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VCE 11 General 2023

5.05 Geometric sequences

Lesson

Geometric sequence definition

A sequence in which each term increases or decreases from the last by multiplying by a constant factor is called a geometric sequence. The constant factor multiplying each term in the sequence to create the next term is referred to as the common ratio, which will result from dividing any two successive terms \dfrac{u_{n+1}}{u_n}.

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

Since, u_2=R\times u_1, u_3=R\times u_2, and so on, any geometric sequence can be written as a recurrence relation, which was found  previously in this chapter  :

u_{n+1}=Ru_n, u_0=a

An explicit generating rule can be found in terms of a and R. This is useful for finding the nth term without listing the sequence or having to use the previous term in the sequence each time to find the next term.

Consider the following table to see the pattern for the explicit formula. For the sequence 5,10,20,40, \ldots, the starting term is 5 and there is a common ratio of 2, that is a=5 and R=2.

A table of the sequence is show below:

nu_{n+1}\text{Pattern}
055 \times 2^0
1105 \times 2^1
2205 \times 2^2
3405 \times 2^3
\ldots
nu_{n+1}5 \times 2^n

By correctly identifying the pattern, the tenth term becomes u_{10}=10\times 2^9 and the one-hundredth term would be u_{100}=5\times 2^{99}. And following the pattern, the explicit formula for the nth term is u_n=5\times 2^{n-1}.

For any geometric progression with starting value a and common ratio R and with terms given by: a,aR,aR^2,aR^3,... , a similar pattern can be observed. Hence, the explicit generating rule for the nth term in any geometric sequence is given by:

u_n=aR^{n-1}

Examples

Example 1

Consider the following sequence.

810, \,270, \,90,\,30,\,...

a

Find an explicit rule for the nth term.

Worked Solution
Create a strategy

Find the common ratio r to substitute into u_n=aR^{n-1}.

Apply the idea

Since 810 \times \dfrac 13=270, 270 \times \dfrac 13=90, and so on, R=\dfrac 13.

The first term is 810, so a=810.

\displaystyle u_n\displaystyle =\displaystyle aR^{n-1}Write the formula
\displaystyle u_n\displaystyle =\displaystyle 810 \times \left(\dfrac{1}{3}\right)^{n-1}Substitute a=810 and R=\dfrac 13
b

Hence, find the 8th term.

Worked Solution
Create a strategy

Substitute n=8 into u_n=810 \times \left(\dfrac{1}{3}\right)^{n-1}.

Apply the idea
\displaystyle u_8\displaystyle =\displaystyle 810 \times \left(\frac{1}{3}\right)^{n-1}Write the formula
\displaystyle =\displaystyle 810 \times \left(\frac{1}{3}\right)^{8-1}Substitute n=8
\displaystyle =\displaystyle 810 \times \left(\frac{1}{3}\right)^7Evaluate the subtraction
\displaystyle =\displaystyle 810 \times \dfrac{1}{2187}Apply the rule \left(\dfrac ab\right)^m=\dfrac{a^m}{b^m}
\displaystyle =\displaystyle \frac{810}{2187}Evaluate multiplication
\displaystyle =\displaystyle \frac{10}{27}Simplify the fraction

Example 2

For the sequence 5, 20, 80, 320,..., find n if the nth term is 327 \,680.

Worked Solution
Create a strategy

Substitute u_n=327 \, 680, a=5, and value of d into u_n=aR^{n-1}.

Apply the idea

Since 5 \times 4=20, 20 \times 4=80, and so on, R=4.

\displaystyle u_n\displaystyle =\displaystyle aR^{n-1}Write the formula
\displaystyle 327 \, 680\displaystyle =\displaystyle 5 \times 4^{n-1}Substitute u_n=327 \, 680, a=5, R=4
\displaystyle \dfrac{327 \, 680}{5}\displaystyle =\displaystyle \dfrac{5 \times 4^{n-1}}{5}Divide both sides by 5
\displaystyle 65 \, 536\displaystyle =\displaystyle 4^{n-1}Evaluate the division
\displaystyle 4^8\displaystyle =\displaystyle 4^{n-1}Write in exponential form
\displaystyle 8\displaystyle =\displaystyle n-1Equate the indices of same bases
\displaystyle 8+1\displaystyle =\displaystyle n-1+1Add 1 to both sides
\displaystyle 9\displaystyle =\displaystyle nEvaluate
\displaystyle n\displaystyle =\displaystyle 9Make n the subject

Example 3

Study the pattern for the following sequence.

-9, \, 8.1, \, -7.29, \, 6.561, \,\ldots

State the common ratio between the terms.

Worked Solution
Create a strategy

Divide the second term by the first term.

Apply the idea
\displaystyle \text{Common ratio}\displaystyle =\displaystyle \dfrac{8.1}{-9}Divide the second term by the first term
\displaystyle =\displaystyle \dfrac{8.1 \times 10}{-9 \times 10}Multiply each by 10
\displaystyle =\displaystyle \dfrac{81}{-90}Evaluate
\displaystyle =\displaystyle -\dfrac{9}{10}Simplify the fraction
Reflect and check

We can also write the common ratio into decimals, that is -\dfrac{9}{10}=-0.9.

Example 4

If a geometric sequence has u_3=12 and u_6=96, find the recurrence relation for the sequence.

Worked Solution
Create a strategy

Use the fact that each equation is equal to u_n=aR^{n-1}.

Then set up the proportion of the ratio of aR^{n-1} terms and u_n values.

Apply the idea

We will get u_3=12 through u_3=aR^{3-1}=12.

We will get u_6=96 through u_6=aR^{6-1}=96.

\displaystyle \dfrac{aR^{6-1}}{aR^{3-1}}\displaystyle =\displaystyle \dfrac{96}{12}Set up the proportion
\displaystyle \dfrac{R^{6-1}}{R^{3-1}}\displaystyle =\displaystyle \dfrac{96}{12}Quotient of same terms is 1
\displaystyle \dfrac{R^{5}}{R^{2}}\displaystyle =\displaystyle \dfrac{96}{12}Evaluate the subtraction
\displaystyle R^{3}\displaystyle =\displaystyle \dfrac{96}{12}Apply the rule \dfrac{a^m}{a^n}=a^{m-n}
\displaystyle R^{3}\displaystyle =\displaystyle 8Evaluate the division
\displaystyle \sqrt[3]{R^{3}}\displaystyle =\displaystyle \sqrt[3]{8}Take the cube root of both sides
\displaystyle R\displaystyle =\displaystyle 2Evaluate

We can substitute u_n=12 where n=3 and R=2 into the explicit formula to work out for a.

\displaystyle u_n\displaystyle =\displaystyle aR^{n-1}Write the explicit formula
\displaystyle 12\displaystyle =\displaystyle a \times 2^{3-1}Substitute u_n=12, R=2, n=3
\displaystyle 12\displaystyle =\displaystyle a \times 2^2Evaluate the subtraction
\displaystyle 12\displaystyle =\displaystyle a \times 4Evaluate the exponent
\displaystyle \dfrac{12}{4}\displaystyle =\displaystyle \dfrac{a \times 4}{4}Divide both sides by 4
\displaystyle 3\displaystyle =\displaystyle aEvaluate
\displaystyle a\displaystyle =\displaystyle 3Make a the subject

Now, we can substitute the values found into the recursive formula u_{n+1}=Ru_n,u_0=a.

\displaystyle u_{n+1}\displaystyle =\displaystyle Ru_n,u_0=aWrite the recursive formula
\displaystyle u_{n+1}\displaystyle =\displaystyle 2u_n,u_0=3Substitute R=2 and a=3

Example 5

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

a

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

Worked Solution
Create a strategy

Multiply T^4 by r steps and equate to T^6.

Apply the idea

From T^4=32 we can multiply r twice to get into T^6=128.

Translating this into an equation will be 32 \times r \times r=128.

\displaystyle 32 \times r \times r\displaystyle =\displaystyle 128Write the equation
\displaystyle 32 \times r^2\displaystyle =\displaystyle 128Evaluate r \times r
\displaystyle \dfrac{32 \times r^2}{32}\displaystyle =\displaystyle \dfrac{128}{32}Divide both sides by 32
\displaystyle r^2\displaystyle =\displaystyle 4Evaluate
\displaystyle \sqrt{r^2}\displaystyle =\displaystyle \sqrt 4Take the square root of both sides
\displaystyle r\displaystyle =\displaystyle 2, -2Evaluate

Since the terms are increasing, r=2.

b

Solve for a, the first term in the progression.

Worked Solution
Create a strategy

Substitute T_n=32, r=2, and n=4 into T_n=aR^{n-1}.

Apply the idea
\displaystyle T_n\displaystyle =\displaystyle aR^{n-1}Write the formula
\displaystyle 32\displaystyle =\displaystyle a \times 2^{4-1}Substitute T_n=32, r=2, n=4
\displaystyle 32\displaystyle =\displaystyle a \times 2^3Evaluate the subtraction
\displaystyle 32\displaystyle =\displaystyle a \times 8Evaluate the exponent
\displaystyle \dfrac{32}{8}\displaystyle =\displaystyle \dfrac{a \times 8}{8}Divide both sides by 8
\displaystyle 4\displaystyle =\displaystyle aEvaluate
\displaystyle a\displaystyle =\displaystyle 4Make a the subject
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 a=4 and R=2 into T_n=aR^{n-1}.

Apply the idea
\displaystyle T_n\displaystyle =\displaystyle aR^{n-1}Write the explicit formula
\displaystyle T_n\displaystyle =\displaystyle 4 \times 2^{n-1}Substitute a=4 and R=2
Idea summary

Any geometric sequence can be expressed in either of the following two forms:

  • Recursive form is a way to express any term in relation to the previous term:
    \displaystyle u_{n+1} = Ru_n, u_0 = a
    \bm{R}
    is the common ratio
  • Explicit form is a way to express any term in relation to the term number:
    \displaystyle u_n = aR^{n-1}
    \bm{a}
    is the first term
    \bm{R}
    is the common ratio

Geometric sequences in tables and graphs

When given a formula for the nth term, a table of values can be generated for the sequence. For example, in the sequence given by the formula u_n=12\times \left(1.5\right)^{n - 1}, by substituting for n appropriately and using a calculator, the following table can be created, listing the first 6 terms of the sequence:

n123456
u_n12182740.560.7591.125

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 (graph can be seen below in the next worked example).

Geometric sequences of the form u_n=aR^{n-1}, where a>0 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 if 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 (|R|<1) zig-zag path - alternating between points on the graph f(n)=a|R|^{n-1} and f(n)=-a|R|^{n-1}, depending on the power being odd or even.

Exploration

Adjust the values of a and R in the applet below to observe the effect on the plotted points.

Loading interactive...

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

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

For the geometric progression with starting value 12 and ratio R = -1.5, we can create a table and plot a graph of the sequence.

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

The new table becomes:

n123456
u_n12-1827-40.560.75-91.125

Checking, for n = 1, we have u_1=12\times \left(-1.5\right)^{1-1}=12 and for n = 2 we have u_2=12\times \left(-1.5\right)^{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.

1
2
3
4
5
6
n
-100
-80
-60
-40
-20
20
40
60
80
100
n\text{th term}

Examples

Example 6

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 each n value into T_n=2\times 3^{n-1}.

Apply the idea
\displaystyle T_n\displaystyle =\displaystyle 2\times 3^{n-1}Write the formula
\displaystyle T_1\displaystyle =\displaystyle 2\times 3^{1-1}Substitute n=1
\displaystyle =\displaystyle 2\times 3^0Evaluate the subtraction
\displaystyle =\displaystyle 2\times 1Apply the zero index law
\displaystyle =\displaystyle 2Evaluate

Repeating the same working to the next n values gives the following complete table.

n123410
T_n26185439 \,366
b

What is the common ratio between consecutive terms?

Worked Solution
Create a strategy

Divide the second term by the first term.

Apply the idea
\displaystyle \text{Common ratio}\displaystyle =\displaystyle \dfrac 62Divide the second term by the first term
\displaystyle =\displaystyle 3Evaluate
c

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

Worked Solution
Apply the idea
1
2
3
4
x
4
8
12
16
20
24
28
32
36
40
44
48
52
56
y

By plotting the values in the table, we have points at (1,2),(2,6),(3,18), and (4,54).

d

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

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

A graph of a geometric sequence is represented by a curved line.

Apply the idea

The graph from part (c) represents the geometric sequence T^n=2 \times 3 ^{n-1}, so the answer is option B.

Idea summary

Because of the quadratic nature of geometric sequence, its graph is represented by a curved line.

Outcomes

U1.AoS2.7

use a given recurrence relation to generate a sequence, deduce the explicit rule, n u from the recursion relation, tabulate, graph and evaluate the sequence

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