The Central Idea
Consider the sequence
\[ 3,\ 6,\ 12,\ 24,\ 48,\ldots \]Each term is twice the previous term:
\[ \frac{6}{3}=2, \] \[ \frac{12}{6}=2, \] \[ \frac{24}{12}=2. \]Because the ratio between consecutive terms is constant, this is a geometric sequence.
The constant ratio is
\[ \boxed{r=2}. \]What Is a Geometric Sequence?
A geometric sequence is a sequence in which each term is obtained by multiplying the preceding term by the same constant.
That constant is called the common ratio.
If the common ratio is \(r\), then
\[ \frac{a_n}{a_{n-1}}=r \]whenever \(a_{n-1}\neq0\).
Equivalently,
\[ \boxed{a_n=ra_{n-1}}. \]The Common Ratio
The constant multiplier from one term to the next is called the common ratio.
It is usually represented by
\[ r. \]When consecutive nonzero terms are available, calculate
\[ \boxed{r=\frac{a_n}{a_{n-1}}}. \]For a geometric sequence, the same ratio appears from one step to the next.
Example: Finding the Common Ratio
Consider
\[ 5,\ 15,\ 45,\ 135,\ldots \]Divide consecutive terms:
\[ \frac{15}{5}=3, \] \[ \frac{45}{15}=3, \] \[ \frac{135}{45}=3. \]Therefore,
\[ \boxed{r=3}. \]The Common Ratio Can Be a Fraction
Geometric sequences do not have to grow.
Consider
\[ 64,\ 32,\ 16,\ 8,\ 4,\ldots \]Then
\[ \frac{32}{64}=\frac12, \] \[ \frac{16}{32}=\frac12. \]Thus,
\[ \boxed{r=\frac12}. \]Repeated multiplication by a positive number between \(0\) and \(1\) produces terms whose magnitudes decrease.
The Common Ratio Can Be Negative
A negative common ratio causes the signs of successive nonzero terms to alternate.
Consider
\[ 2,\ -6,\ 18,\ -54,\ldots \]Then
\[ \frac{-6}{2}=-3, \] \[ \frac{18}{-6}=-3. \]Therefore,
\[ \boxed{r=-3}. \]The Common Ratio Can Be One
A nonzero constant sequence is geometric.
For
\[ 7,\ 7,\ 7,\ 7,\ldots \]we have
\[ \frac77=1. \]Thus,
\[ \boxed{r=1}. \]Multiplying by \(1\) leaves every term unchanged.
What About a Common Ratio of Zero?
If a nonzero term is multiplied by \(0\), every later term becomes zero.
For example,
\[ 5,\ 0,\ 0,\ 0,\ldots \]can be generated recursively by
\[ a_1=5,\qquad a_n=0a_{n-1}. \]However, ratios such as
\[ \frac{0}{0} \]are undefined.
For this reason, the recursive multiplier definition is more reliable than quotient testing when zeros occur.
Recognizing a Geometric Sequence
To determine whether a listed sequence of nonzero terms is geometric:
- divide each term by the term immediately before it;
- compare the resulting ratios;
- if every ratio is the same, the sequence is geometric.
For example,
\[ 4,\ 12,\ 36,\ 108,\ldots \]has ratios
\[ 3,\ 3,\ 3,\ldots \]so it is geometric.
Example: A Sequence That Is Not Geometric
Consider
\[ 2,\ 5,\ 8,\ 11,\ldots \]The ratios are
\[ \frac52, \qquad \frac85, \qquad \frac{11}{8}. \]These ratios are not equal.
Therefore,
\[ \boxed{\text{the sequence is not geometric}}. \]Its constant differences instead identify it as arithmetic.
The Recursive Definition
A geometric sequence is naturally described recursively.
If the first term is \(a_1\) and the common ratio is \(r\), then
\[ \boxed{ a_1=\text{given},\qquad a_n=ra_{n-1}\quad(n\geq2) }. \]The initial condition supplies the starting value, and the recurrence repeatedly multiplies by the same ratio.
Example: Writing a Recursive Definition
Consider
\[ 4,\ 20,\ 100,\ 500,\ldots \]The first term is
\[ a_1=4, \]and the common ratio is
\[ r=5. \]Therefore,
\[ \boxed{ a_1=4,\qquad a_n=5a_{n-1}\quad(n\geq2) }. \]Building the Explicit Formula
Suppose the first term is \(a_1\) and the common ratio is \(r\).
Then
\[ a_2=a_1r. \]The third term is
\[ a_3=a_1r^2. \]The fourth term is
\[ a_4=a_1r^3. \]To reach the \(n\)th term, the ratio has been applied \(n-1\) times.
Therefore,
\[ \boxed{a_n=a_1r^{n-1}}. \]The Explicit Formula
The standard explicit formula for a geometric sequence indexed from \(1\) is
\[ \boxed{a_n=a_1r^{n-1}}. \]Here:
- \(a_n\) is the \(n\)th term;
- \(a_1\) is the first term;
- \(r\) is the common ratio;
- \(n\) is the index.
Why the Exponent Is \(n-1\)
The first term already exists before any multiplication by \(r\).
To reach \(a_2\), multiply by \(r\) once.
To reach \(a_3\), multiply by \(r\) twice.
To reach \(a_n\), multiply by \(r\)
\[ n-1 \]times.
Therefore,
\[ a_n=a_1r^{n-1}. \]Example: Writing an Explicit Formula
Consider
\[ 6,\ 18,\ 54,\ 162,\ldots \]We have
\[ a_1=6 \]and
\[ r=3. \]Therefore,
\[ \boxed{a_n=6\cdot3^{n-1}}. \]Finding a Distant Term
Suppose
\[ a_1=5,\qquad r=2. \]Find \(a_{10}\).
Use
\[ a_n=a_1r^{n-1}. \]Then
\[ a_{10}=5(2)^{10-1} \] \[ =5(2^9) \] \[ =5(512) \] \[ =\boxed{2560}. \]Finding a Term with a Fractional Ratio
Let
\[ a_1=96,\qquad r=\frac12. \]Find \(a_6\).
\[ a_6=96\left(\frac12\right)^5. \]Since
\[ \left(\frac12\right)^5=\frac1{32}, \]we obtain
\[ a_6=96\cdot\frac1{32} \] \[ =\boxed{3}. \]Finding a Term with a Negative Ratio
Let
\[ a_1=2,\qquad r=-3. \]Find \(a_5\).
\[ a_5=2(-3)^4. \]Because
\[ (-3)^4=81, \]we have
\[ \boxed{a_5=162}. \]The even exponent makes this term positive.
Finding the Index of a Known Term
Suppose
\[ a_n=3\cdot2^{n-1}. \]At what index does the value \(96\) occur?
Set
\[ 96=3\cdot2^{n-1}. \]Divide by \(3\):
\[ 32=2^{n-1}. \]Since
\[ 32=2^5, \]we have
\[ n-1=5. \]Therefore,
\[ \boxed{n=6}. \]When Logarithms Are Needed
Sometimes the desired index is not visible from familiar powers.
Suppose
\[ a_n=5(1.2)^{n-1} \]and we want to solve
\[ a_n=20. \]Then
\[ 20=5(1.2)^{n-1}. \]Divide by \(5\):
\[ 4=(1.2)^{n-1}. \]Taking logarithms gives
\[ n-1=\frac{\ln 4}{\ln 1.2}. \]Therefore,
\[ \boxed{ n=1+\frac{\ln 4}{\ln 1.2} }. \]If an integer index is required, the result must be checked against the allowed indices.
Finding a Missing Term
Suppose
\[ 4,\ 12,\ \square,\ 108 \]is geometric.
The common ratio is
\[ \frac{12}{4}=3. \]Therefore, the missing term is
\[ 12(3)=\boxed{36}. \]Geometric Means
A number inserted between two positive values so that the three terms form a geometric sequence is called a geometric mean.
Suppose
\[ a,\ x,\ b \]is geometric.
Then
\[ \frac{x}{a}=\frac{b}{x}. \]Cross-multiplying gives
\[ x^2=ab. \]For positive \(a\) and \(b\), the positive geometric mean is
\[ \boxed{x=\sqrt{ab}}. \]Example: Finding a Geometric Mean
Insert a positive geometric mean between \(4\) and \(36\).
Compute
\[ \sqrt{4\cdot36} \] \[ =\sqrt{144} \] \[ =12. \]Therefore,
\[ \boxed{4,\ 12,\ 36} \]is geometric with common ratio \(3\).
Finding the Common Ratio from Two Terms
Suppose
\[ a_3=20 \]and
\[ a_6=160. \]From index \(3\) to index \(6\), the ratio is applied three times:
\[ 160=20r^3. \]Divide by \(20\):
\[ r^3=8. \]Therefore,
\[ \boxed{r=2}. \]A General Two-Term Formula
For any two terms \(a_m\) and \(a_n\) of a geometric sequence,
\[ \boxed{a_n=a_mr^{n-m}}. \]This is the multiplicative counterpart of the arithmetic-sequence formula
\[ a_n=a_m+(n-m)d. \]The exponent \(n-m\) counts how many multiplicative steps separate the two indices.
Example: Using a Known Later Term
Suppose
\[ a_4=10 \]and
\[ r=2. \]Find \(a_9\).
Use
\[ a_9=a_4r^{9-4}. \]Therefore,
\[ a_9=10(2)^5 \] \[ =10(32) \] \[ =\boxed{320}. \]Finding the First Term from Another Term
Suppose
\[ a_5=162 \]and
\[ r=3. \]Use
\[ a_5=a_1r^4. \]Then
\[ 162=a_1(3^4). \]Since
\[ 3^4=81, \]we have
\[ 162=81a_1. \]Therefore,
\[ \boxed{a_1=2}. \]Finding an Explicit Formula from Two Terms
Suppose
\[ a_2=6 \]and
\[ a_5=162. \]Use
\[ a_5=a_2r^{5-2}. \]Then
\[ 162=6r^3. \]Divide by \(6\):
\[ r^3=27. \]Thus,
\[ r=3. \]Since
\[ a_2=a_1r, \]we have
\[ 6=3a_1, \]so
\[ a_1=2. \]Therefore,
\[ \boxed{a_n=2\cdot3^{n-1}}. \]Geometric Sequences and Exponential Functions
The explicit formula
\[ a_n=a_1r^{n-1} \]contains the index in an exponent.
It can also be written
\[ a_n=\frac{a_1}{r}r^n \]when \(r\neq0\).
This has the same basic form as an exponential function
\[ y=Ab^x. \]Geometric sequences are therefore exponential relationships restricted to a discrete index set.
The Common Ratio Is a Multiplicative Rate
In an arithmetic sequence, equal index steps produce equal differences.
In a geometric sequence, equal index steps produce equal ratios.
That distinction can be summarized as
\[ \boxed{\text{arithmetic: add a constant}} \]versus
\[ \boxed{\text{geometric: multiply by a constant}}. \]This is the discrete version of the distinction between linear and exponential change.
Percent Change and Common Ratio
Repeated percentage change naturally produces a geometric sequence.
If a quantity increases by \(p\%\) each step, then the common ratio is
\[ \boxed{r=1+\frac{p}{100}}. \]If a quantity decreases by \(p\%\) each step, then
\[ \boxed{r=1-\frac{p}{100}}. \]Thus a \(5\%\) increase corresponds to
\[ r=1.05, \]while a \(5\%\) decrease corresponds to
\[ r=0.95. \]Example: Repeated Percentage Growth
Suppose a quantity begins at \(200\) and grows by \(10\%\) each period.
The common ratio is
\[ r=1.10. \]If the first recorded value is \(a_1=200\), then
\[ \boxed{a_n=200(1.10)^{n-1}}. \]The first few terms are
\[ 200,\ 220,\ 242,\ 266.2,\ldots \]Example: Repeated Percentage Decay
Suppose a machine retains \(80\%\) of a quantity after each cycle.
Then
\[ r=0.8. \]If the initial recorded value is \(500\), then
\[ a_n=500(0.8)^{n-1}. \]The first few values are
\[ 500,\ 400,\ 320,\ 256,\ldots \]A Sequence Table
Consider
\[ a_n=2\cdot3^{n-1}. \]| Index \(n\) | Term \(a_n\) |
|---|---|
| \(1\) | \(2\) |
| \(2\) | \(6\) |
| \(3\) | \(18\) |
| \(4\) | \(54\) |
| \(5\) | \(162\) |
The index increases by \(1\), while each output is multiplied by \(3\).
The Discrete Graph
The sequence
\[ a_n=2\cdot3^{n-1} \]produces points such as
\[ (1,2),\ (2,6),\ (3,18),\ (4,54),\ldots \]These points lie on an exponential curve, but the sequence itself contains only the specified indexed points.
As with arithmetic sequences, the sequence is discrete even when a continuous function passes through its points.
Arithmetic Versus Geometric Change
| Arithmetic Sequence | Geometric Sequence |
|---|---|
| Constant difference | Constant ratio |
| Add or subtract the same amount | Multiply by the same factor |
| \(a_n=a_{n-1}+d\) | \(a_n=ra_{n-1}\) |
| \(a_n=a_1+(n-1)d\) | \(a_n=a_1r^{n-1}\) |
| Linear structure | Exponential structure |
Indexing from Zero
If a geometric sequence begins at \(n=0\), then
\[ a_0 \]is the starting term.
The explicit formula becomes
\[ \boxed{a_n=a_0r^n}. \]This is especially convenient because it has exactly the standard exponential form.
Example: Geometric Sequence Beginning at Zero
Let
\[ a_0=4 \]and
\[ r=2. \]Then
\[ a_n=4(2)^n. \]Thus,
\[ a_0=4, \] \[ a_1=8, \] \[ a_2=16, \] \[ a_3=32. \]The sequence begins
\[ \boxed{4,\ 8,\ 16,\ 32,\ldots}. \]A More General Indexing Formula
Suppose a known term is \(a_k\).
Then any other term can be written
\[ \boxed{a_n=a_kr^{n-k}}. \]The exponent \(n-k\) counts the number of repeated multiplications between the two indices.
The standard formula
\[ a_n=a_1r^{n-1} \]is the special case \(k=1\).
Geometric Sequences in Applications
Geometric sequences appear whenever a quantity changes by the same multiplicative factor at each discrete step.
Examples include:
- compound growth over equal periods;
- population growth by a fixed percentage;
- depreciation by a fixed percentage;
- radioactive decay measured at equal intervals;
- repeated dilution or reduction;
- repeated scaling in geometry;
- doubling and halving processes;
- some recursive algorithms and branching processes.
Example: Doubling
Suppose a process begins with \(3\) units and doubles during each cycle.
The sequence is
\[ 3,\ 6,\ 12,\ 24,\ 48,\ldots \]with
\[ r=2. \]The explicit formula is
\[ \boxed{a_n=3\cdot2^{n-1}}. \]Example: Repeated Halving
Suppose a quantity begins at \(160\) and is halved at each step.
Then
\[ 160,\ 80,\ 40,\ 20,\ldots \]has common ratio
\[ r=\frac12. \]Therefore,
\[ \boxed{ a_n=160\left(\frac12\right)^{n-1} }. \]Common Error: Using Differences Instead of Ratios
Geometric sequences are characterized by constant ratios, not constant differences.
For
\[ 2,\ 6,\ 18,\ 54,\ldots \]the differences are
\[ 4,\ 12,\ 36, \]which are not constant.
But the ratios are
\[ 3,\ 3,\ 3. \]Therefore, the sequence is geometric.
Common Error: Reversing the Ratio
The common ratio is
\[ r=\frac{\text{later term}}{\text{earlier term}}. \]For
\[ 20,\ 10,\ 5,\ldots \]we have
\[ r=\frac{10}{20} \] \[ =\boxed{\frac12}. \]Using \(20/10=2\) reverses the direction of the sequence.
Common Error: Using \(r^n\) Instead of \(r^{n-1}\)
Suppose
\[ a_1=5,\qquad r=2. \]The correct formula is
\[ a_n=5(2)^{n-1}. \]At \(n=1\),
\[ a_1=5(2)^0=5. \]If we incorrectly write
\[ 5(2)^n, \]the first term becomes \(10\), so the indexing is shifted.
Common Error: Ignoring a Negative Ratio
Consider
\[ 4,\ -8,\ 16,\ -32,\ldots \]The common ratio is
\[ \frac{-8}{4}=-2. \]Using \(r=2\) would reproduce the magnitudes but not the alternating signs.
Therefore,
\[ \boxed{r=-2}. \]Common Error: Assuming Every Rapidly Growing Sequence Is Geometric
Rapid growth alone does not prove a sequence is geometric.
Consider
\[ 2,\ 5,\ 15,\ 60,\ldots \]The ratios are
\[ \frac52,\qquad3,\qquad4. \]Since the ratios are not constant, the sequence is not geometric.
Common Error: Forgetting the Starting Index
The formula
\[ a_n=a_1r^{n-1} \]assumes that indexing begins at \(1\).
If indexing begins at \(0\), use
\[ a_n=a_0r^n. \]More generally,
\[ a_n=a_kr^{n-k}. \]A Reliable Procedure for Analyzing a Geometric Sequence
- Identify the indexing convention.
- Compare consecutive terms multiplicatively.
- Calculate the common ratio where defined.
- Verify that the same multiplier works throughout.
- Identify a known term such as \(a_1\).
- Write the recursive rule if needed.
- Write the explicit formula if needed.
- Substitute known information to solve for missing quantities.
- Check signs, powers, and index shifts carefully.
Example: Full Geometric-Sequence Workflow
Consider
\[ 81,\ 27,\ 9,\ 3,\ 1,\ldots \]First find the common ratio:
\[ r=\frac{27}{81} \] \[ =\frac13. \]The recursive definition is
\[ a_1=81,\qquad a_n=\frac13a_{n-1}. \]The explicit definition is
\[ \boxed{ a_n=81\left(\frac13\right)^{n-1} }. \]Example: Recovering a Sequence from Two Terms
Suppose
\[ a_3=12 \]and
\[ a_6=96. \]Since there are three multiplicative steps,
\[ 96=12r^3. \]Therefore,
\[ r^3=8 \]and
\[ r=2. \]Now
\[ a_3=a_1r^2, \]so
\[ 12=4a_1. \]Thus,
\[ a_1=3. \]The explicit rule is
\[ \boxed{a_n=3\cdot2^{n-1}}. \]Practice 1: Find the Common Ratio
Find the common ratio of
\[ 7,\ 21,\ 63,\ 189,\ldots \]Practice 2: Fractional Common Ratio
Find the common ratio of
\[ 128,\ 64,\ 32,\ 16,\ldots \]Practice 3: Negative Common Ratio
Find the common ratio of
\[ 5,\ -10,\ 20,\ -40,\ldots \]Practice 4: Write a Recursive Definition
Write a recursive definition for
\[ 3,\ 12,\ 48,\ 192,\ldots \]assuming indexing begins at \(1\).
Practice 5: Write an Explicit Formula
Write an explicit formula for
\[ 10,\ 30,\ 90,\ 270,\ldots \]assuming indexing begins at \(1\).
Practice 6: Find a Distant Term
Let
\[ a_1=2,\qquad r=3. \]Find
\[ a_7. \]Practice 7: Find the Index
Let
\[ a_n=4\cdot2^{n-1}. \]At what index does the value \(128\) occur?
Practice 8: Find the Missing Term
Complete the geometric sequence:
\[ 6,\ 18,\ \square,\ 162. \]Practice 9: Recover the Explicit Formula
Suppose
\[ a_2=10 \]and
\[ a_5=80. \]Find an explicit formula for \(a_n\), assuming a real positive common ratio.
Practice 10: Identify the Error
A student says that
\[ 2,\ 6,\ 18,\ 55,\ldots \]is geometric because the terms are growing rapidly.
Explain the error.
Practice Answers
Practice 1
\[ r=\frac{21}{7} \] \[ =\boxed{3}. \]Practice 2
\[ r=\frac{64}{128} \] \[ =\boxed{\frac12}. \]Practice 3
\[ r=\frac{-10}{5} \] \[ =\boxed{-2}. \]Practice 4
The first term is \(3\), and the common ratio is \(4\).
\[ \boxed{ a_1=3,\qquad a_n=4a_{n-1}\quad(n\geq2) }. \]Practice 5
Here,
\[ a_1=10,\qquad r=3. \]Therefore,
\[ \boxed{a_n=10\cdot3^{n-1}}. \]Practice 6
\[ a_7=2(3)^6 \] \[ =2(729) \] \[ =\boxed{1458}. \]Practice 7
Solve
\[ 4\cdot2^{n-1}=128. \]Divide by \(4\):
\[ 2^{n-1}=32. \]Since
\[ 32=2^5, \]we have
\[ n-1=5. \]Therefore,
\[ \boxed{n=6}. \]Practice 8
The common ratio is
\[ r=\frac{18}{6}=3. \]Therefore,
\[ 18(3)=54. \]The completed sequence is
\[ \boxed{6,\ 18,\ 54,\ 162}. \]Practice 9
From \(a_2\) to \(a_5\) there are three multiplicative steps:
\[ 80=10r^3. \]Therefore,
\[ r^3=8 \]and
\[ r=2. \]Since
\[ a_2=a_1r, \]we have
\[ 10=2a_1, \]so
\[ a_1=5. \]Therefore,
\[ \boxed{a_n=5\cdot2^{n-1}}. \]Practice 10
Rapid growth does not by itself make a sequence geometric.
Check the ratios:
\[ \frac62=3, \] \[ \frac{18}{6}=3, \]but
\[ \frac{55}{18}\neq3. \]Therefore, the sequence does not have a constant common ratio and is
\[ \boxed{\text{not geometric}}. \]What to Remember
A geometric sequence changes by a constant multiplicative factor:
\[ \boxed{r=\frac{a_n}{a_{n-1}}} \]whenever the quotient is defined.
Its recursive form is
\[ \boxed{ a_1=\text{given},\qquad a_n=ra_{n-1} }. \]Its standard explicit form is
\[ \boxed{a_n=a_1r^{n-1}}. \]More generally,
\[ \boxed{a_n=a_kr^{n-k}}. \]- The common ratio may be positive, negative, fractional, or zero.
- A negative common ratio produces alternating signs.
- A ratio between \(0\) and \(1\) produces decreasing positive magnitudes.
- Repeated percentage change produces geometric sequences.
- Geometric sequences are discrete exponential relationships.
- The starting index determines the exponent in the explicit formula.
- Known terms can be used to recover \(r\), \(a_1\), or the general term.
Where We Go Next
Arithmetic sequences have constant first differences, while geometric sequences have constant ratios. Many important sequences follow neither pattern but still have recognizable algebraic structure.
The next lesson, Polynomial Sequences and Finite Differences, develops a systematic way to recognize polynomial patterns by repeatedly taking differences between consecutive terms.