finance application (arithmetic sequence)

Finance applications （arithmetic sequences and recurrence relation)

• use a recurrence relation to model simple interest.

• use a recurrence relation to model flat rate depreciation.

• use a recurrence relation to model unit cost depreciation.

• use a rule to determine the nth term for linear growth or decay.

©

Learning intentions

​Simple interest is usually determined by multiplying the annual interest rate, r%, by the original amount borrowed (invested) called the principal, P, for each year of the loan.

The value after n + 1 years is the value after n years plus the interest in the previous year.

This gives the recurrence relation:

Fill in the Blank

An investment of $3500 pays interest at the rate of 4.2% per annum in the form of simple interest.

Find the amount of interest paid each year.

Fill in the Blank

An investment of $4600 pays interest at the rate of 5.1% per annum in the form of simple interest.

Find the amount of interest paid each year.

.

Model linear growth

The following recurrence relation can be used to model a simple interest investment of $2000, paying interest at the rate of 7.5% per annum:

V0 = 2000, Vn+1 = Vn + 150,

where Vn is the value of the investment after n years.

a. Use the recurrence relation to find the value of the investment after 1, 2 and 3 years.

The following recurrence relation can be used to model a simple interest investment of $2000, paying interest at the rate of 7.5% per annum:

V0 = 2000, Vn+1 = Vn + 150,

where Vn is the value of the investment after n years.

b When will the investment reach $2750 in value?

flat rate depreciation

One way that depreciation is calculated is based on the age of the asset. This is called flat rate or fixed rate depreciation.

The following recurrence relation can be used to model the flat rate depreciation of a car purchased for $18 500, depreciating at a flat rate of 10% per year.

V0 = 18 500, Vn+1 = Vn − 1850

In the recurrence relation, Vn is the value of the car after n years.

Fill in the Blank

Use the recurrence relation to find the value of the car after 1, 2 and 3 years.

V0 = 18 500, Vn+1 = Vn − 1850

,

,

Fill in the Blank

How long will it take for the car’s value to depreciate to zero?

V0 = 18 500, Vn+1 = Vn − 1850

​Unit cost depreciation

A car with a purchase price of $32 000 depreciates at a unit cost of $150 per 1000 kilometres.

a State the recurrence relation for unit cost depreciation of the car, where Vn represents the value of the car after n thousand kilometres.

b Use the recurrence relation to find the value of the car after 1000, 2000 and 3000 kilometres.

c How many kilometres is the car expected to travel before its value depreciates by at least $1000?

Model linear growth or decay

Fill in the Blank

The following recurrence relation can be used to model a simple-interest investment of $4000, paying interest at 6.5% per year.

V0 = 4000, Vn+1 = Vn + 260

a. How much interest is added to the investment each year?

Fill in the Blank

The following recurrence relation can be used to model a simple-interest investment of $4000, paying interest at 6.5% per year.

V0 = 4000, Vn+1 = Vn + 260

b. Use a rule to find the value of the investment after 15 years.

Fill in the Blank

The following recurrence relation can be used to model a simple-interest investment of $4000, paying interest at 6.5% per year.

V0 = 4000, Vn+1 = Vn + 260

c. Use a rule to find when the value of the investment first exceeds $10 000.

Multiple Choice

Which one is linear growth?

• A

• B

• C

• D

Multiple Choice

Which one is linear decay?

• A

• B

• C

• D