisha's investment after `n` months.

Write down a recurrence relation, in terms of `V_0`, `V_n` and `V_(n + 1)`, that would model the change in the value of this investment. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Tisha would like to have a balance of $4500, to the nearest dollar, after 12 months.

What annual interest rate would Tisha require?

Round your answer to two decimal places. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

`$45.33`
`V_0 = 2500, \ V_(n+1) = 1.0034 xx V_n + 150`
`5.87%`

Show Worked Solution

a. `T_1 = 1.0036 xx 2500 = 2509`

`T_2 = 1.0036 xx 2509 = 2518.0324`

`vdots`

`T_5 = 2545.33`

`:. \ text(Total interest) `	`= 2545.33-2500`
	`= $45.33`

b. `text(Monthly interest) = (4.08)/(12) = 0.34%`

`:. \ V_0 = 2500, \ V_(n+1) = 1.0034 xx V_n + 150`

c. `text(By TVM Solver:)`

`N`	`= 12`
`I text(%)`	`=?`
`PV`	`=-2500`
`PMT`	`=-150`
`FV`	`=4500`
`text(PY)`	`= text(CY)=12`

`=> I = 5.87%`

CORE, FUR1 2018 VCAA 21 MC

Which one of the following recurrence relations could be used to model the value of a perpetuity investment, `P_n`, after `n` months?

`P_0 = 120\ 000,qquadP_(n + 1) = 1.0029 × P_n - 356`
`P_0 = 180\ 000,qquadP_(n + 1) = 1.0047 × P_n - 846`
`P_0 = 210\ 000,qquadP_(n + 1) = 1.0071 × P_n - 1534`
`P_0 = 240\ 000,qquadP_(n + 1) = 0.0047 × P_n - 2232`
`P_0 = 250\ 000,qquadP_(n + 1) = 0.0085 × P_n - 2125`

Show Answers Only

`B`

Show Worked Solution

`text(Perpetuity) => P_0 = P_1 = P_2 = … = P_n`

`text(By trial and error, consider option)\ B:`

`P_1`	`= 1.0047 xx 180\ 000 – 846`
	`= 180\ 000`
	`= P_0`

`=> B`

CORE, FUR1 2018 VCAA 17-18 MC

The value of an annuity investment, in dollars, after `n` years, `V_n` , can be modelled by the recurrence relation shown below.

`V_0 = 46\ 000, quadqquadV_(n + 1) = 1.0034V_n + 500`

Part 1

What is the value of the regular payment added to the principal of this annuity investment?

$34.00
$156.40
$466.00
$500.00
$656.40

Part 2

Between the second and third years, the increase in the value of this investment is closest to

$656
$658
$661
$1315
$1975

Show Answers Only

`text(Part 1:)\ D`

`text(Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

`text(Regular payment = $500)`

`=> D`

`text(Part 2)`

`V_1`	`= 1.0034 xx 46\ 000 + 500 = $46\ 656.40`
`V_2`	`= 1.0034 xx 46\ 656.40 + 500 = $47\ 315.03`
`V_3`	`= 1.0034 xx 47\ 315.03 + 500 = $47\ 975.90`

`:.\ text(Increase)`	`= 47\ 975.90 – 47\ 315.03`
	`= $660.87`

`=> C`

CORE, FUR1 SM-Bank 1-2 MC

Part 1

Candy deposits $80 000 into a savings account that earns 5% interest per annum, compounded annually.

At the end of the first year, Candy withdraws $5600.

What is the balance of Candy's savings account at the start of the second year?

`$74\ 400`
`$77\ 600`
`$78\ 400`
`$85\ 600`
`$89\ 600`

Part 2

If `B_n` is the balance of Candy's account at the start of year `n`, a recurrence equation that can be used to model the balance of the account over time is

`B_(n + 1) = 1.5 B_n - 5600\ \ \ \ text(where)\ B_1 = 80\ 000`
`B_(n + 1) = 0.05 B_n - 5600\ \ \ \ text(where)\ B_1 = 80\ 000`
`B_(n + 1) = 1.05 (B_n - 5600)\ \ \ \ text(where)\ B_1 = 80\ 000`
`B_(n + 1) = 0.05 (B_n - 5600)\ \ \ \ text(where)\ B_1 = 80\ 000`
`B_(n + 1) = 1.05 B_n - 5600\ \ \ \ text(where)\ B_1 = 80\ 000`

Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ E`

Show Worked Solution

`text(Part 1)`

`text(Balance at start of 2nd year)`

`= 80\ 000 xx 1.05 – 5600`

`= $78\ 400`

`=> C`

`text(Part 2)`

`B_2`	`= 80\ 000 xx 1.05 – 5600`
	`= 1.05B_1 – 5600`
`B_3`	`= 1.05B_2 – 5600`
`vdots`
`B_(n + 1)`	`= 1.05B_n – 5600`

`=> E`

CORE, FUR2 SM-Bank 2

Spiro is saving for a car. He has an account with $3500 in it at the start of the year.

At the end of each month, Spiro adds another $180 to the account.

The account pays 3.6% interest per annum, compounded monthly.

1. What is the interest rate per month? (1 mark)
  
  --- 1 WORK AREA LINES (style=lined) ---
2. Write a recurrence relation that models Spiro's investment, with `V_n` representing the balance of his account after `n` months. (1 mark)
  
  --- 4 WORK AREA LINES (style=lined) ---
What will be the balance of Spiro's account after 3 months?
Write your answer correct to the nearest cent. (1 mark)
--- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

1. `0.3text(% per month)`
2. `V_0 = 3500,qquadV_(n + 1) = V_n xx 1.003 + 180`
`$4073\ \ (text(nearest $))`

Show Worked Solution

a.i.	`text(Interest rate)`	`= 3.6/12`
		`= 0.3text(% per month)`

a.ii.	`V_0`	`= 3500`
	`V_1`	`= 3500 xx 1.003 + 180`
	`V_2`	`= V_1 xx 1.003 + 180`
	`vdots`
	`V_(n + 1)`	`= V_n xx 1.003 + 180`

`:.\ text(Recurrence relationship:)`

`V_0 = 3500,qquadV_(n + 1) = V_n xx 1.003 + 180`

b.	`V_1`	`= 3500 xx 1.003 + 180 = $3690.50`
	`V_2`	`= 3690.50 xx 1.003 + 180 = $3881.5715`

`V_3`	`= 3881.5715 xx 1.003 + 180`	`= $4073.216…`
		`= $4073.22\ \ text{(nearest cent)}`

CORE, FUR1 2016 VCAA 18 MC

The value of an annuity, `V_n`, after `n` monthly payments of $555 have been made, can be determined using the recurrence relation

`V_0 = 100\ 000,\ \ \ \ \ V_(n + 1) = 1.0025 V_n - 555`

The value of the annuity after five payments have been made is closest to

`$97\ 225`
`$98\ 158`
`$98\ 467`
`$98\ 775`
`$110\ 224`

Show Answers Only

`C`

Show Worked Solution

`V_0`	`= 100\ 000`
`V_1`	`= 1.0025 xx 100\ 000 – 555 = 99\ 695`
`V_2`	`= 1.0025 xx 99\ 695 – 555 = 99\ 389.23…`
`V_3`	`= 1.0025 xx 99\ 389.24 – 555 = 99\ 082.71…`
`V_4`	`= 1.0025 xx V_3 = 98\ 775.41…`
`V_5`	`= 1.0025 xx V_4 = 98\ 467.35…`

`=> C`