There are many real life applications of geometric series and we discuss a few of these here.
Formulae are often developed for many of these applications, particularly when they occur regularly in industry.
For our purposes, it is often best to find solutions to various problems starting from first principles.
A bank client deposits $\$1000$$1000 at the beginning of each year, and is given $7%$7% interest per year for $50$50 years. How much will accrue in the account over that time?
To answer this, we might begin by searching for a pattern by examining what happens in the first few years.
If we set $A_n$An as the amount of money accrued after $n$n years have elapsed, then we have:
$A_0=1000$A0=1000
Then $A_1=1000+1000\times\frac{7}{100}=1000\left(1+\frac{7}{100}\right)=1000\times\left(1.07\right)^1$A1=1000+1000×7100=1000(1+7100)=1000×(1.07)1
This means that the amount accrued after $1$1 year becomes $\$1070$$1070.
By the end of the second year, another $\$1000$$1000 has been added, with interest, but the original $\$1000$$1000 has been boosted by two interest payments.
The total amount is determined as:
$A_2=1000\left(1.07\right)^1+\left[1000\left(1.07\right)\right]\times1.07=1000\left[1.07+1.07^2\right]$A2=1000(1.07)1+[1000(1.07)]×1.07=1000[1.07+1.072]
By the end of the third year, the total accrual becomes:
$A_3=1000\left[1.07+1.07^2+1.07^3\right]$A3=1000[1.07+1.072+1.073]
A pattern is emerging, and so by the end of $50$50 years, the total accrued becomes:
$A_{50}=1000\left[1.07+1.07^2+1.07^3+...+1.07^{50}\right]$A50=1000[1.07+1.072+1.073+...+1.0750].
Inside the square brackets is a geometric series with first term and common ratio both equal to $1.07$1.07.
Recalling the formula for the sum of a geometric sequence as $S_n=\frac{a\left(r^n-1\right)}{r-1}$Sn=a(rn−1)r−1 we have for this series:
$A_n=1000\times\frac{1.07\left(1.07^{50}-1\right)}{1.07-1}=434985.96$An=1000×1.07(1.0750−1)1.07−1=434985.96
Hence the amount accrued in the account will be approximately $\$434986$$434986.
For example $1$1 above, devise a formula that a banker might use for any client wishing to deposit an amount $P$P at the beginning of ever year for $n$n years where an interest rate of $r%$r% p.a. is applied. Use the formula to find the accrued amount of the regular annual payment of $\$1000$$1000 after $50$50 years where $8%$8% is applied each year.
From our solution to example $1$1, and calling $R=1+\frac{r}{100}$R=1+r100, we have the generalised formula given by:
$A_n$An | $=$= | $P\times\frac{R\left(R^n-1\right)}{R-1}$P×R(Rn−1)R−1 |
Hence, at $8%$8%, we have:
$A_n=1000\times\frac{1.08\left(1.08^{50}-1\right)}{1.08-1}=573770.16$An=1000×1.08(1.0850−1)1.08−1=573770.16
This means that one extra percentage in interest each year makes a difference in total accrual of approximately $\$138784$$138784.
Show that the repeating decimal $N=0.2323232323...$N=0.2323232323... is a rational number. That is to say, the number can be put in the form $\frac{p}{q}$pq where $p$p and $q$q are integers and $q\ne0$q≠0.
The repeating decimal can be written:
$N=0.23+0.0023+0.000023+...$N=0.23+0.0023+0.000023+... which is an infinite geometric series whose first term is given by $a=0.23$a=0.23 and whose common ratio is given by $r=0.001$r=0.001.
The limiting sum becomes:
$N$N | $=$= | $\frac{a}{1-r}$a1−r |
$=$= | $\frac{0.23}{1-0.001}$0.231−0.001 | |
$=$= | $\frac{\frac{23}{100}}{1-\frac{1}{100}}$231001−1100 | |
$=$= | $\frac{\frac{23}{100}}{\frac{100-1}{100}}$23100100−1100 | |
$=$= | $\frac{23}{99}$2399 | |
The same strategy can be applied to any repeating decimal.
The recipe for making a Koch snowflake is as follows;
The emerging figure has an infinite perimeter but we can show that is has a finite area as follows.
The total area $A_T$AT of the snow flake becomes:
$A_T$AT | $=$= | $A+3\left(\frac{A}{9}\right)+12\left(\frac{A}{9^2}\right)+48\left(\frac{A}{9^3}\right)+192\left(\frac{A}{9^4}\right)+...$A+3(A9)+12(A92)+48(A93)+192(A94)+... |
$=$= | $A\left[1+\left(\frac{3}{9}\right)+\left(\frac{12}{9^2}\right)+\left(\frac{48}{9^3}\right)+\left(\frac{192}{9^4}\right)+...\right]$A[1+(39)+(1292)+(4893)+(19294)+...] | |
The expression on the right and within the main bracket begins with the number $1$1, but all the other terms form a geometric series with first term $\frac{1}{3}$13 and common ratio $\frac{4}{9}$49. We can see this because the numerator of each fraction is increasing by a factor of $4$4 and the denominator is increasing by a factor of $9$9.
Since $|\frac{4}{9}|$|49| is less than $1$1, the series sums to $\frac{a}{1-r}=\frac{\frac{1}{3}}{1-\frac{4}{9}}=\frac{3}{5}$a1−r=131−49=35. Adding the extra $1$1 at the beginning, we see that the total sum within the main brackets is $\frac{8}{5}$85.
This means that Koch snowflake has a total area given by $A_T=\frac{8A}{5}$AT=8A5.
The recurring decimal $0.8888\dots$0.8888… can be expressed as a fraction when viewed as an infinite geometric series.
Express the first decimal place, $0.8$0.8 as an unsimplified fraction.
Express the second decimal place, $0.08$0.08 as an unsimplified fraction.
Hence write, using fractions, the first five terms of the geometric sequence representing $0.8888\dots$0.8888…
State the values of $a$a, the first term, and $r$r, the common ratio, of this sequence.
$a$a$=$=$\editable{}$
$r$r$=$=$\editable{}$
If we add up infinitely many terms of this sequence, we will have the fraction equivalent of our recurring decimal. Calculate the infinite sum of the sequence as a fraction.
At the start of 2014 Pauline deposits $\$5000$$5000 into an investment account. At the end of each quarter she makes an extra deposit of $\$700$$700. By looking at the pattern investment, Pauline realises she can use her knowledge of geometric series to find the balance in the account at some point in the future.
The table below shows the first few quarters of 2014. All values in the table are in dollars.
Quarter | Opening Balance | Interest | Deposit | Closing Balance |
---|---|---|---|---|
Jan-Mar | $5000$5000 | $200$200 | $700$700 | $5900$5900 |
Apr-Jun | $5900$5900 | $236.00$236.00 | $700$700 | $6836.00$6836.00 |
Jul-Sep | $6836.00$6836.00 | $273.44$273.44 | $700$700 | $7809.44$7809.44 |
Use the numbers for the January quarter to calculate the quarterly interest rate.
Write an expression for the amount in the account at the end of the first quarter.
Do not evaluate the expression.
$\editable{}\times\editable{}+\editable{}$×+
Using $5000\times1.04+700$5000×1.04+700 as the starting balance, write an expression for the amount in the account at the end of the second quarter.
$\editable{}\times\left(\editable{}\right)^2+\editable{}\times\editable{}+\editable{}$×()2+×+
Given that the amount in the account at the end of the second quarter can be expressed as $5000\times\left(1.04\right)^2+700\times1.04+700$5000×(1.04)2+700×1.04+700, write a similar expression for the amount in the account at the end of the third quarter.
$\editable{}\times\left(\editable{}\right)^3+\editable{}\times\left(\editable{}\right)^2+\editable{}\times\editable{}+\editable{}$×()3+×()2+×+
The amount in the account after $n$n quarters can be expressed as a term of a geometric sequence plus the sum of a geometric sequence.
Write an expression for the amount in the investment account after $n$n quarters.
Hence determine the total amount in Pauline’s account at the beginning of 2016 to the nearest dollar.
Rochelle invests $\$190000$$190000 at a rate of $7%$7% per annum compounded annually, and wants to work out how much she can withdraw each year to ensure the investment lasts $20$20 years.
We will use geometric sequences and series to determine what Rochelle's annual withdrawal amount should be if she wants the investment to last $20$20 years.
The amount in the account after $n$n years can be expressed as the $n$nth term of a geometric sequence minus the sum of a different geometric sequence.
Write an expression for the amount in the investment account after $n$n years.
Use $x$x to represent the amount to be withdrawn each year.
Hence determine Rochelle's annual withdrawal amount, correct to the nearest cent.
Use arithmetic and geometric sequences and series
Apply sequences and series in solving problems