The Central Idea
A recurrence can describe repeated change.
If the same amount is added at every step, then the sequence has a constant difference:
\[ \boxed{a_n=a_{n-1}+d}. \]If the same factor is multiplied at every step, then the sequence has a constant ratio:
\[ \boxed{a_n=ra_{n-1}}. \]These are the recursive forms of arithmetic and geometric sequences.
Arithmetic Recurrences
An arithmetic recurrence has the form
\[ \boxed{a_n=a_{n-1}+d}, \]where \(d\) is a fixed number called the common difference.
The recurrence says that each term is obtained by adding the same amount to the previous term.
Example: Positive Common Difference
Let
\[ a_1=4, \] \[ a_n=a_{n-1}+3\qquad(n\ge2). \]Then
\[ a_2=4+3=7, \] \[ a_3=7+3=10, \] \[ a_4=10+3=13. \]Therefore,
\[ \boxed{4,\ 7,\ 10,\ 13,\ldots}. \]The Common Difference
In an arithmetic sequence, consecutive terms differ by the same amount.
For
\[ 8,\ 13,\ 18,\ 23,\ldots \]the consecutive differences are
\[ 13-8=5,\qquad 18-13=5,\qquad 23-18=5. \]Thus,
\[ \boxed{d=5}. \]A recursive description is
\[ a_1=8,\qquad a_n=a_{n-1}+5. \]Arithmetic Recurrences Can Decrease
The common difference may be negative.
Suppose
\[ a_1=20, \] \[ a_n=a_{n-1}-4. \]Then
\[ 20,\ 16,\ 12,\ 8,\ldots \]has common difference
\[ \boxed{d=-4}. \]A Zero Common Difference
If \(d=0\), then
\[ a_n=a_{n-1}. \]Every term equals the preceding term.
For example,
\[ a_1=6,\qquad a_n=a_{n-1} \]gives
\[ \boxed{6,\ 6,\ 6,\ 6,\ldots}. \]From Arithmetic Recurrence to Explicit Form
Suppose
\[ a_1=A \]and
\[ a_n=a_{n-1}+d. \]To reach \(a_n\) from \(a_1\), the amount \(d\) is added \(n-1\) times. Therefore,
\[ \boxed{a_n=A+(n-1)d}. \]Why the Arithmetic Formula Works
Starting with \(a_1=A\), repeated substitution gives
\[ a_2=A+d, \] \[ a_3=A+2d, \] \[ a_4=A+3d. \]The pattern is
\[ \boxed{a_n=A+(n-1)d}. \]The linear dependence on \(n\) reflects repeated addition.
Example: Convert Arithmetic Recursion to Explicit Form
Let
\[ a_1=11, \] \[ a_n=a_{n-1}+6. \]Here, \(A=11\) and \(d=6\). Therefore,
\[ a_n=11+6(n-1). \]Simplifying,
\[ \boxed{a_n=6n+5}. \]From Explicit Arithmetic Form to Recurrence
Suppose
\[ a_n=4n+7. \]First find the initial term:
\[ a_1=4(1)+7=11. \]The coefficient of \(n\) is the constant difference:
\[ d=4. \]Therefore, an equivalent recursive definition is
\[ \boxed{a_1=11,\qquad a_n=a_{n-1}+4}. \]Finding the Common Difference
If two consecutive terms are known,
\[ \boxed{d=a_n-a_{n-1}}. \]For example, if \(a_7=32\) and \(a_8=37\), then
\[ d=37-32=5. \]Finding the Difference from Nonconsecutive Terms
Suppose an arithmetic sequence satisfies
\[ a_3=10,\qquad a_7=26. \]There are \(7-3=4\) equal steps, and the total change is
\[ 26-10=16. \]Therefore,
\[ d=\frac{16}{4}=4. \]Hence,
\[ \boxed{d=4}. \]Arithmetic Recurrences and Linear Change
An arithmetic sequence changes by the same amount at each step. Its explicit form
\[ a_n=a_1+(n-1)d \]is linear in \(n\). If the points \((n,a_n)\) are graphed, they lie on a straight line, although the sequence includes only the discrete indexed points.
Geometric Recurrences
A geometric recurrence has the form
\[ \boxed{a_n=ra_{n-1}}, \]where \(r\) is a fixed number called the common ratio.
Each term is obtained by multiplying the preceding term by the same factor.
Example: Common Ratio Greater Than One
Let
\[ a_1=3, \] \[ a_n=2a_{n-1}. \]Then
\[ a_2=6,\qquad a_3=12,\qquad a_4=24. \]Therefore,
\[ \boxed{3,\ 6,\ 12,\ 24,\ldots}, \]with
\[ \boxed{r=2}. \]The Common Ratio
In a geometric sequence, the ratio of consecutive nonzero terms is constant.
For
\[ 5,\ 15,\ 45,\ 135,\ldots \]we have
\[ \frac{15}{5}=3,\qquad \frac{45}{15}=3,\qquad \frac{135}{45}=3. \]Thus,
\[ \boxed{r=3}. \]Geometric Recurrences Can Shrink
If \(0<r<1\), repeated multiplication reduces positive terms.
For example,
\[ a_1=16,\qquad a_n=\frac12a_{n-1} \]gives
\[ \boxed{16,\ 8,\ 4,\ 2,\ 1,\ldots}. \]Negative Common Ratios
Suppose
\[ a_1=2, \] \[ a_n=-3a_{n-1}. \]Then
\[ 2,\ -6,\ 18,\ -54,\ldots \]alternates sign while its magnitude is multiplied by \(3\) each step. The common ratio is
\[ \boxed{r=-3}. \]Ratios of One and Zero
If \(r=1\), the sequence remains constant:
\[ a_1=9,\qquad a_n=a_{n-1} \]gives
\[ 9,\ 9,\ 9,\ldots \]If \(r=0\), then every term after the first recursive step is zero:
\[ a_1=7,\qquad a_n=0\cdot a_{n-1} \]gives
\[ 7,\ 0,\ 0,\ldots \]From Geometric Recurrence to Explicit Form
Suppose
\[ a_1=A \]and
\[ a_n=ra_{n-1}. \]To reach \(a_n\) from \(a_1\), the factor \(r\) is applied \(n-1\) times. Therefore,
\[ \boxed{a_n=Ar^{n-1}}. \]Why the Geometric Formula Works
Starting with \(a_1=A\), repeated substitution gives
\[ a_2=Ar, \] \[ a_3=Ar^2, \] \[ a_4=Ar^3. \]Therefore,
\[ \boxed{a_n=Ar^{n-1}}. \]The exponential dependence on \(n\) reflects repeated multiplication.
Example: Convert Geometric Recursion to Explicit Form
Let
\[ a_1=5, \] \[ a_n=4a_{n-1}. \]Here, \(A=5\) and \(r=4\). Therefore,
\[ \boxed{a_n=5\cdot4^{n-1}}. \]From Explicit Geometric Form to Recurrence
Suppose
\[ a_n=7\cdot3^{n-1}. \]The first term is \(a_1=7\), and each increase of the index introduces another factor of \(3\).
Therefore,
\[ \boxed{a_1=7,\qquad a_n=3a_{n-1}}. \]Finding the Common Ratio
If \(a_{n-1}\neq0\), then
\[ \boxed{r=\frac{a_n}{a_{n-1}}}. \]For example, if \(a_5=24\) and \(a_6=72\), then
\[ r=\frac{72}{24}=3. \]Finding the Ratio from Nonconsecutive Terms
Suppose
\[ a_2=6,\qquad a_5=162. \]From position \(2\) to position \(5\), the ratio is applied three times:
\[ 162=6r^3. \]Therefore,
\[ r^3=27, \]so
\[ \boxed{r=3}. \]Arithmetic versus Geometric Change
| Structure | Recursive Form | Repeated Change | Explicit Form |
|---|---|---|---|
| Arithmetic | \(a_n=a_{n-1}+d\) | Add a fixed amount | \(a_n=a_1+(n-1)d\) |
| Geometric | \(a_n=ra_{n-1}\) | Multiply by a fixed factor | \(a_n=a_1r^{n-1}\) |
Difference versus Ratio Tests
To test for arithmetic structure, compare consecutive differences:
\[ a_n-a_{n-1}. \]To test for geometric structure, compare consecutive ratios where defined:
\[ \frac{a_n}{a_{n-1}}. \]A constant difference indicates arithmetic structure. A constant ratio indicates geometric structure.
Example: Identify Arithmetic Structure
Consider
\[ 12,\ 17,\ 22,\ 27,\ldots \]The differences are \(5,5,5\), so the sequence is arithmetic.
\[ \boxed{a_1=12,\qquad a_n=a_{n-1}+5}. \]Example: Identify Geometric Structure
Consider
\[ 2,\ 10,\ 50,\ 250,\ldots \]The ratios are \(5,5,5\), so the sequence is geometric.
\[ \boxed{a_1=2,\qquad a_n=5a_{n-1}}. \]A Sequence Can Be Neither
Consider
\[ 1,\ 4,\ 9,\ 16,\ldots \]Its consecutive differences are \(3,5,7,\ldots\), which are not constant, and its ratios are not constant either. Therefore, it is neither arithmetic nor geometric.
Indexing from Zero
If an arithmetic sequence begins with \(a_0=A\), then
\[ \boxed{a_n=A+nd}. \]If a geometric sequence begins with \(a_0=A\), then
\[ \boxed{a_n=Ar^n}. \]Starting at index \(0\) changes the number of recursive transitions from \(n-1\) to \(n\).
Fixed Amount versus Fixed Percentage
Adding a fixed amount each period gives an arithmetic recurrence:
\[ B_n=B_{n-1}+100. \]Increasing by a fixed percentage gives a geometric recurrence:
\[ B_n=1.10B_{n-1}. \]A fixed amount and a fixed percentage are fundamentally different kinds of repeated change.
Example: Fixed-Dollar Growth
Suppose an annual salary begins at \(\$50{,}000\) and increases by \(\$2{,}000\) each year. Let \(S_n\) be the salary after \(n\) increases.
\[ S_0=50000, \] \[ \boxed{S_n=S_{n-1}+2000}. \]The explicit form is
\[ \boxed{S_n=50000+2000n}. \]Example: Percentage Growth
Suppose instead that the same salary increases by \(4\%\) each year.
\[ S_0=50000, \] \[ \boxed{S_n=1.04S_{n-1}}. \]The explicit form is
\[ \boxed{S_n=50000(1.04)^n}. \]The yearly dollar increase is not constant because the percentage is applied to a changing base.
Repeated Percentage Decay
If a quantity loses \(20\%\) of its current value each period, it retains \(80\%\) of the previous value.
\[ \boxed{a_n=0.8a_{n-1}}. \]Percentage decrease is multiplicative, so it produces geometric rather than arithmetic change.
Arithmetic First Differences
If
\[ a_n=a_{n-1}+d, \]then
\[ a_n-a_{n-1}=d. \]Thus, the first-difference sequence is constant:
\[ \boxed{d,\ d,\ d,\ldots}. \]Difference equations will generalize this viewpoint later in the chapter.
Geometric Multiplicative Change
If
\[ a_n=ra_{n-1} \]and \(a_{n-1}\neq0\), then
\[ \frac{a_n}{a_{n-1}}=r. \]Geometric sequences preserve a constant multiplicative factor from one nonzero term to the next.
Arithmetic Growth versus Geometric Growth
Compare
\[ a_1=2,\qquad a_n=a_{n-1}+2 \]with
\[ b_1=2,\qquad b_n=2b_{n-1}. \]The arithmetic sequence is
\[ 2,\ 4,\ 6,\ 8,\ 10,\ldots \]while the geometric sequence is
\[ 2,\ 4,\ 8,\ 16,\ 32,\ldots \]Repeated addition produces linear growth. Repeated multiplication by a factor greater than \(1\) produces exponential growth.
Recursive Rules Reveal Mechanism
The explicit formula
\[ a_n=500(1.03)^n \]makes direct evaluation easy.
The recursive form
\[ a_0=500,\qquad a_n=1.03a_{n-1} \]makes the mechanism visible: each new value is \(103\%\) of the preceding value. Both representations are useful.
Common Error: Confusing Difference and Ratio
Consider
\[ 3,\ 6,\ 12,\ 24,\ldots \]The differences are not constant, but the ratios are
\[ 2,\ 2,\ 2. \]Therefore, the sequence is geometric, not arithmetic.
Common Error: Treating a Percentage as an Additive Amount
If a quantity grows by \(5\%\), we do not add \(0.05\) to the quantity. We multiply the previous value by
\[ 1.05. \]Thus, repeated percentage growth is modeled by
\[ \boxed{a_n=1.05a_{n-1}}. \]Common Error: Forgetting the Initial Term
The recurrence \(a_n=a_{n-1}+4\) describes a family of arithmetic sequences. It does not identify a unique sequence until an initial value is supplied.
Common Error: Using the Wrong Step Count
If the first term is \(a_1=A\), then
\[ a_n=A+(n-1)d \]and
\[ a_n=Ar^{n-1}. \]The first term has undergone zero recursive steps, so \(n-1\), not \(n\), is the appropriate number of transitions.
Common Error: Dividing by Zero in the Ratio Test
The ratio
\[ \frac{a_n}{a_{n-1}} \]is defined only when \(a_{n-1}\neq0\).
A recurrence may still be geometric even when zero terms occur, but an ordinary ratio calculation cannot divide by zero.
A Reliable Procedure for Arithmetic Recurrences
- Identify the initial value and starting index.
- Identify the fixed additive change \(d\).
- Write the recurrence \(a_n=a_{n-1}+d\).
- Generate terms by repeated addition when needed.
- For indexing from \(1\), use \(a_n=a_1+(n-1)d\) for explicit form.
- Check consecutive differences to verify the pattern.
A Reliable Procedure for Geometric Recurrences
- Identify the initial value and starting index.
- Identify the fixed multiplicative factor \(r\).
- Write the recurrence \(a_n=ra_{n-1}\).
- Generate terms by repeated multiplication when needed.
- For indexing from \(1\), use \(a_n=a_1r^{n-1}\) for explicit form.
- Check consecutive nonzero ratios to verify the pattern.
A Reliable Procedure for Classifying a Sequence
- Compute several consecutive differences.
- If the differences are constant, classify the sequence as arithmetic.
- If not, compute consecutive ratios where defined.
- If the ratios are constant, classify the sequence as geometric.
- If neither test is constant, classify the sequence as neither.
Practice 1: Arithmetic Recurrence
Let
\[ a_1=7,\qquad a_n=a_{n-1}+4. \]List the first six terms.
Practice 2: Arithmetic Explicit Form
Convert
\[ a_1=12,\qquad a_n=a_{n-1}-3 \]to an explicit formula.
Practice 3: Identify the Common Difference
Find the common difference of
\[ 31,\ 25,\ 19,\ 13,\ldots \]and write a recursive definition.
Practice 4: Geometric Recurrence
Let
\[ b_1=4,\qquad b_n=3b_{n-1}. \]List the first five terms.
Practice 5: Geometric Explicit Form
Convert
\[ b_1=10,\qquad b_n=\frac12b_{n-1} \]to an explicit formula.
Practice 6: Identify the Common Ratio
Find the common ratio of
\[ 81,\ 27,\ 9,\ 3,\ldots \]and write a recursive definition.
Practice 7: Arithmetic or Geometric?
Classify
\[ 6,\ 11,\ 16,\ 21,\ldots \]as arithmetic, geometric, or neither.
Practice 8: Arithmetic or Geometric?
Classify
\[ 2,\ -6,\ 18,\ -54,\ldots \]as arithmetic, geometric, or neither.
Practice 9: Model Fixed Change
A tank contains \(900\) liters of water and loses \(35\) liters per hour. Let \(V_n\) be the amount after \(n\) hours. Write a recurrence and explicit formula.
Practice 10: Model Percentage Change
An investment begins at \(\$2000\) and increases by \(6\%\) per period. Let \(B_n\) be the value after \(n\) periods. Write a recurrence and explicit formula.
Practice Answers
Practice 1
\[ \boxed{7,\ 11,\ 15,\ 19,\ 23,\ 27}. \]Practice 2
\[ a_n=12-3(n-1), \]so
\[ \boxed{a_n=15-3n}. \]Practice 3
\[ 25-31=-6,\qquad 19-25=-6. \]Therefore,
\[ \boxed{d=-6}, \]and
\[ \boxed{a_1=31,\qquad a_n=a_{n-1}-6}. \]Practice 4
\[ \boxed{4,\ 12,\ 36,\ 108,\ 324}. \]Practice 5
\[ \boxed{b_n=10\left(\frac12\right)^{n-1}}. \]Practice 6
\[ \frac{27}{81}=\frac13,\qquad \frac{9}{27}=\frac13. \]Therefore,
\[ \boxed{r=\frac13}, \]and
\[ \boxed{a_1=81,\qquad a_n=\frac13a_{n-1}}. \]Practice 7
The consecutive differences are \(5,5,5\). Therefore,
\[ \boxed{\text{arithmetic with }d=5}. \]Practice 8
The consecutive ratios are \(-3,-3,-3\). Therefore,
\[ \boxed{\text{geometric with }r=-3}. \]Practice 9
\[ V_0=900, \] \[ \boxed{V_n=V_{n-1}-35}, \]and because indexing begins at \(0\),
\[ \boxed{V_n=900-35n}. \]Practice 10
A \(6\%\) increase gives a growth factor of \(1.06\). Therefore,
\[ B_0=2000, \] \[ \boxed{B_n=1.06B_{n-1}}, \]and
\[ \boxed{B_n=2000(1.06)^n}. \]What to Remember
Arithmetic recurrence:
\[ \boxed{a_n=a_{n-1}+d} \]with explicit form
\[ \boxed{a_n=a_1+(n-1)d}. \]Geometric recurrence:
\[ \boxed{a_n=ra_{n-1}} \]with explicit form
\[ \boxed{a_n=a_1r^{n-1}}. \]- Arithmetic sequences have constant consecutive differences.
- Geometric sequences have constant consecutive ratios where defined.
- Repeated fixed amounts are additive.
- Repeated fixed percentages are multiplicative.
- Indexing determines whether the step count is \(n\) or \(n-1\).
- Recursive form emphasizes the repeated mechanism.
- Explicit form gives direct access to any indexed term.
Where We Go Next
Arithmetic and geometric recurrences use especially simple rules: add a constant or multiply by a constant. More general recurrences can combine previous terms with coefficients and additional structure.
The next lesson, Linear Recurrence Relations, introduces recurrence equations in which each new term is built from linear combinations of earlier terms, creating a broader framework that includes many important discrete models.