Topics
6. Loans, Investments, & Annuities
topic badge
AustraliaAU
Year 12

6.01 Simple and compound interest using sequences

Lesson
Worksheet
Practice
Lesson

Introduction

Interest is the extra money that banks and lenders charge us to borrow money. It may also refer to additional money you earn from depositing money, such as in a savings account. There are two different types of interest simple interest and compound interest.

We can solve problems involving simple and compound interest in three ways:

  • Using the simple interest or compound interest formula.

  • Using sequences - recursive or explicit forms.

  • Using the financial application of your CAS calculator.

Simple interest formula

Simple interest is a method where the interest amount is fixed (i.e. it doesn't change). This fixed interest charge is based on the original amount, which is called the principal. Simple interest can be calculated using the simple interest formula: I=PRT, where P is the principal (the initial amount borrowed or invested), R is the interest rate per time period, expressed as a decimal or fraction and T is the number of time periods (the duration of the loan or deposit). Note that this formula calculates the interest, not the final balance.

Examples

Example 1

Calculate the simple interest earned on an investment of \$5440 at 6\% p.a. for 566 days.

Assume that a year has 365 days and write your answer to the nearest cent.

Worked Solution
Create a strategy

Use the simple interest formula given by I=PRT.

Apply the idea

We are given: P=\$5440, \, R=6\%=0.06 and T=566 \text{ days}

Since our rate is per annum, we will need to convert our number of days into some number of years. So we need to divide the time period by 365 to get: T= \dfrac{566}{365} .

\displaystyle I\displaystyle =\displaystyle 5440\times 0.06 \times \dfrac{566}{365}Substitute P, \, R, \, T
\displaystyle =\displaystyle \$506.14Evaluate
Idea summary

Simple interest formula:

\displaystyle I=PRT
\bm{P}
is the principal (the initial amount borrowed or invested)
\bm{R}
is the interest rate per time period, expressed as a decimal or fraction
\bm{T}
is the number of time periods (the duration of the loan or deposit)

Simple interest and sequences

With simple interest the balance is increased or decreased by adding or subtracting the same amount every time, therefore simple interest problems can also be modelled using an  arithmetic sequence  . You can think of it as "next value equals the value before plus the simple interest".

Consider the following problem: James invests \$15\, 000 into an investment account that pays simple interest of 3.2\% per annum. The table below shows the value of the investment over the first four years.

\text{Month}01234
\text{Balance } \left(V_{n} \right)15\, 00015\, 48015\, 98016\, 44016\, 920

We can see this is an arithmetic sequence with a=15\,000 and d=480.

The recursive form is V_{n+1} = V_{n}+480, where V_{0}=15\,000.

Note: When creating a sequence to generate the value of the investment at the end of each period we let V_{0} equal to the initial amount so that V_{1} is then the amount at the end of the first instalment period. This makes it easier to answer questions involving the value of the investment over time.

For a principal investment/loan, P, at the simple interest rate of r per period, the sequence of the value of the investment over time forms an arithmetic sequence with a starting value of P=a and a common d=a\times r.

The sequence which generates the value, V_{n}, of the investment/loan at the end of each instalment period is:

  • Recursive form: V_{n}=V_{n-1}+d, where V_{0}=a

  • Explicit form: V_{n}=a + nd

The sequence which generates the value, V_{n}, of the investment/loan at the beginning of each instalment period is:

  • Recursive form: V_{n}=V_{n-1}+d, where V_{1}=a

  • Explicit form: V_{n}=a + \left(n-1 \right)d

Examples

Example 2

Manpreet lives in India and invests 46 \,000 INR into an investment account that pays 6.6\% simple interest per annum.

a

By what amount will the account increase each year?

Worked Solution
Create a strategy

Use the simple interest formula given by I=PRT.

Apply the idea

We are given: P=\$46\,000, \, R=6.6\%=0.066 and T=1 \text{ year}

\displaystyle I\displaystyle =\displaystyle PRTUse the formula
\displaystyle =\displaystyle 46\,000 \times 0.066 \times 1 Substitute the values
\displaystyle =\displaystyle 3036 \text{ INR}Evaluate

The account will increase by the interest of 3036 \text{ INR} each year.

b

Complete the recurrence relation for Manpreet's situation, where t_{n} is the balance at the end of the nth year and t_{0} is the initial investment:

t_{n+1}=t_{n} + ⬚, where t_{0}=⬚.

Worked Solution
Create a strategy

Use the initial amount and the amount we are increasing by each year.

Apply the idea

We are given an initial investment of t_{0}=46\,000, and we know from part (a) that we are adding 3036 \text{ INR} each year to the previous amount.

So the recurrence relation is:

t_{n+1}=t_{n} + 3036, where t_{0}=46\,000.

c

Complete and then simplify the explicit rule that can be used to find the balance at the end of n years.

t_{n}=⬚+(n-1) \times ⬚, which simplifies to t_n=⬚.

Worked Solution
Create a strategy

Use the formula t_n=a+\left(n-1 \right)d where a=t_1.

Apply the idea

Since the balance is increasing by 3036 each year, d=3036. The initial balance was 46\,000. But a=t_1 since it is the value of the account after 1 year. So we can add t_0 and d to find a.

\displaystyle a\displaystyle =\displaystyle t_0+dAdd the initial value and d
\displaystyle =\displaystyle 46\,000+3036Substitute the values
\displaystyle =\displaystyle 49\,036Evaluate
\displaystyle t_{n}\displaystyle =\displaystyle a+\left(n-1 \right)dUse the explicit rule
\displaystyle =\displaystyle 49\,036 + \left(n-1 \right) \times 3036Substitute the values
\displaystyle =\displaystyle 46\,036 + 3036n -3036Expand the brackets
\displaystyle =\displaystyle 46\,000 + 3036nSimplify

The completed statement is:

t_{n}=49\,036+(n-1) \times 3036, which simplifies to t_n=46\,000 + 3036n.

d

Determine the balance after 9 years.

Worked Solution
Create a strategy

Use the equation found in part (c).

Apply the idea

We need to find the balance after 9 years so we will be using n=9.

\displaystyle t_{9}\displaystyle =\displaystyle 46\,000 + 3036 \times 9Substitute n=9
\displaystyle =\displaystyle 73\,324 \text{ INR}Evaluate
e

Determine how many whole years it takes for the balance to exceed 86\,986 INR.

Worked Solution
Create a strategy

Use the equation found in part (c).

Apply the idea

We want the expression for t_n to be greater than 86\,986. So we need to solve the inequality 6\,000 + 3036n \gt 86\,986.

\displaystyle 46\,000 + 3036n\displaystyle >\displaystyle 86\,986Write the inequality
\displaystyle 3036n\displaystyle >\displaystyle 40\,986Subtract 46\,000 from both sides
\displaystyle n\displaystyle >\displaystyle \dfrac{40\,986}{3036}Divide both sides by 3036
\displaystyle >\displaystyle 13.5 Evaluate

Since we want the number of whole years, the answer is 14 years.

Idea summary

The sequence which generates the value, V_{n}, of the investment/loan at the end of each instalment period is:

  • Recursive form: V_{n}=V_{n-1}+d, where V_{0}=a
  • Explicit form: V_{n}=a + nd

The sequence which generates the value, V_{n}, of the investment/loan at the beginning of each instalment period is:

  • Recursive form: V_{n}=V_{n-1}+d, where V_{1}=a
  • Explicit form: V_{n}=a + \left(n-1 \right)d

Compound interest formula

Most of the time, when banks and financial institutions calculate interest, they are using compound interest.

Compound interest is calculated at the end of each compounding period, which is typically a day, month, quarter, or year. At the end of each compounding period, the total amount (principal plus interest) from previous compounding periods is used to calculate the new quantity of interest. We multiply the total amount by the interest and then add it to the total. Note that the compound interest formula calculates the final balance, or amount. To find the amount of interest we need to subtract the principal from the final balance.

The compound interest formula is: A=P \left(1+\dfrac{r}{n} \right)^{nt} where A is the final amount of money (principal and interest together), P is the principal (the initial amount of money invested), r is the interest rate per year, expressed as a decimal, t is the number of years and n is the number of compounding periods in one year (eg: quarterly means n=4).

Note: This formula is often written in the form A=P \left(1+r\right)^{t}. In this form r refers to the rate per period and n refers to the total number of periods.

Examples

Example 3

A \$3710 investment earns interest at 4.8\% p.a. compounded quarterly over 13 years. Use the compound interest formula to calculate the value of this investment to the nearest cent.

Worked Solution
Create a strategy

Use the compound interest formula given by A= P \times \left(1+\dfrac{r}{n}\right)^{nt}.

Apply the idea

We are given that P=3710, \, r=4.8\%=0.048 and t=13.

The investment is compounded quarterly so n=4.

\displaystyle A\displaystyle =\displaystyle P\times \left(1+\dfrac{r}{n} \right)^{nt}Write the formula
\displaystyle =\displaystyle 3710 \times \left(1+ \dfrac{0.048}{4} \right)^{4\times13}Substitute the values
\displaystyle =\displaystyle \$6898.59Evaluate
Idea summary

The compound interest formula:

\displaystyle A=P \left(1+\dfrac{r}{n} \right)^{nt}
\bm{A}
is the final amount of money
\bm{P}
is the principal
\bm{r}
is the interest rate per year, expressed as a decimal
\bm{t}
is the number of years
\bm{n}
is the number of compounding periods in one year

Alternative formula:

\displaystyle A=P \left(1+r \right)^{n}
\bm{A}
is the final amount of money
\bm{P}
is the principal
\bm{r}
is the interest rate per period
\bm{n}
is the number of compounding periods

Compound interest using sequences

With compound interest the balance is increased by multiplying the same amount every time, therefore compound interest problems can be modelled using a  geometric sequence  . The "next term" is made by increasing the term before by the interest rate percentage. Therefore you can think of it as "next value equals the value before multiplied by (1 + the interest rate as a decimal)".

Consider the following problem:

Emma puts \$5000 into an investment account paying a compound interest rate of 4.2\% p.a. The table below shows the value of the investment over the first four years.

\text{Year } \left(n\right)\text{Calculation}\text{Amount } \left(V_n\right)
05000
15000 \times 1.042 =5210
25210 \times 1.042 =5428.82
35428.82 \times 1.042 =5656.83
45656.83 \times 1.042 =5984.42

We can see that the next value is equal to the previous value multiplied by 1.042. Therefore we have a geometric sequence and the recursive rule for this investment is V_{n+1}=1.042 \times V_{n}, where V_{0}=5000.

For a principal investment/loan, P, at the compound interest rate of r per period, the sequence of the value of the investment over time forms a geometric sequence with a starting value of P=a and a common ratio of \left(1+r \right).

The sequence which generates the value, V_{n}, of the investment/loan at the end of each instalment period is:

  • Recursive form: V_{n}=V_{n-1} \times \left(1+r \right), where V_{0}=a

  • Explicit form: a\left(1+r \right)^{n}

The sequence which generates the value, V_{n}, of the investment/loan at the beginning of each instalment period is:

  • Recursive form: V_{n}=V_{n-1} \times \left(1+r \right), where V_{1}=a

  • Explicit form: V_{n}=a \left(1+r \right)^{n-1}

Examples

Example 4

The balance of an investment, in dollars, at the end of each year where interest is compounded annually is given by A_{n}=1.05A_{n-1},\, A_{0}=30\,000.

a

State the annual interest rate.

Worked Solution
Create a strategy

Use the recursive form: V_{n}=V_{n-1} \times \left(1+r \right), where V_{0}=a

Apply the idea

Comparing the given expresion for A_{n}=1.05A_{n-1} to the recursive form V_{n}=V_{n-1} \times \left(1+r \right), we can see that the multipliers must be equal: 1+r =1.05.

\displaystyle 1+r\displaystyle =\displaystyle 1.05Write the equation
\displaystyle r\displaystyle =\displaystyle 1.05-1Subtract 1 from both sides
\displaystyle =\displaystyle 0.05Evaluate
\displaystyle =\displaystyle 5\%Write as a percentage
b

State the amount invested.

Worked Solution
Create a strategy

Use the recursive form: V_{n}=V_{n-1} \times \left(1+r \right), where V_{0}=a

Apply the idea

The amount invested is A_0 = \$30\,000.

c

Determine the balance at the end of the first year.

Worked Solution
Create a strategy

Substitute the value of n into the given recursive equation.

Apply the idea

We need to determine the balance after 1 year so n=1.

\displaystyle A_{n}\displaystyle =\displaystyle 1.05A_{n-1}Use the recursive rule
\displaystyle A_{1}\displaystyle =\displaystyle 1.05A_{0}Substitute n=1
\displaystyle =\displaystyle 1.05 \times 30\,000Substitute A_{0}=30\,000
\displaystyle =\displaystyle \$31\,5000Evaluate
d

Use the sequences facility on your calculator to calculate the balance at the end of 15 years. Round your answer to the nearest cent.

Worked Solution
Create a strategy

Use the sequences facility on your calculator and enter the recursive rule and value of n.

Apply the idea

Enter the recursive rule A_{n}=1.05A_{n-1}, \, A_0=30\,000, and n=15 to get: \$62\,367.85

Example 5

The following table shows the balance (in dollars) in a savings account in 2014, where interest is compounded monthly.

Balance at the beginning of the monthInterestBalance at the end of month
July8000160X
August8160163.208323.20
September8323.20Y8489.66
OctoberZ169.798659.45
November8659.45173.198832.64
a

Calculate the value of X.

Worked Solution
Create a strategy

Add the beginning balance and the interest.

Apply the idea

The balance at the beginning of July was \$8000 and the interest accrued during July was \$160.

\displaystyle X\displaystyle =\displaystyle 8000+160Add the balance and interest
\displaystyle =\displaystyle \$8160Evaluate
b

Use the numbers for July to calculate the monthly interest rate.

Worked Solution
Create a strategy

Divide the interest by the beginning balance.

Apply the idea
\displaystyle \text{Interest rate}\displaystyle =\displaystyle \dfrac{160}{8000}Divide the interest by the balance
\displaystyle =\displaystyle 0.02Evaluate
\displaystyle =\displaystyle 2\%Write as a percentage
Reflect and check

We could have also used the compound interest formula A=P(1+r)^n, where A=8160, \, P=8000, and n=1 period.

\displaystyle 8160\displaystyle =\displaystyle 8000(1+r)^1Substitute the values
\displaystyle 8160\displaystyle =\displaystyle 8000+8000rExpand the brackets
\displaystyle 160\displaystyle =\displaystyle 8000rSubtract 8000 from both sides
\displaystyle r\displaystyle =\displaystyle \dfrac{160}{8000}Divide both sides by 8000
\displaystyle =\displaystyle 0.02Evaluate
c

Calculate the value of Y.

Worked Solution
Create a strategy

Find the difference between the balance at the end of the month and the balance at the beginning of the month.

Apply the idea

The beginning balance plus Y should equal the end balance. So to find Y we can subtract the beginning balance from the end balance.

\displaystyle Y\displaystyle =\displaystyle 8489.66-8323.20Subtract the balances
\displaystyle =\displaystyle \$166.46Evaluate
d

Calculate the value of Z.

Worked Solution
Create a strategy

Copy the balance at the end of September.

Apply the idea

The balance at the beginning of October is the same as the balance at the end of September which is \$8489.66. So:

\displaystyle Z\displaystyle =\displaystyle \$8489.66Copy the balance
e

Write a recursive rule, B_{n}, that gives the balance at the end of the nth month, with July being the first month.

Write both parts of the rulw including B_0.

Worked Solution
Create a strategy

Use the recursive form: V_{n}=V_{n-1} \times \left(1+r \right), where V_{0}=a.

Apply the idea

The balance at the beginning of July was \$8000 which will be B_0. From part (b) we found the rate to be 0.02.

\displaystyle B_n\displaystyle =\displaystyle B_{n-1}\times (1+0.02)Substitute r=0.02
\displaystyle =\displaystyle 1.02B_{n-1}Simplify

So the recursive rule is: B_{n}=1.02B_{n-1}, \, B_{0}=8000

f

Write an explicit rule for B_{n}, the balance at the end of the nth month, with July being the first month.

Worked Solution
Create a strategy

Use the explicit form V_n=V_0(1+r)^n.

Apply the idea

From the previous part we know that B_0=8000 and r=0.02.

\displaystyle B_{n}\displaystyle =\displaystyle B_{0} \left(1+r \right)^{n}Write the rule
\displaystyle B_{n}\displaystyle =\displaystyle 8000 \left(1.02 \right)^{n}Substitute the values
Idea summary

The sequence which generates the value, V_{n}, of the investment/loan at the end of each instalment period is:

  • Recursive form: V_{n}=V_{n-1}+d, where V_{0}=a

  • Explicit form: V_{n}=a + nd

The sequence which generates the value, V_{n}, of the investment/loan at the beginning of each instalment period is:

  • Recursive form: V_{n}=V_{n-1}+d, where V_{1}=a

  • Explicit form: V_{n}=a + \left(n-1 \right)d

Outcomes

ACMGM094

use a recurrence relation to model a compound interest loan or investment, and investigate (numerically or graphically) the effect of the interest rate and the number of compounding periods on the future value of the loan or investment

ACMGM096

with the aid of a calculator or computer-based financial software, solve problems involving compound interest loans or investments; for example, determining the future value of a loan, the number of compounding periods for an investment to exceed a given value, the interest rate needed for an investment to exceed a given value

What is Mathspace

About Mathspace