13.03 Geometric sequences & series

Geometric sequences

Geometric sequences all start with a first term and then either increase or decrease by a constant factor called the common ratio. We denote the first term by $a_1$a1​ and the common ratio by the letter $r$r. For example, the sequence $4,8,16,32\dots$4,8,16,32… is geometric with $a_1=4$a1​=4 and $r=2$r=2. The sequence $100,-50,25,-12.5,\dots$100,−50,25,−12.5,… is geometric with $a_1=100$a1​=100 and $r=-\frac{1}{2}$r=−12​.

The size and sign of the geometric ratio plays an important role in how the sequence grows. Geometric ratios greater than $r=1$r=1 will cause terms in the sequence to get larger. Ratios between $0$0 and $1$1 will cause the terms to get smaller. Negative ratios will cause sign changes across consecutive terms, just like the last example mentioned in the previous paragraph.

A great example of a geometric sequence concerns animal cell division, where a single cell divides into two ‘daughter’ cells through biological processes known as mitosis and cytokinesis. The daughter cells divide again and again, each time creating two new daughter cells so that the number of daughters in each new generation form the geometric sequence $2,4,8,16,32,\dots$2,4,8,16,32,…

Clearly the first term is $a_1=2$a1​=2 and the common ratio $r=2$r=2.

Another example concerns radioactive decay – a process whereby half of a certain amount of radioactive material disappears in a specified time known as a “half-life”. If for example the half-life of a certain radioactive material is $10$10 years, then a quantity of $120$120 grams of the material reduces to $60$60 grams in the first $10$10 years, then to $30$30 grams in the next $10$10 years, and then to $15$15 grams in the next $10$10 years, and so on. Over $50$50 years, we see the quantity reduce according to the geometric sequence $120,60,30,15,7.5$120,60,30,15,7.5.

In general terms, every geometric sequence begins as $a_1,a_1*r,a_1*r^2,a_1*r^3,...$a1​,a1​*r,a1​*r2,a1​*r3,... so that the $n$nth term is given by:

$a_n=a_1*r^{n-1}$an​=a1​*rn−1

Having a formula for the $n$nth term allows us to quickly calculate the value of any term. For example, in the geometric sequence beginning $12,18,27,\dots$12,18,27,… we might want to know what the value of the $7$7th term is. Using the formula, and noticing that $a_1=12$a1​=12 and $r=\frac{3}{2}$r=32​ we can show that $a_7=12\times\left(\frac{3}{2}\right)^6$a7​=12×(32​)6 or $136.6875$136.6875. 

This applet will allow you to visualize the geometric side of geometric sequences. Play with the values of $a_1$a1​ and $r$r. What happens when $r$r is less than $1$1? What about when $a_1$a1​ is negative? What else do you notice?

 

Practice questions

QUESTION 1

QUESTION 2

QUESTION 3

 

Writing the recursive formula

It is generally known as a recursive relationship and for geometric sequences, the recursive formula is given by:

$a_{n+1}=r\times a_n,a_1='thefirstterm'$an+1​=r×an​,a1​=′thefirstterm′  

The equation states that the $\left(n+1\right)$(n+1)th term is $r$r times the $n$nth term with the first term equal to $a_1$a1​.

Thus the second term, $a_2$a2​ is $r$r times the first term $a_1$a1​, or $a_1*r$a1​*r. 

The third term $a_3$a3​ is $r$r times $a_2$a2​ or  $a_1*r^2$a1​*r2. 

The fourth term $a_4$a4​ is $r$r times $a_3$a3​, or $a_1*r^3$a1​*r3, and so on.

Hence, step by step, the sequence is revealed as $a_1$a1​, $a_1*r$a1​*r, $a_1*r^2$a1​*r2, $a_1*r^3...$a1​*r3... , $a_1*r^{n-1}$a1​*rn−1. 

Take for example the recursive relationship given as $a_{n+1}=\frac{a_n}{2}$an+1​=an​2​ with $a_1=64$a1​=64. From this formula, we see that 

$a_2=\frac{a_1}{2}=32$a2​=a1​2​=32 and  

$a_3=\frac{a_2}{2}=16$a3​=a2​2​=16, and so on.

This means that the sequence becomes $64,32,16,8,...$64,32,16,8,... which is clearly geometric with $a_1=64$a1​=64 and $r=\frac{1}{2}$r=12​. 

Consider the recurrence relationship given as $a_{n+1}=3a_n+2$an+1​=3an​+2 with $a_1=5$a1​=5.

To test whether or not the relationship is geometric, we can evaluate the first three terms.

$a_1=5$a1​=5,

$a_2=3\times5+2=17$a2​=3×5+2=17 

$a_3=3\times17+2=53$a3​=3×17+2=53.

Thus, the sequence begins $5,17,53,...$5,17,53,... and we immediately see that $\frac{53}{17}$5317​ is not the same fraction as $\frac{17}{5}$175​, and thus the recursive relationship is not geometric. In fact the only way the relationship given by $a_{n+1}=ra_n+k$an+1​=ran​+k is geometric is when the constant term $k$k is zero.

 

Practice questions

Question 4

Question 5

Question 6

 

Solving for terms

Sometimes we are asked to find the common ratio of a certain geometric sequence, and then use this to find a group of terms or simply a specific term. We need to remember that the  $n$nth term of a GP is given by:

$a_n=a_1r^{n-1}$an​=a1​rn−1

Note that there are two variables in the formula – the first term $a_1$a1​ and the common ratio $r$r and any set of three consecutive terms will allow us to evaluate $r$r. 

Take for example the geometric sequence beginning $3,12,48,...$3,12,48,... . The common ratio is simply the ratio $\frac{a_2}{a_1}=\frac{a_3}{a_2}=4$a2​a1​​=a3​a2​​=4.

This means that we can write down a formula for the $n$nth term as 

$a_n$an​	$=$=	$a_1r^{n-1}$a1​rn−1
	$=$=	$3\left(4\right)^{n-1}$3(4)n−1

and this in turn allows us to determine any term or sequence of terms we like. For example, we see that the $5$5th term is given by $a_5=3\times4^4=768$a5​=3×44=768. 

As another example, we might wonder whether the sequence $\sqrt{3},6,12\sqrt{3}$√3,6,12√3 is geometric. It may not be immediately obvious, but we can show that the numbers are in geometric sequence in two ways. In the first method, we note that both 

$\frac{6}{\sqrt{3}}$6√3​	$=$=	$\frac{6}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}$6√3​×√3√3​
	$=$=	$2\sqrt{3}$2√3

and that $\frac{12\sqrt{3}}{6}=2\sqrt{3}$12√36​=2√3, and so it is immediately geometric with $r=2\sqrt{3}$r=2√3.

In the second method we note that in any geometric sequence, $\frac{t_2}{t_1}=\frac{t_3}{t_2}$t2​t1​​=t3​t2​​ and by "cross" multiplication, we see that $\left(t_2\right)^2=t_1\times t_3$(t2​)2=t1​×t3​ or that the middle term is always the square root of the product of terms on each side of it. The middle term is defined to be the geometric mean of the outer two terms so that $t_2=\sqrt{t_1\times t_3}$t2​=√t1​×t3​ . 

In a final example consider the three terms $2,x,32$2,x,32 where the middle term is unknown. If this sequence is geometric then we can find $x$x, for we know that $\frac{x}{2}=\frac{32}{x}$x2​=32x​ and so $x^2=64$x2=64 and this means that there are two possible solutions for $x$x as $x=8$x=8 or $x=-8$x=−8. The two sequences are thus $2,8,32$2,8,32 or $2,-8,32$2,−8,32.

 

Practice questions

QUESTION 7

QUESTION 8

QUESTION 9

 

Graphs and tables of geometric sequences

Having a formula for the $n$nth term allows us to quickly generate a table of values for the sequence. For example in the sequence $12,18,27,\dots$12,18,27,… the first term is $12$12 and the common ratio is $1.5$1.5 and so the general term is given by the formula $a_n=12\times\left(1.5\right)^{n-1}$an​=12×(1.5)n−1 . By substituting for $n$n appropriately and using a scientific calculator, we can quickly generate the following table of the first $7$7 terms of the sequence:

n	1	2	3	4	5	6
tn	12	18	27	40.5	60.75	91.125

Perhaps more interestingly though is the different types of graphs that geometric sequences correspond to. Usually the graphs are not linear like arithmetic sequences. Graphs of geometric sequences are best known as rising or reducing graphs where the rate of rising continually changes, resulting in a curved growth or decay path. This happens whenever the common ratio is positive like the geometric sequence depicted in the above table. However, when the common ratio is negative, the values of successive terms flip their sign so that the graph is depicted as either a growing or diminishing zig-zag path. Think, for example, about the geometric sequence that that is identical to the one in the table, but has a negative ratio $r=-1.5$r=−1.5 so that its $n$nth term is given by $a_n=12\times\left(-1.5\right)^{n-1}$an​=12×(−1.5)n−1 . The new table becomes:

n	1	2	3	4	5	6
tn	12	-18	27	-40.5	60.75	-91.125

Checking, for $n=1$n=1, we have $a_1=12\times\left(-1.5\right)^{1-1}=12$a1​=12×(−1.5)1−1=12  and for $n=2$n=2 we have $a_2=12\times\left(-1.5\right)^{2-1}=-18$a2​=12×(−1.5)2−1=−18 so even numbered terms become negative and odd numbered terms become positive. 

Here is a graph of the two geometric sequences depicted in both tables. Note that the odd terms of the zig-zag graph coincide with the terms of the first geometric sequence. 

Note also that had the absolute value of the ratios of both geometric sequences been less than 1, then the the absolute value of the terms in both sequences would be reducing in size. 

 

Practice questions

Question 10

QUESTION 11

QUESTION 12

Geometric series

Suppose we wish to add the first $n$n terms of a geometric sequence. This is known as a series, or alternatively the partial sum of the sequence. We could write the sum as:

$S_n=a_1+a_1r+a_1r^2+a_1r^3+...+a_1r^{n-1}$Sn​=a1​+a1​r+a1​r2+a1​r3+...+a1​rn−1

If we multiply both sides of this equation by the common ratio $r$r we see that:

$rS_n=a_1r+a_1r^2+a_1r^3+...+a_1r^{n-1}+a_1r^n$rSn​=a1​r+a1​r2+a1​r3+...+a1​rn−1+a1​rn

Then, by carefully subtracting  $rS_n$rSn​ from $S_n$Sn​ term by term, we see that all of the middle terms disappear:

$S_n-rS_n=a_1+\left(a_1r-a_1r\right)+\left(a_1r^2-a_1r^2\right)+...+\left(a_1r^{n-1}-a_1r^{n-1}\right)-a_1r^n$Sn​−rSn​=a1​+(a1​r−a1​r)+(a1​r2−a1​r2)+...+(a1​rn−1−a1​rn−1)−a1​rn

This means that $S_n-rS_n=a_1-a_1r^n$Sn​−rSn​=a1​−a1​rn and when common factors are taken out on both sides of this equation, we find $S_n\left(1-r\right)=a_1\left(1-r^n\right)$Sn​(1−r)=a1​(1−rn) . Finally, by dividing both sides by $\left(1-r\right)$(1−r)  (excluding the trivial case of $r=1$r=1) we reveal the geometric sum formula:

 

$S_n=\frac{a_1\left(1-r^n\right)}{1-r}$Sn​=a1​(1−rn)1−r​

 

An extra step, multiplying the numerator and denominator by $-1$−1 , reveals a slightly different form for  $S_n$Sn​ that is easier to manage when the common ratio is greater than $r=1$r=1 :

$S_n=\frac{a_1\left(r^n-1\right)}{r-1}$Sn​=a1​(rn−1)r−1​

As an example, adding up the first $10$10 terms of the geometric progression that begins $96+48+12+...$96+48+12+...  is straightforward. Since $a=96$a=96 and $r=\frac{1}{2}$r=12​ we have:

$S_{10}=\frac{96\times\left(1-\left(\frac{1}{2}\right)^{10}\right)}{1-\left(\frac{1}{2}\right)}=191.8125$S10​=96×(1−(12​)10)1−(12​)​=191.8125

To add up the first $20$20 terms of the sequence $2,6,18,...$2,6,18,... we might use the second version of the formula to reveal:

$S_{10}=\frac{2\times\left(3^{20}-1\right)}{3-1}=3486784400$S10​=2×(320−1)3−1​=3486784400

You might be wondering why the sum formulas exclude the case for $r=1$r=1. This is not an issue, for if $r=1$r=1, then the series becomes:

 $S_n=a_1+a_1+a_1+...+a_1=a_1\times n$Sn​=a1​+a1​+a1​+...+a1​=a1​×n

Practice problems

Question 1

QUESTION 2

QUESTION 3

 

Infinite geometric series

Suppose a rare species of frog can jump up to $2$2 meters in one bound. One such frog with an interest in mathematics sits $4$4 meters from a wall. The curious amphibian decides to jump $2$2 meters toward it in a single bound. After it completes the feat, it jumps again, but this time only a distance of $1$1 meter. Again the frog jumps, but only $\frac{1}{2}$12​ a meter, then jumps again, and again, each time halving the distance it jumps. Will the frog get to the wall?

The total distance traveled toward the wall after the frog jumps $n$n times is given by $S_n=\frac{2\left(1-\left(\frac{1}{2}\right)^n\right)}{1-\frac{1}{2}}$Sn​=2(1−(12​)n)1−12​​ .

So after the tenth jump, the total distance is given by $S_{10}=4\left(1-\left(\frac{1}{2}\right)^{10}\right)=4\left(1-\frac{1}{1024}\right)$S10​=4(1−(12​)10)=4(1−11024​)

If we look carefully at the last expression, we realize that, as the frog continues jumping toward the wall, the quantity inside the square parentheses approaches the value of $1$1 but will always be less than $1$1. This is the key observation that needs to be made. Therefore the entire sum must remain less than $4$4, but the frog can get as close to $4$4 as it likes simply by continuing to jump according to the geometric pattern described.

Whenever we have a geometric progression with its common ratio within the interval $-1−1<r<1 then, no matter how many terms we add together, the sum will never exceed some number $L$L called the limiting sum. We sometimes refer to it as the infinite sum of the geometric series. An infinite geometric series with a limiting sum is said to converge, whereas an infinite geometric series that does not have a limiting sum is sometimes called a diverging series.

Since for any geometric series, $S_n=\frac{a_1\left(1-r^n\right)}{1-r}$Sn​=a1​(1−rn)1−r​ , if the common ratio is within the interval $-1−1<r<1 then, as more terms are added, the quantity $\left(1-r^n\right)$(1−rn) will become closer and closer to $1$1. This means that the sum will get closer and closer to:

$S_{\infty}=\frac{a_1}{1-r}$S∞​=a1​1−r​ 

Checking our frogs progress, $S_{\infty}=\frac{2}{1-\left(\frac{1}{2}\right)}=4$S∞​=21−(12​)​=4 is the limiting sum.

As another example, the limiting sum of the geometric series $108+36+12\dots$108+36+12… is simply $S_{\infty}=\frac{108}{1-\left(\frac{1}{3}\right)}=162$S∞​=1081−(13​)​=162. 

 

Practice problems

QUESTION 13

QUESTION 14

QUESTION 15

 

Applications of geometric series

There are many real-life applications of geometric series. Formulas 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 basic principles and the information given in the problem. 

 

Worked examples

These examples will walk us through developing these formulas.

Question 16

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?

Think: We might begin by searching for a pattern by examining what happens in the first few years.

Do: 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 parentheses 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 $S_n=\frac{a_1\left(r^n-1\right)}{r-1}$Sn​=a1​(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. 

 

Reflect: Now let's 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% per year is applied. 

From our solution  above, and calling $R=1+\frac{r}{100}$R=1+r100​, we have the generalized formula given by:

$A_n$An​	$=$=	$P\times\frac{R\left(R^n-1\right)}{R-1}$P×R(Rn−1)R−1​

Question 17

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. 

Think:  Can you we write the decimal as an infinite series?

The repeating decimal can be written: $N=0.23+0.0023+0.000023+...$N=0.23+0.0023+0.000023+... 

The first term is given by  $a_1=0.23$a1​=0.23 and the common ratio is $r=0.001$r=0.001.

Do: The sum becomes:

$N$N	$=$=	$\frac{a_1}{1-r}$a1​1−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​

Reflect: Can this  same strategy can be applied to any repeating decimal?

Question 18

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 it has a finite area. Develop a formula for that area.

Think: Can we use a series to describe the area?

Solve: 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 parentheses is $\frac{8}{5}$85​.

This means that Koch snowflake has a total area given by $A_T=\frac{8A}{5}$AT​=8A5​.

Reflect: Now use your formula to find the area of a Koch snowflake whose equilateral triangles have an area $A=\frac{1}{3}$A=13​.

 

Practice problems

QUESTION 19

QUESTION 20

QUESTION 21