3.05 Geometric progressions - calculator free

 

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 $a$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​.

To describe the rule in words we say "next term is $r$r multiplied by previous term". Therefore we can write any geometric sequence as the recurrence relation:

$t_n=rt_{n-1},t_1=a$tn​=rtn−1​,t1​=a or alternatively $t_{n+1}=rt_n$tn+1​=rtn​, where $t_1=a$t1​=a

Note that it is also possible to define the initial term as $t_0$t0​, this is particularly useful in financial applications of sequences.

We can also find an explicit formula in terms of $a$a 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 a starting term of $5$5 and a common ratio of $2$2, that is $a=5$a=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 is $t_{10}=5\times2^9$t10​=5×29  and the one-hundredth term is $t_{100}=5\times2^{99}$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 $a$a and common ratio $r$r has the terms given by: $a,ar,ar^2,ar^3,...$a,ar,ar2,ar3,... We see a similar pattern to our previous table and can write down the formula for the $n$nth term:

$t_n=ar^{n-1}$tn​=arn−1

 

 

Worked examples

Example 1

For the sequence  $15,30,60,120...$15,30,60,120..., find an explicit rule for the $n$nth term and hence, find the $8$8th term. 

Think: Check that the sequence is geometric, is the next term made by multiplying the previous term by a constant factor? Then write down the starting value $a$a and common ratio $r$r and substitute these into the general form: $t_n=ar^{n-1}$tn​=arn−1

Do: Dividing the second term by the first we get, $\frac{t_2}{t_1}=\frac{2}{1}$t2​t1​​=21​. Checking the ratio between the successive pairs we also get $\frac{2}{1}$21​.  So we have a geometric sequence with: $a=15$a=15 and $r=2$r=2. The general formula for this sequence is: $t_n=15\left(2\right)^{n-1}$tn​=15(2)n−1.

Hence, the $8$8th term is: $t_8=15\left(2\right)^7=15\times128=1920$t8​=15(2)7=15×128=1920.

Example 2

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. There are two methods we can use to do this.

Method 1: 

Think: Determine how many times $t_3$t3​ is multiplied by $r$r to make $t_6$t6​

Do:

$t_3\times r\times r\times r$t3​×r×r×r	$=$=	$t_6$t6​
$12\times r^3$12×r3	$=$=	$96$96
$r^3$r3	$=$=	$8$8
$\therefore r$∴r	$=$=	$2$2

To find $a$a think: if we divide $t_3$t3​ by $r$r we get $t_2$t2​. Then divide $t_2$t2​ by $r$r to find $t_1$t1​ or $a$a. 

Therefore $a=12\div2\div2=3.$a=12÷​2÷​2=3. 

Think: We have a GP with $a=3$a=3 and $r=2$r=2. In words we can say "next term is two times previous term, with first term of $3$3".

Do: Write  - The recurrence relation for this sequence is:  

$t_n=2t_{n-1},t_1=3$tn​=2tn−1​,t1​=3

Method 2:

Think: As we have two terms we can set up two equations in terms of $a$a and $r$r using $t_n=ar^{n-1}$tn​=arn−1.

Do:

$t_3$t3​:

$ar^2=12$ar2=12

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

and

$t_6$t6​:

$ar^5=96$ar5=96

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

 

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

$\frac{ar^5}{ar^2}$ar5ar2​	$=$=	$\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) $a\times2^2=12$a×22=12 and so $a$a is $3$3. The recurrence relation for this sequence is given by:  

$t_n=2t_{n-1},t_1=3$tn​=2tn−1​,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 $6$6 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:

Sequences of the form $t_n=ar^{n-1}$tn​=arn−1 are exponential graphs, and where $a>0$a>0 they 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 $a$a is negative the path will be reflected about the $x$x-axis

What happens when $r$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($-1−1<r<0) zig-zag path - alternating between points on the graph $f(n)=ar^{n-1}$f(n)=arn−1 and $f(n)=-ar^{n-1}$f(n)=−arn−1, depending on the power being odd or even.

 

Worked example

Example 3

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 $a$a and $r$r in the applet below to observe the effect on the plotted points.

Created with Geogebra

 

Practice questions

Question 4

Question 5

Question 6