8.04 Geometric sequences

A sequence in which each term increases or decreases from the last by a constant factor is called a geometric sequence. We refer to the constant factor the terms are changing by as the common ratio, which will result from dividing any two successive terms $\left(\frac{t_{n+1}}{t_n}\right)$(tn+1​tn​​).

We denote the first term in the sequence by the letter $t_1$t1​ 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 $t_1=4$t1​=4 and $r=2$r=2. The sequence $100,-50,25,-12.5,\dots$100,−50,25,−12.5,… is geometric with $t_1=100$t1​=100 and $r=-\frac{1}{2}$r=−12​.

Since, $t_2=r\times t_1$t2​=r×t1​, $t_3=r\times t_2$t3​=r×t2​ and so on, we can write any geometric sequence as the recurrence relation:

$t_{n+1}=rt_n$tn+1​=rtn​

We can also find an explicit formula in terms of $t_1$t1​ and $r$r, this is useful for finding the $n$nth term without listing the sequence.

Let's look at a table of a concrete example to see the pattern for the explicit formula. For the sequence $5,10,20,40,\ldots$5,10,20,40,…, we have starting term of $5$5 and a common ratio of $2$2, that is $t_1=5$t1​=5 and $r=2$r=2. A table of the sequence is show below:

$n$n	$t_n$tn​	Pattern
$1$1	$5$5	$5\times2^0$5×20
$2$2	$10$10	$5\times2^1$5×21
$3$3	$20$20	$5\times2^2$5×22
$4$4	$40$40	$5\times2^3$5×23
...
$n$n	$t_n$tn​	$5\times2^{n-1}$5×2n−1

The pattern starts to become clear and we could guess that the tenth term becomes $t_{10}=5\times2^9$t10​=5×29  and the one-hundredth term $t_{100}=5\times2^{99}$t100​=5×299. And following the pattern, the explicit formula for the $n$nth term is $t_n=5\times2^{n-1}$tn​=5×2n−1.

For any geometric progression with starting value $t_1$t1​ and common ratio $r$r has the terms given by: $t_1,t_1r,t_1r^2,t_1r^3,...$t1​,t1​r,t1​r2,t1​r3,... We see a similar pattern to our previous table and can write down the formula for the $n$nth term:

$t_n=t_1r^{n-1}$tn​=t1​rn−1

 

Worked examples

Example 1

For the sequence  $810,270,90,30...$810,270,90,30..., find an explicit rule for the $n$nth term and hence, find the $8$8th term. 

Think: Check that the sequence is geometric, does each term differ from the last by a constant factor? Then write down the the starting value $t_1$t1​ and common ratio $r$r and substitute these into the general form: $t_n=t_1r^{n-1}$tn​=t1​rn−1

Do: Dividing the second term by the first we get, $\frac{t_2}{t_1}=\frac{1}{3}$t2​t1​​=13​. Checking the ratio between the successive pairs we also get $\frac{1}{3}$13​.  So we have a geometric sequence with: $t_1=810$t1​=810 and $r=\frac{1}{3}$r=13​. The general formula for this sequence is: $t_n=810\left(\frac{1}{3}\right)^{n-1}$tn​=810(13​)n−1.

Hence, the $8$8th term is: $t_8=810\left(\frac{1}{3}\right)^7=\frac{10}{27}$t8​=810(13​)7=1027​.

Example 2

For the sequence $5,20,80,320,...$5,20,80,320,..., find $n$n if the $n$nth term is $327680$327680.

Think: Find a general rule for the sequence, substitute in $327680$327680 for $t_n$tn​ and rearrange for $n$n.

Do: This is a geometric sequence with $t_1=5$t1​=5 and common ratio $r=4$r=4. Hence, the general rule is: $t_n=5\left(4\right)^{n-1}$tn​=5(4)n−1, substituting $t_n=327680$tn​=327680, we get:

$327680$327680	$=$=	$5\left(4\right)^{n-1}$5(4)n−1
$\therefore4^{n-1}$∴4n−1	$=$=	$65536$65536
$4^{n-1}$4n−1	$=$=	$4^8$48	Solve by guess and check, technology or logarithms.
Hence, $n-1$n−1	$=$=	$8$8
$n$n	$=$=	$9$9

Hence, the $9$9th term in the sequence is $327680$327680.

Example 3

If a geometric sequence has $t_3=12$t3​=12 and $t_6=96$t6​=96, find the recurrence relation for the sequence. 

Think: To find the recurrence relation we need the starting value and common ratio. As we have two terms we can set up two equations in terms of $t_1$t1​ and $r$r using $t_n=t_1r^{n-1}$tn​=t1​rn−1.

Do:

$t_3$t3​:

$t_1r^2=12$t1​r2=12

$.....\left(1\right)$.....(1)

and

$t_6$t6​:

$t_1r^5=96$t1​r5=96

$.....\left(2\right)$.....(2)

 

If we now divide equation $\left(2\right)$(2) by equation $\left(1\right)$(1), we obtain the following:

$\frac{t_1r^5}{t_1r^2}$t1​r5t1​r2​	$=$=	$\frac{96}{12}$9612​
$r^3$r3	$=$=	$8$8
$\therefore r$∴r	$=$=	$2$2

With the common ratio found to be $2$2, then we know that, using equation $\left(1\right)$(1) $t_1\times2^2=12$t1​×22=12 and so $t_1$t1​ is $3$3. The recurrence relation for this sequence is given by:  

$t_{n+1}=2t_n,t_1=3$tn+1​=2tn​,t1​=3

Practice questions

QUESTION 1

QUESTION 2

QUESTION 3

 

Geometric sequences in tables and graphs

When given a formula for the $n$nth term we can generate a table of values for the sequence. For example in the sequence 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 calculator, we can generate the following table of the first $7$7 terms of the sequence:

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

Perhaps more interesting though is the different types of graphs that geometric sequences correspond to. The graphs are not linear like arithmetic progressions, except for the trivial case of $r=1$r=1. The path of points plotted from a geometric sequence follow an exponential curve for positive values of $r$r:

Recall from exponential graphs, sequences of the form $t_n=t_1r^{n-1}$tn​=t1​rn−1, where $a>0$a>0  will follow:

• The path of an exponential growth function for $r>1$r>1
• The path of an exponential decay function for $00<r<1
• If $t_1$t1​ is negative the path will be reflected about the $x$x-axis

What we haven't previously looked at in the exponential chapter was what if $r$r is negative? The values of successive terms flip their sign so that the graph is depicted as either a growing($|r|>1$|r|>1) or diminishing($|r|<1$|r|<1) zig-zag path - alternating between points on the graph$f(n)=t_1|r|^{n-1}$f(n)=t1​|r|n−1 and $f(n)=-t_1|r|^{n-1}$f(n)=−t1​|r|n−1, depending on the power being odd or even.

 

Worked examples

Example 4

For the geometric progression with starting value $12$12 and ratio $r=-1.5$r=−1.5, create a table and plot a graph of the sequence.

Think: This is the same as the example in the previous table but the ratio is now negative. The $n$nth term is given by $t_n=12\times\left(-1.5\right)^{n-1}$tn​=12×(−1.5)n−1, the table will be the same but the sign of the terms will alternate. 

Do: The new table becomes:

$n$n	$1$1	$2$2	$3$3	$4$4	$5$5	$6$6
$t_n$tn​	$12$12	$-18$−18	$27$27	$-40.5$−40.5	$60.75$60.75	$-91.125$−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, continuing on 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 progression. 

Try adjusting the values of $a=t_1$a=t1​ and $r$r in the applet below to observe the effect on the plotted points.

Created with Geogebra

For $|r|>1$|r|>1 the values will diverge as $n$n increases, the terms will keep getting larger in size without bound.

For $|r|<1$|r|<1 the values will converge as $n$n increases, each term getting smaller and smaller and approaching a limit of $0$0.  

Practice questions

QUESTION 4

QUESTION 5

QUESTION 6

 

Geometric sequences on CAS calculator

Select the brand of calculator you use below to work through an example of using a calculator for geometric sequences.

 

Casio Classpad

 

TI Nspire

 

Applications of geometric sequences

When looking at exponential models you have encountered many applications of geometric sequences such as compound interest, exponential growth of bacteria, exponential decay of radioactive elements. If we have an amount increasing or decreasing by a constant factor at set time periods, then you can consider that process as being geometric.

Worked example

Example 5

After receiving $\$500$$500 for her birthday, Hayley decides to spend $10%$10% of this money each week.

a) Find a model for $B_n$Bn​, the amount of birthday money she has left at the start of the $n$nth week.

At the beginning of the second week, Hayley will have spent $\$50$$50, and will only have $\$450$$450 or $90%$90% of the birthday money left. During the second week she will spend slightly less -  $10%$10% of her remaining $\$450$$450 or $\$45$$45, so at the beginning of the third week $\$405$$405 will remain. The sequence of birthday dollars that remain at the beginning of the first five weeks is given as $500,450,405,364.5,328.05$500,450,405,364.5,328.05 and we can see here that this sequence is geometric with the first term $t_1=500$t1​=500 and the common ratio $r=0.9$r=0.9.  

Hence, $B_n=500(0.9)^{n-1}$Bn​=500(0.9)n−1.

b) If instead of spending $10%$10%, Hayley decides to double her savings by setting aside an additional $10%$10% in savings each week. How long before she reaches her savings goal?

Firstly, she will add $10%$10% to the birthday money, so that at the beginning of the second week, the additional $\$50$$50 will take the total to $\$550$$550. At the beginning of week $3$3 she will add a further $10%$10% of this accrued $\$550$$550, so that the new total becomes $\$605$$605. She will keep adding $10%$10% of what ever is there at the beginning of each week until she reaches or exceeds $\$1000$$1000.

The geometric progression for this plan becomes $500,550,605,665.5,732.05,...$500,550,605,665.5,732.05,...

This progression has $t_1=500$t1​=500 and $r=\frac{550}{500}=1.1$r=550500​=1.1, so that the $n$nth term of the progression is given by $t_n=500\left(1.1\right)^{n-1}$tn​=500(1.1)n−1.

To solve for when the balance reaches her goal of $\$1000$$1000, solve for $n$n when $t_n=1000$tn​=1000; 

$500\left(1.1\right)^{n-1}$500(1.1)n−1	$=$=	$1000$1000
$1.1^{n-1}$1.1n−1	$=$=	$2$2	Dividing both sides by 2
$\therefore(n-1)$∴(n−1)	$\approx$≈	$7.27$7.27	Solve by guess and check, technology or logarithms.
$n$n	$\approx$≈	$8.27$8.27

So $n$n must be greater than $8.27$8.27 for Hayley to have achieved her goal, so at the beginning of the $9$9th week her savings will exceed $\$1000$$1000.

Practice questions

QUESTION 7

QUESTION 8

Compound interest using sequences

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

We can see that the next value is equal to the previous value multiplied by $1.042$1.042. Therefore we have a geometric sequence and the recursive rule for this investment is  $V_{n+1}=1.042\times V_n$Vn+1​=1.042×Vn​, where $V_0=5000$V0​=5000

 

Practice questions

QUESTION 9

QUESTION 10