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:
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 formula below. Note that this formula calculates the interest, not the final balance.
$I=Prt$I=Prt
where $P$P is the principal (the initial amount borrowed or invested)
$r$r is the interest rate per time period, expressed as a decimal or fraction
$t$t is the number of time periods (the duration of the loan or deposit)
Calculate the simple interest earned on an investment of $\$5440$$5440 at $6%$6% p.a. for $566$566 days.
Assume that a year has $365$365 days and write your answer to the nearest cent.
With simple interest the balance is increased or decreased by adding or subtracting the same amount every time, therefore simple interest problems can also be modelled using an arithmetic sequence. You can think of it as "next value equals the value before plus the simple interest".
Consider the following problem: James invests $\$15000$$15000 into an investment account that pays simple interest of $3.2%$3.2% per annum. The table below shows the value of the investment over the first four years.
Month ($n$n) | 0 | 1 | 2 | 3 | 4 |
---|---|---|---|---|---|
Balance ($V_n$Vn) | $15000$15000 | $15480$15480 | $15960$15960 | $16440$16440 | $16920$16920 |
We can see this is an arithmetic sequence with $a=15000$a=15000 and $d=480$d=480.
The recursive form is $V_{n+1}=V_n+480$Vn+1=Vn+480, where $V_0=15000$V0=15000.
Note: When creating a sequence to generate the value of the investment at the end of each period we let $V_0$V0 equal to the initial amount so that $V_1$V1 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, $P$P, at the simple interest rate of $r$r per period, the sequence of the value of the investment over time forms an arithmetic sequence with a starting value of $P=a$P=a and a common difference $d=a\times r$d=a×r.
The sequence which generates the value, $V_n$Vn, of the investment/loan at the end of each instalment period is:
$V_n=V_{n-1}+d$Vn=Vn−1+d, where $V_0=a$V0=a
$V_n=a+nd$Vn=a+nd
The sequence which generates the value, $V_n$Vn, of the investment/loan at the beginning of each instalment period is:
$V_n=V_{n-1}+d$Vn=Vn−1+d, where $V_1=a$V1=a
$V_n=a+\left(n-1\right)d$Vn=a+(n−1)d
Select the brand of calculator you use below to work through an example of using a calculator for simple interest problems involving sequences.
Casio Classpad
How to use the CASIO Classpad to complete the following tasks regarding sequences in a simple interest context.
Consider an investment of $\$1000$$1000 at $5%$5% simple interest.
Using a recursive relationship, generate the value of the investment at the end of each year for $10$10 years.
Using an explicit rule, generate the value of the investment at the end of each year for $10$10 years.
Find the balance in the account at the end of $15$15 years.
Find when the investment will reach $\$2000$$2000.
Graph the balance of the account over the first ten years.
TI Nspire
How to use the TI Nspire to complete the following tasks regarding sequences in a simple interest context.
Consider an investment of $\$1000$$1000 at $5%$5% simple interest.
Using a recursive relationship, generate the value of the investment at the end of each year for $10$10 years.
Using an explicit rule, generate the value of the investment at the end of each year for $10$10 years.
Find the balance in the account at the end of $15$15 years.
Find when the investment will reach $\$2000$$2000.
Graph the balance of the account over the first ten years.
Manpreet lives in India and invests $58000$58000 INR into an investment account that pays $6.6%$6.6% simple interest per annum.
By what amount will the account increase each year?
Complete the recurrence relation for Manpreet's situation, where $t_n$tn is the balance at the end of the $n$nth year and $t_0$t0 is the initial investment.
$t_{n+1}=t_n$tn+1=tn$+$+$\editable{}$, where $t_0$t0$=$=$\editable{}$.
Complete and then simplify the explicit rule that can be used to find the balance at the end of $n$n years.
$t_n=$tn=$\editable{}$$+\left(n-1\right)\times$+(n−1)×$\editable{}$ which simplifies to $t_n=$tn=$\editable{}$.
Determine the balance after $5$5 years.
Determine how many whole years it takes for the balance to exceed $102022$102022 INR.
Most of the time, when banks and financial institutions calculate interest, they are using compound interest.
Compound interest is calculated at the end of each compounding period, which is typically a day, month, quarter, or year. At the end of each compounding period, the total amount (principal plus interest) from previous compounding periods is used to calculate the new quantity of interest. We multiply the total amount by the interest and then add it to the total. Note that the compound interest formula calculates the final balance, or amount. To find the amount of interest we need to subtract the principal from the final balance.
$A=P\times(1+\frac{r}{n})^{nt}$A=P×(1+rn)nt
where:
$A$A is the final amount of money (principal and interest together)
$P$P is the principal (the initial amount of money invested)
$r$r is the interest rate per year, expressed as a decimal
$t$t is the number of years
$n$n is the number of compounding periods in one year (eg: quarterly means $n=4$n=4)
A $\$3710$$3710 investment earns interest at $4.8%$4.8% p.a. compounded quarterly over $13$13 years.
Use the compound interest formula to calculate the value of this investment to the nearest cent.
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$1 + the interest rate as a decimal)".
Consider the following problem: Emma puts $\$5000$$5000 into an investment account paying a compound interest rate of $4.2%$4.2% pa. The table below shows the value of the investment over the first four years.Year ($n$n) | 0 | 1 | 2 | 3 | 4 |
---|---|---|---|---|---|
$5000\times1.042=$5000×1.042= | $5210\times1.042=$5210×1.042= | $5428.82\times1.042=$5428.82×1.042= | $5656.83\times1.042=$5656.83×1.042= | ||
Amount ($V_n$Vn) | $5000$5000 | $5210$5210 | $5428.82$5428.82 | $5656.83$5656.83 | $5894.42$5894.42 |
For a principal investment/loan, $P$P, at the compound interest rate of $r$r per period, the sequence of the value of the investment over time forms a geometric sequence with a starting value of $P=a$P=a and a common ratio of $(1+r)$(1+r).
The sequence which generates the value, $V_n$Vn, of the investment/loan at the end of each instalment period is:
$V_n=V_{n-1}\times(1+r)$Vn=Vn−1×(1+r), where $V_0=a$V0=a
$V_n=a(1+r)^n$Vn=a(1+r)n
The sequence which generates the value, $V_n$Vn, of the investment/loan at the beginning of each instalment period is:
$V_n=V_{n-1}\times(1+r)$Vn=Vn−1×(1+r), where $V_1=a$V1=a
$V_n=a(1+r)^{n-1}$Vn=a(1+r)n−1
Select the brand of calculator you use below to work through an example of using a calculator for compound interest problems involving sequences.
Casio Classpad
How to use the CASIO Classpad to complete the following tasks regarding sequences in a compound interest context.
Consider an investment of $\$1500$$1500 at $4%$4% p.a. compounded monthly.
Using a recursive relationship, generate the value of the investment at the end of each month for the first $6$6 months.
Using an explicit rule, generate the value of the investment at the end of each month for the first $6$6 months.
Find the balance in the account at the end of $1$1 year.
Find when the investment will reach $\$2000$$2000.
Sketch a graph of the balance of the account over the first four years.
TI Nspire
How to use the TI Nspire to complete the following tasks regarding sequences in a compound interest context.
Consider an investment of $\$1500$$1500 at $4%$4% p.a. compounded monthly.
Using a recursive relationship, generate the value of the investment at the end of each month for the first $6$6 months.
Using an explicit rule, generate the value of the investment at the end of each month for the first $6$6 months.
Find the balance in the account at the end of $1$1 year.
Find when the investment will reach $\$2000$$2000.
Sketch a graph of the balance of the account over the first four years.
$\$3000$$3000 is invested at the beginning of the year in an account that earns $12%$12% per annum interest, compounded quarterly.
Write a recursive rule, $V_n$Vn, in terms of $V_{n-1}$Vn−1, that gives the balance in the account at the end of the $n$nth quarter.
Write both parts of the rule (including for $V_0$V0) on the same line, separated by a comma.
How much money is in the account at the end of the first year?
Round your answer to the nearest cent.
The balance of an investment, in dollars, at the end of each year where interest is compounded annually is given by $A_n=1.05A_{n-1}$An=1.05An−1, $A_0=30000$A0=30000.
State the annual interest rate.
State the amount invested.
Determine the balance at the end of the first year.
Use the sequences facility on your calculator to calculate the balance at the end of $15$15 years.
Round your answer to the nearest cent.
The following spreadsheet shows the balance (in dollars) in a savings account in 2014, where interest is compounded monthly.
A | B | C | D | |
1 | Month | Balance at beginning of month | Interest | Balance at end of month |
2 | July | $8000$8000 | $160$160 | $X$X |
3 | August | $8160$8160 | $163.20$163.20 | $8323.20$8323.20 |
4 | September | $8323.20$8323.20 | $Y$Y | $8489.66$8489.66 |
5 | October | $Z$Z | $169.79$169.79 | $8659.45$8659.45 |
6 | November | $8659.45$8659.45 | $173.19$173.19 | $8832.64$8832.64 |
Calculate the value of $X$X.
Use the numbers for July to calculate the monthly interest rate.
Calculate the value of $Y$Y.
Calculate the value of $Z$Z.
Write a recursive rule, $B_n$Bn, that gives the balance at the end of the $n$nth month, with July being the first month.
Write both parts of the rule (including for $B_0$B0) on the same line, separated by a comma.
Write an explicit rule for $B_n$Bn, the balance at the end of the $n$nth month, with July being the first month.
Your CAS calculator and the sequences application are useful tools for investigating the effects of changing interest rates and compounding periods.
Select the brand of calculator you use below to work through an example of using a calculator to compare two different investments through graphing.
Casio Classpad
How to use the CASIO Classpad to complete the following tasks regarding comparison of investments from graphs.
Consider the following two investment options:
Option A. $\$1500$$1500 at $5%$5% compounded annually.
Option B. $\$1200$$1200 at $6.5%$6.5% compounded annually.
Sketch graphs to compare the two investments over the first $10$10 years.
Find the difference in the balance of the two investments at the end of $10$10 years.
At the end of which year does the balance for Option B first exceed that of Option A.
TI Nspire
How to use the TI Nspire to complete the following tasks regarding comparison of investments from graphs.
Consider the following two investment options:
Option A. $\$1500$$1500 at $5%$5% compounded annually.
Option B. $\$1200$$1200 at $6.5%$6.5% compounded annually.
Sketch graphs to compare the two investments over the first $10$10 years.
Find the difference in the balance of the two investments at the end of $10$10 years.
At the end of which year does the balance for Option B first exceed that of Option A.