The Central Idea
Consider the arithmetic series
\[ 4+7+10+13+16. \]Because the terms have a constant difference, we can use the arithmetic-sum formula
\[ \boxed{S_n=\frac{n}{2}(a_1+a_n)}. \]Here,
\[ n=5,\qquad a_1=4,\qquad a_5=16, \]so
\[ S_5=\frac52(4+16)=\boxed{50}. \]Now consider the geometric series
\[ 3+6+12+24+48. \]Its terms have common ratio \(2\), so we can use the finite geometric-sum formula
\[ \boxed{S_n=a_1\frac{1-r^n}{1-r}}. \]Therefore,
\[ S_5=3\frac{1-2^5}{1-2}=\boxed{93}. \]Arithmetic Series
An arithmetic series is the sum of terms from an arithmetic sequence.
If
\[ a_k=a_1+(k-1)d, \]then the first \(n\) terms form the finite series
\[ \boxed{S_n=\sum_{k=1}^{n}a_k}. \]Geometric Series
A geometric series is the sum of terms from a geometric sequence.
If
\[ a_k=a_1r^{k-1}, \]then
\[ \boxed{S_n=\sum_{k=1}^{n}a_1r^{k-1}}. \]The Arithmetic-Sum Formula
A finite arithmetic series can be summed by
\[ \boxed{S_n=\frac{n}{2}(a_1+a_n)}. \]This says
\[ \boxed{ \text{sum} = \text{number of terms} \times \text{average of first and last terms} }. \]Why the Arithmetic Formula Works
Write the series forward:
\[ S_n=a_1+a_2+a_3+\cdots+a_{n-1}+a_n. \]Write the same series in reverse:
\[ S_n=a_n+a_{n-1}+a_{n-2}+\cdots+a_2+a_1. \]Adding the two equations term by term produces \(n\) identical pairs, each equal to \(a_1+a_n\):
\[ 2S_n=n(a_1+a_n). \]Therefore,
\[ \boxed{S_n=\frac{n}{2}(a_1+a_n)}. \]The Average-Term View
Arithmetic terms are evenly spaced numerically, so their average is the midpoint of the first and last terms:
\[ \text{average term}=\frac{a_1+a_n}{2}. \]Multiplying by the number of terms gives
\[ \boxed{S_n=n\left(\frac{a_1+a_n}{2}\right)}. \]Arithmetic Sum Using the Common Difference
Since
\[ a_n=a_1+(n-1)d, \]substitute this expression into the arithmetic-sum formula:
\[ S_n=\frac{n}{2}\left[a_1+a_1+(n-1)d\right]. \]Therefore,
\[ \boxed{S_n=\frac{n}{2}\left[2a_1+(n-1)d\right]}. \]Two Equivalent Arithmetic-Sum Formulas
The standard forms are
\[ \boxed{S_n=\frac{n}{2}(a_1+a_n)} \]and
\[ \boxed{S_n=\frac{n}{2}\left[2a_1+(n-1)d\right]}. \]Choose the form that best matches the information given.
Example: Arithmetic Sum from First and Last Terms
Find
\[ 5+9+13+\cdots+41. \]Here,
\[ a_1=5,\qquad d=4,\qquad a_n=41. \]First find the number of terms:
\[ 41=5+4(n-1). \] \[ 36=4(n-1), \] \[ n=10. \]Then
\[ S_{10}=\frac{10}{2}(5+41)=\boxed{230}. \]Example: Arithmetic Sum from \(a_1\), \(d\), and \(n\)
Suppose
\[ a_1=12,\qquad d=3,\qquad n=20. \]Then
\[ S_{20} = \frac{20}{2}\left[2(12)+19(3)\right]. \] \[ =10(24+57) \] \[ =\boxed{810}. \]Sum of the First \(n\) Positive Integers
The sequence
\[ 1,2,3,\ldots,n \]is arithmetic. Therefore,
\[ \boxed{1+2+\cdots+n=\frac{n(n+1)}{2}}. \]Sum of the First \(n\) Odd Positive Integers
The first \(n\) odd positive integers are
\[ 1,3,5,\ldots,2n-1. \]Use the arithmetic-sum formula:
\[ S_n=\frac{n}{2}[1+(2n-1)]. \]Thus,
\[ \boxed{1+3+5+\cdots+(2n-1)=n^2}. \]Sum of the First \(n\) Even Positive Integers
The first \(n\) even positive integers are
\[ 2,4,6,\ldots,2n. \]Therefore,
\[ S_n=\frac{n}{2}(2+2n) \] \[ =\boxed{n(n+1)}. \]Arithmetic Sums in Sigma Notation
If
\[ a_k=a_1+(k-1)d, \]then
\[ \sum_{k=1}^{n}[a_1+(k-1)d] \]is an arithmetic series. It may be evaluated by expansion, sigma properties, or an arithmetic-sum formula.
Example: Arithmetic Sum from Sigma Notation
Evaluate
\[ \sum_{k=1}^{15}(4k+1). \]The first and last terms are
\[ a_1=5 \]and
\[ a_{15}=61. \]Therefore,
\[ S_{15}=\frac{15}{2}(5+61)=\boxed{495}. \]The Finite Geometric-Sum Formula
For
\[ a_1+a_1r+a_1r^2+\cdots+a_1r^{n-1}, \]with \(r\neq1\),
\[ \boxed{S_n=a_1\frac{1-r^n}{1-r}}. \]An equivalent form is
\[ \boxed{S_n=a_1\frac{r^n-1}{r-1}}. \]Deriving the Geometric-Sum Formula
Begin with
\[ S_n=a_1+a_1r+a_1r^2+\cdots+a_1r^{n-1}. \]Multiply by \(r\):
\[ rS_n=a_1r+a_1r^2+\cdots+a_1r^{n-1}+a_1r^n. \]Subtract:
\[ S_n-rS_n=a_1-a_1r^n. \]Factor:
\[ S_n(1-r)=a_1(1-r^n). \]For \(r\neq1\),
\[ \boxed{S_n=a_1\frac{1-r^n}{1-r}}. \]Why \(r=1\) Is a Special Case
If \(r=1\), every term equals \(a_1\):
\[ a_1+a_1+\cdots+a_1. \]With \(n\) terms,
\[ \boxed{S_n=na_1}. \]The standard geometric formula cannot be used directly because its denominator becomes zero.
Choosing a Geometric-Sum Form
The forms
\[ a_1\frac{1-r^n}{1-r} \]and
\[ a_1\frac{r^n-1}{r-1} \]are algebraically equivalent.
When \(r>1\), the second form often keeps the numerator and denominator
positive. When \(0
Example: Geometric Sum with Ratio Greater Than One
Evaluate
\[ 2+6+18+54+162. \]Here,
\[ a_1=2,\qquad r=3,\qquad n=5. \]Therefore,
\[ S_5=2\frac{3^5-1}{3-1} \] \[ =2\frac{242}{2} \] \[ =\boxed{242}. \]Example: Geometric Sum with a Fractional Ratio
Evaluate
\[ 64+32+16+8+4. \]Here,
\[ a_1=64,\qquad r=\frac12,\qquad n=5. \]Then
\[ S_5 = 64\frac{1-(1/2)^5}{1-1/2} \] \[ =\boxed{124}. \]Example: Geometric Sum with a Negative Ratio
Evaluate
\[ 3-6+12-24+48. \]Here,
\[ a_1=3,\qquad r=-2,\qquad n=5. \]Therefore,
\[ S_5=3\frac{1-(-2)^5}{1-(-2)} \] \[ =3\frac{33}{3} \] \[ =\boxed{33}. \]Geometric Sums in Sigma Notation
A geometric series may be written
\[ \boxed{\sum_{k=1}^{n}a_1r^{k-1}} \]or, after shifting the index,
\[ \boxed{\sum_{k=0}^{n-1}a_1r^k}. \]Both expressions contain the same terms.
Example: Geometric Sum from Sigma Notation
Evaluate
\[ \sum_{k=0}^{6}5(2)^k. \]The index values \(0\) through \(6\) give seven terms, with first term \(5\) and ratio \(2\).
\[ S=5\frac{2^7-1}{2-1} \] \[ =5(127) \] \[ =\boxed{635}. \]Counting Terms Carefully
If an index runs from \(m\) through \(n\) in unit steps, the number of terms is
\[ \boxed{n-m+1}. \]Thus
\[ \sum_{k=0}^{6}5(2)^k \]contains
\[ 6-0+1=7 \]terms, not six.
Finding the Number of Arithmetic Terms
If \(a_1\), \(d\), and \(a_n\) are known, use
\[ a_n=a_1+(n-1)d \]to solve for \(n\) before using the sum formula.
Finding the Number of Geometric Terms
If \(a_1\), \(r\), and \(a_n\) are known, use
\[ a_n=a_1r^{n-1} \]to solve for \(n\).
Familiar powers may make the result immediate. Otherwise, logarithms may be needed.
Example: Finding the Number of Geometric Terms
Consider
\[ 3+6+12+\cdots+384. \]Since
\[ 384=3(2)^{n-1}, \]we have
\[ 128=2^{n-1}=2^7. \]Therefore,
\[ n=8. \]The sum is then
\[ S_8=3\frac{2^8-1}{2-1}=\boxed{765}. \]Arithmetic Accumulation Produces Quadratic Structure
An arithmetic sequence has a linear explicit formula. Summing its first \(n\) terms introduces a factor of \(n\), so the partial-sum formula is generally quadratic in \(n\).
For example,
\[ 1+3+5+\cdots+(2n-1)=n^2. \]Accumulating constant first differences therefore produces quadratic partial-sum structure.
Geometric Accumulation Retains Exponential Structure
For a geometric sequence,
\[ a_n=a_1r^{n-1}. \]Its finite sum contains
\[ r^n, \]so the accumulated total retains exponential structure:
\[ \boxed{S_n=a_1\frac{1-r^n}{1-r}}. \]Comparing Arithmetic and Geometric Sums
| Arithmetic | Geometric |
|---|---|
| Constant difference | Constant ratio |
| \(a_n=a_1+(n-1)d\) | \(a_n=a_1r^{n-1}\) |
| \(S_n=\frac{n}{2}(a_1+a_n)\) | \(S_n=a_1\frac{1-r^n}{1-r}\) |
| Linear term pattern | Exponential term pattern |
| Quadratic partial-sum structure | Exponential partial-sum structure |
Application: Increasing Weekly Deposits
Suppose a person deposits \(20\) dollars during the first week and increases the weekly deposit by \(5\) dollars each week for \(12\) weeks.
Then
\[ a_1=20,\qquad d=5,\qquad n=12. \]The final weekly deposit is
\[ a_{12}=20+11(5)=75. \]The total deposited is
\[ S_{12}=\frac{12}{2}(20+75) \] \[ =\boxed{570}. \]Application: Repeated Doubling
Suppose a process contributes
\[ 1,2,4,8,\ldots \]units over ten stages.
Here,
\[ a_1=1,\qquad r=2,\qquad n=10. \]Therefore,
\[ S_{10}=\frac{2^{10}-1}{2-1}=\boxed{1023}. \]Application: Rows of Seats
Suppose an auditorium has \(18\) seats in the first row and each successive row has \(2\) more seats. There are \(25\) rows.
The last row has
\[ a_{25}=18+24(2)=66. \]Total seating is
\[ S_{25}=\frac{25}{2}(18+66) \] \[ =\boxed{1050}. \]Application: Repeated Percentage Change
Suppose an amount of \(100\) occurs in the first period and each later amount is \(90\%\) of the preceding one.
Then
\[ a_1=100,\qquad r=0.9. \]The total over eight periods is
\[ \boxed{ S_8=100\frac{1-(0.9)^8}{1-0.9} }. \]Application: Rebound Distances
Suppose a ball rebounds to \(60\%\) of its previous height. The rebound heights form a geometric sequence.
A total-distance problem may count some rebound heights twice: once upward and once downward. The geometric sequence provides the distances, but the physical interpretation determines how those distances enter the series.
Common Error: Using the Wrong Family
Before choosing a formula, inspect the terms.
The series
\[ 2+5+8+11+\cdots \]is arithmetic because its differences are constant.
The series
\[ 2+6+18+54+\cdots \]is geometric because its ratios are constant.
Common Error: Confusing \(a_n\) with \(S_n\)
The symbol \(a_n\) represents a term. The symbol \(S_n\) represents an accumulated sum.
For
\[ 2+4+6+8, \]we have
\[ a_4=8 \]but
\[ S_4=20. \]Common Error: Forgetting to Find \(n\)
In
\[ 5+9+13+\cdots+41, \]the value \(41\) is the last term, not the number of terms.
We must solve
\[ 41=5+4(n-1) \]to obtain
\[ n=10. \]Common Error: Using \(r^{n-1}\) in the Sum Formula
The \(n\)th term formula is
\[ a_n=a_1r^{n-1}. \]But the finite geometric-sum formula contains \(r^n\):
\[ \boxed{S_n=a_1\frac{1-r^n}{1-r}}. \]Common Error: Ignoring the \(r=1\) Case
If \(r=1\), the standard fraction would divide by zero.
Instead, every term equals \(a_1\), so
\[ \boxed{S_n=na_1}. \]Common Error: Losing a Negative Ratio
For
\[ 4-8+16-32, \]the common ratio is
\[ \boxed{r=-2}, \]not \(2\).
The negative sign determines the alternating pattern and must remain inside powers such as \((-2)^n\).
A Reliable Procedure for Arithmetic Sums
- Verify that the terms form an arithmetic sequence.
- Identify \(a_1\).
- Identify \(d\) if needed.
- Identify or solve for \(n\).
- Find \(a_n\) if needed.
- Choose the sum formula that matches the known information.
- Substitute carefully.
- Check whether the total is reasonable.
A Reliable Procedure for Geometric Sums
- Verify that the terms form a geometric sequence.
- Identify \(a_1\).
- Identify the common ratio \(r\).
- Count or solve for the number of terms \(n\).
- If \(r=1\), use \(S_n=na_1\).
- If \(r\neq1\), use a finite geometric-sum formula.
- Keep negative and fractional ratios inside parentheses when exponentiating.
- Check the result against direct addition when practical.
Example: Full Arithmetic-Sum Workflow
Find
\[ 14+19+24+\cdots+99. \]First,
\[ a_1=14,\qquad d=5,\qquad a_n=99. \]Find \(n\):
\[ 99=14+5(n-1). \] \[ 85=5(n-1), \] \[ n=18. \]Then
\[ S_{18}=\frac{18}{2}(14+99) \] \[ =9(113) \] \[ =\boxed{1017}. \]Example: Full Geometric-Sum Workflow
Find
\[ 5+15+45+\cdots+3645. \]Here,
\[ a_1=5,\qquad r=3. \]Find \(n\):
\[ 3645=5(3)^{n-1}. \] \[ 729=3^{n-1}=3^6, \]so
\[ n=7. \]Therefore,
\[ S_7=5\frac{3^7-1}{3-1} \] \[ =5\frac{2186}{2} \] \[ =\boxed{5465}. \]Practice 1: Arithmetic Sum
Find
\[ 3+7+11+15+19. \]Practice 2: Arithmetic Sum Formula
An arithmetic sequence has
\[ a_1=8,\qquad a_{20}=65. \]Find \(S_{20}\).
Practice 3: Arithmetic Sum from \(d\)
Suppose
\[ a_1=5,\qquad d=6,\qquad n=12. \]Find \(S_{12}\).
Practice 4: Find the Number of Terms
Find the number of terms in
\[ 7+11+15+\cdots+63. \]Practice 5: Geometric Sum
Find
\[ 2+4+8+16+32. \]Practice 6: Fractional Geometric Ratio
Find
\[ 81+27+9+3+1. \]Practice 7: Negative Geometric Ratio
Find
\[ 2-4+8-16+32. \]Practice 8: Geometric Sigma Notation
Evaluate
\[ \sum_{k=0}^{5}3(2)^k. \]Practice 9: Identify the Formula
A series has first term \(10\), common difference \(4\), and \(15\) terms. Find the sum.
Practice 10: Identify the Error
A student evaluates a five-term geometric series using
\[ S_5=a_1\frac{1-r^4}{1-r}. \]Explain the error.
Practice Answers
Practice 1
\[ S_5=\frac52(3+19)=\boxed{55}. \]Practice 2
\[ S_{20}=\frac{20}{2}(8+65)=\boxed{730}. \]Practice 3
\[ S_{12} = \frac{12}{2}[2(5)+11(6)] \] \[ =6(76) \] \[ =\boxed{456}. \]Practice 4
\[ 63=7+4(n-1). \] \[ 56=4(n-1), \] \[ \boxed{n=15}. \]Practice 5
\[ S_5=2\frac{2^5-1}{2-1}=\boxed{62}. \]Practice 6
\[ 81+27+9+3+1=\boxed{121}. \]Practice 7
\[ S_5=2\frac{1-(-2)^5}{1-(-2)} \] \[ =2\frac{33}{3} \] \[ =\boxed{22}. \]Practice 8
The index values \(0\) through \(5\) produce six terms.
\[ S=3\frac{2^6-1}{2-1} \] \[ =3(63) \] \[ =\boxed{189}. \]Practice 9
Use the arithmetic-sum formula:
\[ S_{15} = \frac{15}{2}[2(10)+14(4)] \] \[ =\frac{15}{2}(76) \] \[ =\boxed{570}. \]Practice 10
The finite geometric-sum formula uses \(r^n\), not \(r^{n-1}\). For five terms,
\[ \boxed{S_5=a_1\frac{1-r^5}{1-r}} \]when \(r\neq1\).
What to Remember
For a finite arithmetic series,
\[ \boxed{S_n=\frac{n}{2}(a_1+a_n)} \]or equivalently,
\[ \boxed{S_n=\frac{n}{2}[2a_1+(n-1)d]}. \]For a finite geometric series with \(r\neq1\),
\[ \boxed{S_n=a_1\frac{1-r^n}{1-r}} \]or equivalently,
\[ \boxed{S_n=a_1\frac{r^n-1}{r-1}}. \]If \(r=1\),
\[ \boxed{S_n=na_1}. \]- Arithmetic series come from constant differences.
- Geometric series come from constant ratios.
- The number of terms \(n\) must be known or determined.
- The arithmetic formula uses the average of the first and last terms.
- The geometric formula comes from scaling and subtracting aligned sums.
- Arithmetic accumulation typically produces quadratic partial-sum structure.
- Geometric accumulation retains exponential structure.
- Negative and fractional ratios must be handled carefully.
The central comparison is
\[ \boxed{ \text{constant additive change} \longrightarrow \text{arithmetic sum} } \]versus
\[ \boxed{ \text{constant multiplicative change} \longrightarrow \text{geometric sum}. } \]Where We Go Next
Arithmetic and geometric formulas let us calculate many terms or totals without manually repeating every step. But repeated step-by-step procedures remain essential when a pattern is defined algorithmically or recursively.
The next lesson, Iteration and Recurrence Algorithms, turns recursive rules into explicit computational procedures and develops the connection between mathematical recurrence, state updates, loops, and algorithmic thinking.