7.04 Geometric sequences

Introduction to 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 the letter $a$a 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 $a=4$a=4 and $r=2$r=2. The sequence $100,-50,25,-12.5,\dots$100,−50,25,−12.5,… is geometric with $a=100$a=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 $a=2$a=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 $a,ar,ar^2,ar^3,...$a,ar,ar2,ar3,... so that the $n$nth term is given by:

$t_n=ar^{n-1}$tn​=arn−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 $a=12$a=12 and $r=\frac{3}{2}$r=32​ we can show that $t_7=12\times\left(\frac{3}{2}\right)^6$t7​=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 $a$a and $r$r. What happens when $r$r is less than $1$1? What about when $a$a 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 recurrence relationship and for geometric sequences, the recurrence formula is given by:

$t_{n+1}=r\times t_n,t_1=a$tn+1​=r×tn​,t1​=a  

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 $a$a.

Thus the second term, $t_2$t2​ is $r$r times the first term $t_1$t1​, or $ar$ar. 

The third term $t_3$t3​ is $r$r times $t_2$t2​ or  $ar^2$ar2. 

The fourth term $t_4$t4​ is $r$r times $t_3$t3​, or $ar^3$ar3, and so on.

Hence, step by step, the sequence is revealed as $a$a, $ar$ar, $ar^2$ar2, $ar^3...$ar3... , $ar^{n-1}$arn−1. 

Take for example the recursive relationship given as $t_{n+1}=\frac{t_n}{2}$tn+1​=tn​2​ with $t_1=64$t1​=64. From this formula, we see that 

$t_2=\frac{t_1}{2}=32$t2​=t1​2​=32 and  

$t_3=\frac{t_2}{2}=16$t3​=t2​2​=16, and so on.

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

Consider the recurrence relationship given as $t_{n+1}=3t_n+2$tn+1​=3tn​+2 with $t_1=5$t1​=5.

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

$t_1=5$t1​=5,

$t_2=3\times5+2=17$t2​=3×5+2=17 

$t_3=3\times17+2=53$t3​=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 $t_{n+1}=rt_n+k$tn+1​=rtn​+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:

$t_n=ar^{n-1}$tn​=arn−1

Note that there are two variables in the formula – the first term $a$a 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 $\frac{t_2}{t_1}=\frac{t_3}{t_2}=4$t2​t1​​=t3​t2​​=4.

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

$t_n$tn​	$=$=	$ar^{n-1}$arn−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 $t_5=3\times4^4=768$t5​=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 $t_n=12\times\left(1.5\right)^{n-1}$tn​=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 $t_n=12\times\left(-1.5\right)^{n-1}$tn​=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 $t_1=12\times\left(-1.5\right)^{1-1}=12$t1​=12×(−1.5)1−1=12  and for $n=2$n=2 we have $t_2=12\times\left(-1.5\right)^{2-1}=-18$t2​=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