topic badge

3.04 Geometric sequences

Lesson

Forms of geometric sequence

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(\dfrac{t_{n+1}}{t_n}\right).

We denote the first term in the sequence by the letter a and the common ratio by the letter r. For example, the sequence 4,8,16,32\ldots is geometric with a=4 and r=2. The sequence 100,-50,25,-12.5,\ldots is geometric with a=100 and r=\dfrac{-1}{2}.

To describe the rule in words we say "next term is r multiplied by previous term". Therefore we can write any geometric sequence as the recurrence relation: t_n=rt_{n-1},t_1=a or alternatively t_{n+1}=rt_n, where t_1=a.

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

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

nt_{n}\text{Pattern}
155\times 2^0
2105\times 2^1
3205\times 2^2
4405\times 2^3
...
nt_n5\times 2^{n-1}

The table shows the sequence 5,10,20,40,\ldots , we have a starting term of 5 and a common ratio of 2, that is a=5 and r=2.

The pattern becomes clear and we could guess that the tenth term is t_{10}=5\times 2^9 and the one hundredth term is t_{100}=5\times 2^{99}. And following the pattern, the explicit formula for the nth term is t_n=5\times 2^{n-1}.

For any geometric progression with starting value a and common ratio r has the terms given by: a,ar,ar^2,ar^3, \ldots. We see a similar pattern to our previous table and can write down the formula for the nth term: t_n=ar^{n-1}.

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

  • Recursive form: t_n=rt_{n-1}, where t_1=a or alternatively t_{n+1}=rt_n, where t_1=a

  • Explicit form: t_n=ar^{n-1}

Examples

Example 1

Study the pattern for the following geometric sequence, and write down the next two terms.4,\, 12, \,36, \,\ldots

Worked Solution
Create a strategy

To find the common ratio, divide the second term by the first term. Use the common ratio to determine the next two terms.

Apply the idea

To get the common ratio r, we divide the second term, 12, by the first term, 4.

\displaystyle r\displaystyle =\displaystyle \dfrac{12}{4}Divide 12 by 4
\displaystyle =\displaystyle 3Evaluate

Calculating the next two terms, we have:

\displaystyle 4 \text{th term}\displaystyle =\displaystyle 36 \times 3Multiply 36 by r=3
\displaystyle =\displaystyle 108Evaluate
\displaystyle 5 \text{th term}\displaystyle =\displaystyle 108 \times 3Multiply 108 by r=3
\displaystyle =\displaystyle 324Evaluate

So, the geometric sequence is 4, 12,36,108,324.

Example 2

In a geometric progression, T_4=32 and T_6=128.

a

Solve for r, the common ratio in the sequence.

Worked Solution
Create a strategy

Multiply the 4th term by r^2 to give us the 6th term.

Apply the idea

T_5=r\times T_4 and T_6=r\times T_5. So:

\displaystyle T_6\displaystyle =\displaystyle r\times \left(r\times T_4\right)Substitute T_5=r\times T_4
\displaystyle =\displaystyle r^2 \times T_4Simplify
\displaystyle 128\displaystyle =\displaystyle 32 \times r^2Substitute T_6=128,T_4=32
\displaystyle r^2\displaystyle =\displaystyle \dfrac{128}{32}Divide both sides by 32
\displaystyle r^2\displaystyle =\displaystyle 4Evaluate the division
\displaystyle r\displaystyle =\displaystyle \pm2Square root both sides
b

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

Worked Solution
Create a strategy

Use the formula: T_n=ar^{n-1}.

Apply the idea

We are given T_4=32 and r=2.

\displaystyle T_4\displaystyle =\displaystyle a\times r^{4-1}Substitute n=4
\displaystyle 32\displaystyle =\displaystyle a\times 2^3Simplify and substitute r=2, T_4=32
\displaystyle 32\displaystyle =\displaystyle 8aEvaluate the power
\displaystyle a\displaystyle =\displaystyle \dfrac{32}{8}Divide both sides by 8
\displaystyle =\displaystyle 4Evaluate
c

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

Worked Solution
Create a strategy

Substitute the values of a, and r, into the formula: T_n=ar^{n-1}.

Apply the idea

We have found that a=4 and r=2.

\displaystyle T_n\displaystyle =\displaystyle a \times r^{n-1}Write the formula
\displaystyle =\displaystyle 4 \times 2^{n-1}Substitute the values
Idea summary

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

  • Recursive form: t_n=rt_{n-1}, where t_1=a or alternatively t_{n+1}=rt_n, where t_1=a

  • Explicit form: t_n=ar^{n-1}

Geometric sequences in tables and graphs

When given a formula for the 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}, by substituting for n appropriately and using a calculator, we can generate the following table of the first 6 terms of the sequence:

n123456
T_n12182740.560.75 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. The path of points plotted from a geometric sequence follow an exponential curve for positive values of r.

Sequences of the form t_n=ar^{n-1} are exponential graphs, and where a>0 they will follow:

  • The path of an exponential growth function for r>1.

  • The path of an exponential decay function for 0<r<1.

  • If a is negative the path will be reflected about the x-axis.

What happens when r is negative? The values of successive terms flip their sign so that the graph is depicted as either a growing (r<-1) or diminishing (-1<r<0) zig-zag path - alternating between points on the graph f(n)=ar^{n-1} and f(n)=-ar^{n-1}, depending on the power being odd or even.

Consider the geometric progression with starting value 12 and ratio r=-1.5. This is the same as the example in the previous table but the ratio is now negative. The nth term is given by t_n=12 \times (-1.5)^{n-1}, the table will be the same but the sign of the terms will alternate.

The new table becomes:

n123456
T_n12-1827-40.560.75 -91.125

Below is the comparison of the graphs of T_n=12 \times 1.5^{n-1} and T_n=12\times (-1.5)^{n-1}. The graph in red illustrates T_n=12 \times 1.5^{n-1} and the zig-zag graph illustrates T_n=12\times (-1.5)^{n-1}.

The graph of two geometric sequences. Ask your teacher for more information.

Based on the graph above, it can be observed that the odd terms of the zig-zag graph coincide with the terms of the first geometric progression.

Exploration

Try adjusting the values of a and r in the applet below to observe the effect on the plotted points.

Loading interactive...

If r \gt 0, the points trend upwards along an exponential function. If r \lt 0, the points zig-zag about the x-axis.

If a \gt 0, the plotted points trend upwards. If a \lt 0, the plotted points trend downwards.

Examples

Example 3

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

a

Complete the table of values:

n123410
T_n
Worked Solution
Create a strategy

Substitute the appropriate values of n into the equation and evaluate.

Apply the idea
\displaystyle T_{1}\displaystyle =\displaystyle 2 \times 3^{1-1}Substitute n=1
\displaystyle =\displaystyle 2Evaluate
\displaystyle T_{2}\displaystyle =\displaystyle 2 \times 3^{2-1}Substitute n=2
\displaystyle =\displaystyle 6Evaluate
\displaystyle T_{3}\displaystyle =\displaystyle 2 \times 3^{3-1}Substitute n=3
\displaystyle =\displaystyle 18Evaluate
\displaystyle T_{4}\displaystyle =\displaystyle 2 \times 3^{4-1}Substitute n=4
\displaystyle =\displaystyle 2 \times 27Evaluate the power
\displaystyle =\displaystyle 54Evaluate
\displaystyle T_{10}\displaystyle =\displaystyle 2 \times 3^{10-1}Substitute n=10
\displaystyle =\displaystyle 39\,366Evaluate

So, the complete table of values is given by:

n123410
T_n26185439\,366
b

What is the common ratio between consecutive terms?

Worked Solution
Create a strategy

To find the common ratio, divide the second term by the first term. Use the table from part (a).

Apply the idea

To get the common ratio r, we divide the second term, 6, by the first term, 2.

\displaystyle r\displaystyle =\displaystyle \dfrac{6}{2}Divide 6 by 2
\displaystyle =\displaystyle 3Evaluate
c

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

Worked Solution
Create a strategy

Use the table from part (a) to graph the appropriate points.

Apply the idea
1
2
3
4
n
5
10
15
20
25
30
35
40
45
50
55
T_n

From the table, the first 4 points have coordinates (1,2),\,(2,6),\,(3,18),\,(4,54).

d

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

A
a straight line
B
a curve
Worked Solution
Create a strategy

Draw a continuous function through the points.

Apply the idea
1
2
3
4
n
5
10
15
20
25
30
35
40
45
50
55
T_n

We can see that when we draw a function through the points, we get a curve.

The correct answer is Option B.

Example 4

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

1
2
3
4
5
x
5
10
15
20
25
30
35
40
45
50
55
y

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

Worked Solution
Create a strategy

Use the y-coordinates of the points as the terms of the sequence.

Apply the idea

The y-coordinates of the terms in order are: 54, \,18, \,6,\, 2,\ldots So this is the sequence with first term a=54.

We must first check that the sequence is geometric by dividing each term by the term before it to check that there is a common ratio.

\displaystyle \dfrac{T_2}{T_1}\displaystyle =\displaystyle \dfrac{18}{54}Divide T_2 by T_1
\displaystyle =\displaystyle \dfrac{1}{3}Evaluate
\displaystyle \dfrac{T_3}{T_2}\displaystyle =\displaystyle \dfrac{6}{18}Divide T_3 by T_2
\displaystyle =\displaystyle \dfrac{1}{3}Evaluate
\displaystyle \dfrac{T_4}{T_3}\displaystyle =\displaystyle \dfrac{2}{6}Divide T_4 by T_3
\displaystyle =\displaystyle \dfrac{1}{3}Evaluate

Since each ratio is \dfrac{1}{3} this is the value of r.

By substituting the above values of a and r into the recursive rule T_n=r T_{n-1},\,T_1=a we get::T_n=\frac{1}{3}T_{n-1},\,T_1=54

Example 5

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

n1234
T_n7-2163-189
a

Identify r, the common ratio between consecutive terms.

Worked Solution
Create a strategy

To find the common ratio, divide the second term by the first term based on the given table of values.

Apply the idea

To get the common ratio r, we divide the second term, -21 by the first term, 7.

\displaystyle r\displaystyle =\displaystyle \dfrac{-21}{7}Divide -21 by 7
\displaystyle =\displaystyle -3Evaluate
b

Write a simplified expression for the general nth term of the sequence, T_n.

Worked Solution
Create a strategy

Substitute the values of a, the first term in sequence, and r, the common ratio, into the formula for finding the nth term: T_n=ar^{n-1}.

Apply the idea

We are given a=7 from the table where n=1, and we now know that r=-3.

\displaystyle T_n\displaystyle =\displaystyle a \times r^{n-1}Write the formula
\displaystyle =\displaystyle 7 \times (-3)^{n-1}Substitute a and r
c

Find the 12th term of the sequence.

Worked Solution
Create a strategy

Substitute n=12 into the equation for T_n found from part (b).

Apply the idea
\displaystyle T_{12}\displaystyle =\displaystyle 7\times (-3)^{12-1}Substitute n=12
\displaystyle =\displaystyle -1\,240\,029Evaluate
Idea summary

We can represent geometric sequences in both tables and graphs.

From a graph of points, the sequence can be read from the y-coordinates.

Outcomes

ACMGM071

use recursion to generate a geometric sequence

ACMGM072

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

ACMGM073

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

ACMGM074

use geometric sequences to model and analyse (numerically, or graphically only) practical problems involving geometric growth and decay; for example, analysing a compound interest loan or investment, the growth of a bacterial population that doubles in size each hour, the decreasing height of the bounce of a ball at each bounce; or calculating the value of office furniture at the end of each year using the declining (reducing) balance method to depreciate

What is Mathspace

About Mathspace