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3.05 Geometric progressions - calculator free

Lesson

 

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+1tn).

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=rtn1,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}$5×2n1

 

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=5×2n1.

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=arn1

 

Forms of geometric sequences

For any geometric sequence with starting value $a$a and common ratio $r$r, we can express it in either of the following two forms:

  • Recursive form is a way to express any term in relation to the previous term:

$t_n=rt_{n-1}$tn=rtn1, where $t_1=a$t1=a or alternatively $t_{n+1}=rt_n$tn+1=rtn, where $t_1=a$t1=a

  • Explicit form is a way to express any term in relation to the term number

$t_n=ar^{n-1}$tn=arn1

 

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=arn1

Do: Dividing the second term by the first we get, $\frac{t_2}{t_1}=\frac{2}{1}$t2t1=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)n1.

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=2tn1,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=arn1.

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=2tn1,t1=3

 

Practice questions

Question 1

Study the pattern for the following geometric sequence, and write down the next two terms.

  1. $3$3, $15$15, $75$75, $\editable{}$, $\editable{}$

Question 2

Consider the first four terms in this geometric sequence: $-8$8, $-16$16, $-32$32, $-64$64

  1. If $T_n$Tn is the $n$nth term, evaluate $\frac{T_2}{T_1}$T2T1.

  2. Evaluate $\frac{T_3}{T_2}$T3T2

  3. Evaluate $\frac{T_4}{T_3}$T4T3

  4. Hence find the value of $T_5$T5.

Question 3

In a geometric progression, $T_4=54$T4=54 and $T_6=486$T6=486.

  1. Solve for $r$r, the common ratio in the sequence. Write both solutions on the same line separated by a comma.

  2. For the case where $r=3$r=3, solve for $a$a, the first term in the progression.

  3. Consider the sequence in which the first term is positive. Find an expression for $T_n$Tn, the general $n$nth term of this sequence.

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)n1, 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=arn1 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($-11<r<0) zig-zag path - alternating between points on the graph $f(n)=ar^{n-1}$f(n)=arn1 and $f(n)=-ar^{n-1}$f(n)=arn1, 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)n1, 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)11=12  and for $n=2$n=2 we have $t_2=12\times\left(-1.5\right)^{2-1}=-18$t2=12×(1.5)21=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.

 

Practice questions

Question 4

The $n$nth term of a geometric progression is given by the equation $T_n=2\times3^{n-1}$Tn=2×3n1.

  1. Complete the table of values:

    $n$n $1$1 $2$2 $3$3 $4$4 $10$10
    $T_n$Tn $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
  2. What is the common ratio between consecutive terms?

  3. Plot the points in the table that correspond to $n=1$n=1, $n=2$n=2, $n=3$n=3 and $n=4$n=4.

    Loading Graph...

  4. If the plots on the graph were joined they would form:

    a straight line

    A

    a curved line

    B

Question 5

Consider the following graph of the first $4$4 terms of a sequence.

  1. Write a recursive rule for $T_n$Tn in terms of $T_{n-1}$Tn1, including the initial term $T_1$T1.

    Enter both parts on the same line, separated by a comma.

Question 6

The given table of values represents terms in a geometric sequence.

$n$n $1$1 $2$2 $3$3 $4$4
$T_n$Tn $7$7 $-21$21 $63$63 $-189$189
  1. Identify $r$r, the common ratio between consecutive terms.

  2. Write a simplified expression for the general $n$nth term of the sequence, $T_n$Tn.

  3. Find the $12$12th term of the sequence.

Outcomes

3.2.5

use recursion to generate a geometric sequence

3.2.6

display the terms of a geometric sequence in both tabular and graphical form and demonstrate that geometric sequences can be used to model exponential growth and decay in discrete situations

3.2.7

deduce a rule for the nth term of a particular geometric sequence from the pattern of the terms in the sequence, and use this rule to make predictions

3.2.8

use geometric sequences to model and analyse (numerically, or graphically only) practical problems involving geometric growth and decay

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