Applications of Geometric Series

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. 

 

Example 1

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. 

 

Example 2

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.

 

Example 3 

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}}$23100​1−1100​​
	$=$=	$\frac{\frac{23}{100}}{\frac{100-1}{100}}$23100​100−1100​​
	$=$=	$\frac{23}{99}$2399​

 The same strategy can be applied to any repeating decimal. 

 

Example 4

The recipe for making a Koch snowflake is as follows;

1. Draw an equilateral triangle with area $A$A as shown in figure $(a)$(a) below.
2. To each of the three sides of figure $(a)$(a) add an equilateral triangle with sides of length $\frac{1}{3}$13​ that of the original triangle, as shown in figure $(b)$(b). Note that the area of each of the new triangles is $\frac{A}{9}$A9​.
3. To each of the $12$12 sides of figure $(b)$(b) add an equilateral triangle with sides of length $\frac{1}{3}$13​ that of the triangles formed in step $2$2. The area of each of these $12$12 new triangles becomes $\frac{A}{9^2}$A92​.
4. To each of the $48$48 sides of figure $(c)$(c) add an equilateral triangle with sides of length $\frac{1}{3}$13​ of the $12$12 triangles formed in step $3$3.
5. Continue in this manner indefinitely to form what becomes the Koch snowflake.

 

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​=13​1−49​​=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​.

Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3