Sequences and Series in Precalculus: Arithmetic, Geometric, Recursive Rules and Sigma Notation
A practical guide to nth-term formulas, finite sums, infinite geometric series, worked examples and exam-style practice for US and UK mathematics students.
Two job offers reveal the difference between arithmetic and geometric growth
Imagine you receive two job offers. Both pay $30,000 (or £30,000) in the first year.
Offer A: Your salary increases by $1,500 every year.
Offer B: Your salary increases by 4% every year.
Which offer pays more?
Offer A follows a fixed addition rule. Offer B follows a fixed multiplication rule. The difference becomes clearer when you calculate the salaries.
| Year | Offer A: Add $1,500 | Offer B: Multiply by 1.04 |
|---|---|---|
| 1 | $30,000 | $30,000 |
| 5 | $36,000 | $35,096 |
| 10 | $43,500 | $42,699 |
| 13 | $48,000 | $48,031 |
| 20 | $58,500 | $63,205 |
Offer A pays a higher annual salary during the first 12 years. Offer B pulls ahead in year 13.
But what happens when you add every year's salary together?
After 10 years, Offer A pays $367,500 in total, compared with approximately $360,183 from Offer B. After 20 years, Offer A pays $885,000, while Offer B pays approximately $893,342.
The percentage-based offer eventually wins on both annual salary and cumulative earnings, although the crossover years differ.
This example introduces three fundamental ideas in Precalculus:
An arithmetic sequence increases or decreases by a constant amount.
A geometric sequence changes by a constant multiplier.
A series is the sum of the terms in a sequence.
Understanding these patterns helps you solve nth-term problems, calculate arithmetic and geometric series, interpret sigma notation and model situations involving repeated growth or decay.
By the end of this article, you will know how to identify a sequence, write explicit and recursive formulas, calculate finite and infinite sums, and avoid common errors in Precalculus and advanced high school mathematics.
1. What is a sequence in mathematics?
A sequence is an ordered list of numbers generated according to a rule.
For example:
3, 6, 9, 12, 15, …
Each number is called a term. The position of a term is indicated by a subscript.
a₁ = the first term
a₂ = the second term
a₃ = the third term
aₙ = the term in position n
The letter used for a sequence can vary. Some UK textbooks use uₙ or Tₙ, while many US textbooks use aₙ. The mathematics remains the same.
How is a sequence related to a function?
A sequence can be viewed as a function whose inputs are positive integers, usually 1, 2, 3, and so on.
For example, if
aₙ = 2n + 1,
then:
a₁ = 3
a₂ = 5
a₃ = 7
a₄ = 9
The input n determines the position, and the output aₙ gives the corresponding term.
When you graph a sequence, you plot individual points at integer positions. You do not automatically join the points into a continuous curve because the sequence is defined only at its permitted integer inputs.
Indexing matters: If a problem begins at n = 0 rather than n = 1, the first term is a₀. Always check the starting index before using a formula.
2. How to identify a sequence: the Step, Start and Shortcut method
Before memorising a formula, ask three questions.
STEP: What changes between consecutive terms?
START: What is the first term, and where does the indexing begin?
SHORTCUT: Does the problem ask for a term, a position or a sum?
Consider:
5, 10, 20, 40, …
Subtract consecutive terms:
10 − 5 = 5
20 − 10 = 10
40 − 20 = 20
The differences are not constant.
Now divide consecutive terms:
10 ÷ 5 = 2
20 ÷ 10 = 2
40 ÷ 20 = 2
The ratio is constant, so the sequence is geometric.
Its first term is 5 and its common ratio is 2. Therefore,
aₙ = 5 × 2ⁿ⁻¹
The three-question method prevents you from choosing an arithmetic formula simply because the numbers appear to follow a pattern.
A useful qualification: a few terms can suggest a rule without uniquely determining it. In formal mathematics, the stated rule or additional information establishes which sequence is intended.
3. Arithmetic sequences: finding the nth term
An arithmetic sequence has a constant difference between consecutive terms. This difference is called the common difference, represented by d.
The defining relationship is:
d = aₙ₊₁ − aₙ
Examples:
3, 7, 11, 15, … has d = 4.
20, 17, 14, 11, … has d = −3.
5, 5, 5, 5, … has d = 0.
An arithmetic sequence can increase, decrease or remain constant.
Arithmetic sequence formula
The nth-term formula is:
aₙ = a₁ + (n − 1)d
Here:
a₁ is the first term.
n is the required position.
d is the common difference.
Why is the multiplier n − 1 rather than n?
To reach the first term, you take zero steps. To reach the fifth term, you take four steps. Every move to the next term adds one more common difference.
Worked example 1: Find a distant term
Find the 40th term of:
7, 11, 15, 19, …
The first term is 7 and the common difference is 4.
a₄₀ = 7 + (40 − 1) × 4
a₄₀ = 7 + 156
Answer: a₄₀ = 163
Worked example 2: Find the position of a number
Which term of the sequence 7, 11, 15, 19, … is equal to 99?
Use the nth-term formula:
7 + 4(n − 1) = 99
4(n − 1) = 92
n − 1 = 23
n = 24
Answer: 99 is the 24th term.
If the calculated position is not a positive integer within the sequence's indexing, the target number is not a term of that sequence.
Worked example 3: Find the first term and common difference
An arithmetic sequence has a₃ = 17 and a₉ = 41. Find a₁ and d.
There are six steps between positions 3 and 9.
d = (41 − 17) ÷ (9 − 3)
d = 24 ÷ 6 = 4
Since a₃ is two steps beyond a₁:
a₁ = 17 − 2 × 4
Answer: a₁ = 9 and d = 4.
Check: 9, 13, 17, 21, 25, 29, 33, 37, 41.
Worked example 4: Insert arithmetic means
Insert three numbers between 2 and 18 so that all five numbers form an arithmetic sequence.
Five terms create four equal steps.
d = (18 − 2) ÷ 4 = 4
Starting at 2 and adding 4 each time gives:
2, 6, 10, 14, 18
Answer: 6, 10 and 14.
Recursive formula for an arithmetic sequence
An explicit formula calculates any term directly. A recursive formula uses an earlier term to calculate the next one.
For the sequence 7, 11, 15, 19, …:
a₁ = 7
aₙ = aₙ₋₁ + 4, for n ≥ 2
The starting value is essential. Without it, the rule does not identify one unique sequence.
4. Geometric sequences: constant ratios and exponential growth
A geometric sequence has a constant ratio between consecutive nonzero terms. The common ratio is represented by r.
When division is defined,
r = aₙ₊₁ ÷ aₙ
Examples:
3, 6, 12, 24, … has r = 2.
81, 27, 9, 3, … has r = 1/3.
4, −8, 16, −32, … has r = −2.
Unlike an arithmetic sequence, which repeatedly adds the same amount, a geometric sequence repeatedly multiplies by the same number.
Geometric sequence formula
The explicit formula is:
aₙ = a₁ × rⁿ⁻¹
The exponent is n − 1 because the first term requires zero multiplications by the common ratio.
Worked example 5: Find the tenth term
Find a₁₀ for the geometric sequence:
3, 6, 12, 24, …
Here a₁ = 3 and r = 2.
a₁₀ = 3 × 2⁹
a₁₀ = 3 × 512
Answer: a₁₀ = 1,536.
Worked example 6: Find a missing starting value
A geometric sequence has a₂ = 12 and a₅ = 96. Find its first term and common ratio.
From position 2 to position 5 there are three multiplication steps.
a₅ ÷ a₂ = r³
96 ÷ 12 = r³
r³ = 8
Therefore, r = 2.
Since a₂ = a₁ × r:
12 = 2a₁
Answer: a₁ = 6 and r = 2.
Check: 6, 12, 24, 48, 96.
If an equation gives an even power of r, both positive and negative ratios may be possible. For example, r² = 16 gives r = 4 or r = −4. Use the other information in the problem to decide which values are valid.
Worked example 7: Geometric sequences and bouncing balls
A ball is dropped from a height of 10 feet. Each rebound reaches 60% of the previous height. Find the height of the fifth rebound.
The first rebound is:
10 × 0.6 = 6 feet
The rebound heights form the sequence:
6, 3.6, 2.16, 1.296, …
This is geometric, with a₁ = 6 and r = 0.6.
The fifth rebound is:
a₅ = 6 × 0.6⁴
a₅ = 6 × 0.1296
Answer: 0.7776 feet.
The original drop is not counted as a rebound. The first rebound is the first term in this sequence.
What does the common ratio tell you?
For a geometric sequence with a nonzero first term:
r > 1: the absolute values of the terms grow.
0 < r < 1: the absolute values shrink towards zero.
r < −1: the absolute values grow while the signs alternate.
−1 < r < 0: the terms alternate in sign and approach zero.
r = 1: the sequence is constant.
r = −1: the terms alternate between two values.
r = 0: all terms after the first are zero.
These rules describe the usual real-valued geometric sequence. If a term is zero, the ratio test by division may not be available, so use the defining multiplication rule instead.
5. Recursive sequences: when the next term depends on earlier terms
A recursive sequence specifies how to generate later terms from one or more earlier terms.
For an arithmetic sequence:
a₁ = 7
aₙ = aₙ₋₁ + 4
For a geometric sequence:
a₁ = 2
aₙ = 3aₙ₋₁
The difference between recursive and explicit formulas is important in Precalculus.
An explicit formula gives the term directly from its position. A recursive formula describes how the sequence develops step by step.
A recursive sequence that is neither arithmetic nor geometric
Consider:
a₁ = 3
aₙ = 2aₙ₋₁ − 1, for n ≥ 2
Calculate the first five terms:
a₁ = 3
a₂ = 2 × 3 − 1 = 5
a₃ = 2 × 5 − 1 = 9
a₄ = 2 × 9 − 1 = 17
a₅ = 2 × 17 − 1 = 33
The sequence is:
3, 5, 9, 17, 33, …
Its consecutive differences are 2, 4, 8 and 16. Those differences double each time.
This suggests the explicit formula:
aₙ = 2ⁿ + 1
Testing the first four terms confirms that the formula matches the recursive rule. More importantly, it can be verified algebraically: if aₙ₋₁ = 2ⁿ⁻¹ + 1, then
aₙ = 2(2ⁿ⁻¹ + 1) − 1 = 2ⁿ + 1.
That proves the formula works for every positive integer n.
Fibonacci sequences
Some recursive sequences depend on two previous terms rather than one.
The Fibonacci sequence is defined by:
F₁ = 1, F₂ = 1
Fₙ = Fₙ₋₁ + Fₙ₋₂, for n ≥ 3
Its terms are:
1, 1, 2, 3, 5, 8, 13, 21, …
Two starting values are required because the rule refers to two earlier positions.
How do you convert a recursive rule into an explicit formula?
Use this process:
Calculate several initial terms.
Test whether consecutive differences are constant.
If they are not, test consecutive ratios where division is defined.
Look for another pattern if neither test works.
Propose an explicit formula and verify it against the recurrence.
This is a method for discovering a candidate formula, not a guarantee that every recurrence has a simple closed-form expression.
Convert an explicit formula into a recursive rule
Given:
aₙ = 5 + 3(n − 1)
The first term is 5 and the common difference is 3.
Therefore:
a₁ = 5
aₙ = aₙ₋₁ + 3, for n ≥ 2
For the geometric sequence
aₙ = 2 × 3ⁿ⁻¹,
the recursive rule is:
a₁ = 2
aₙ = 3aₙ₋₁, for n ≥ 2
6. What is a series? Arithmetic series formulas
A sequence is a list of terms. A series is the sum of those terms.
Sequence:
2, 4, 6, 8
Series:
2 + 4 + 6 + 8 = 20
The sum of the first n terms is written as Sₙ.
Sₙ = a₁ + a₂ + a₃ + … + aₙ
The arithmetic series formula
To understand the formula, consider adding the integers from 1 to 100.
Pair the first and last numbers:
1 + 100 = 101
2 + 99 = 101
3 + 98 = 101
Every pair has the same sum, and there are 50 pairs. The total is 50 × 101 = 5,050.
This pairing idea leads to the arithmetic series formula:
Sₙ = n(a₁ + aₙ) ÷ 2
In words, multiply the number of terms by the average of the first and last terms.
Because
aₙ = a₁ + (n − 1)d,
you can also write:
Sₙ = n[2a₁ + (n − 1)d] ÷ 2
Both formulas give the same answer.
Worked example 8: Sum the first 40 terms
Find the sum of the first 40 terms of:
7, 11, 15, 19, …
From the earlier calculation, a₄₀ = 163.
S₄₀ = 40(7 + 163) ÷ 2
S₄₀ = 20 × 170
Answer: S₄₀ = 3,400.
Worked example 9: Total earnings from the fixed-raise offer
Offer A begins at $30,000 and increases by $1,500 each year. Find total earnings over 20 years.
The 20th year's salary is:
a₂₀ = 30,000 + 19 × 1,500 = 58,500
The cumulative total is:
S₂₀ = 20(30,000 + 58,500) ÷ 2
Answer: $885,000.
This is the total of 20 annual salaries, not the salary earned in the twentieth year alone.
7. Geometric series: the formula for adding repeated multiplication
A finite geometric series has the form:
Sₙ = a₁ + a₁r + a₁r² + … + a₁rⁿ⁻¹
To derive its sum formula, let S represent the sum:
S = a₁ + a₁r + a₁r² + … + a₁rⁿ⁻¹
Multiply by r:
rS = a₁r + a₁r² + … + a₁rⁿ
Subtract the second equation from the first. Most terms cancel, leaving:
S − rS = a₁ − a₁rⁿ
Factor both sides:
S(1 − r) = a₁(1 − rⁿ)
For r ≠ 1, divide by 1 − r.
Finite geometric series formula:
Sₙ = a₁(1 − rⁿ) ÷ (1 − r)
An equivalent form is:
Sₙ = a₁(rⁿ − 1) ÷ (r − 1)
When r = 1, every term equals a₁, so the sum is simply Sₙ = na₁.
Worked example 10: Add ten geometric terms
Find the sum:
3 + 6 + 12 + 24 + …
for ten terms.
Here a₁ = 3, r = 2 and n = 10.
S₁₀ = 3(1 − 2¹⁰) ÷ (1 − 2)
S₁₀ = 3(1 − 1,024) ÷ (−1)
Answer: S₁₀ = 3,069.
Worked example 11: Total earnings from the percentage-raise offer
Offer B starts at $30,000 and increases by 4% each year. Find total earnings over 20 years.
The salary sequence is geometric, with a₁ = 30,000 and r = 1.04.
S₂₀ = 30,000(1.04²⁰ − 1) ÷ 0.04
Using 1.04²⁰ ≈ 2.1911:
S₂₀ ≈ $893,342
Offer B therefore pays approximately $8,342 more over 20 years in this simplified model.
The example assumes that each year's salary is paid once and that the percentage increase compounds annually. It excludes taxes, bonuses, inflation and other employment conditions.
8. Infinite geometric series and the sum to infinity
Can you add infinitely many numbers and obtain a finite answer?
Yes, provided the geometric terms become small enough.
For a geometric series with first term a₁ and common ratio r, the sum to infinity is:
S∞ = a₁ ÷ (1 − r), provided |r| < 1
The condition means that r lies strictly between −1 and 1.
Why does it work?
The finite sum is:
Sₙ = a₁(1 − rⁿ) ÷ (1 − r)
When |r| < 1, the value of rⁿ approaches zero as n increases. The finite sum therefore approaches a fixed limit.
If |r| ≥ 1 and a₁ is nonzero, the geometric series does not converge to a finite sum.
Worked example 12: Find the sum to infinity
Find the sum:
12 + 6 + 3 + 1.5 + …
The first term is 12 and the ratio is 1/2.
Since |1/2| < 1, the sum to infinity exists.
S∞ = 12 ÷ (1 − 1/2)
S∞ = 12 ÷ 0.5
Answer: S∞ = 24.
Worked example 13: Convert a repeating decimal into a fraction
Write 0.333… as a fraction.
Expand the decimal:
0.333… = 0.3 + 0.03 + 0.003 + …
This is a geometric series with first term 0.3 and common ratio 0.1.
S∞ = 0.3 ÷ (1 − 0.1)
S∞ = 0.3 ÷ 0.9
Answer: 0.333… = 1/3.
Similarly:
0.999… = 0.9 + 0.09 + 0.009 + …
S∞ = 0.9 ÷ (1 − 0.1) = 1
Therefore, 0.999… equals 1 exactly, not merely a number close to 1.
Worked example 14: Calculate the total distance travelled by a bouncing ball
A ball falls from a height of 10 feet. Each rebound reaches 60% of the previous height. Assume the ball continues rebounding indefinitely according to the same rule.
The first fall covers 10 feet.
The rebound heights are:
6, 3.6, 2.16, …
The sum of all rebound heights is:
S∞ = 6 ÷ (1 − 0.6) = 15 feet
Each rebound height is travelled twice: once on the way up and once on the way down.
Total distance = 10 + 2 × 15
Answer: 40 feet.
The key distinction is between the sum of the rebound heights and the total distance travelled. The latter includes both upward and downward travel after the first fall.
9. Sigma notation: how to read and evaluate a sum
Sigma notation is a compact way to represent repeated addition. The symbol Σ is the Greek capital letter sigma.
For example:
Σₖ₌₁⁵ (2k + 1)
This means to substitute k = 1, 2, 3, 4 and 5 into the expression 2k + 1, then add the results.
The expansion is:
3 + 5 + 7 + 9 + 11 = 35
The parts of a sigma expression have specific meanings:
k is the summation index, or counter.
1 is the lower limit, where counting begins.
5 is the upper limit, where counting ends.
2k + 1 is the expression that generates each term.
How many terms are in a sigma sum?
For integer limits L and U, inclusive of both endpoints:
Number of terms = U − L + 1
For example, the sum from k = 3 to k = 10 contains:
10 − 3 + 1 = 8 terms.
It does not contain seven terms because both 3 and 10 are included.
Sigma notation with a geometric expression
Consider:
Σₖ₌₀⁴ 3 × 2ᵏ
The index starts at zero and ends at four. That produces five terms:
3, 6, 12, 24, 48
Their sum is 93.
The geometric series formula gives the same answer:
S₅ = 3(1 − 2⁵) ÷ (1 − 2)
S₅ = 3 × 31 = 93
Notice that the number of terms is five even though the upper limit is four. The lower limit is zero.
Three useful properties of sigma notation
For matching summation limits:
A constant can be taken outside the sum.
Σ ck = cΣk
A sum can be separated into individual sums.
Σ(aₖ + bₖ) = Σaₖ + Σbₖ
A constant added n times sums to n times that constant.
Σₖ₌₁ⁿ c = nc
These are direct consequences of ordinary algebra and repeated addition.
Three summation formulas worth learning
For k = 1 to n:
Sum of the first n positive integers
Σk = n(n + 1) ÷ 2
Sum of the squares of the first n positive integers
Σk² = n(n + 1)(2n + 1) ÷ 6
Sum of the cubes of the first n positive integers
Σk³ = [n(n + 1) ÷ 2]²
These formulas are especially useful in advanced Precalculus, introductory calculus and A-level mathematics.
Worked example 15: Evaluate a polynomial sum
Find:
Σₖ₌₁¹⁰ (3k² − 2)
Separate the expression:
3Σk² − Σ2
For n = 10:
Σk² = 10 × 11 × 21 ÷ 6 = 385
The constant 2 is added ten times, so:
Σ2 = 20
Therefore:
3 × 385 − 20 = 1,155 − 20
Answer: 1,135.
How do you write a series in sigma notation?
Write the sum
4 + 7 + 10 + 13 + … + 31
in sigma notation.
The sequence is arithmetic, with first term 4 and common difference 3.
The nth term is:
aₙ = 4 + 3(n − 1)
aₙ = 3n + 1
To find the last position, solve:
3n + 1 = 31
n = 10
Therefore, the sigma notation is:
Σₖ₌₁¹⁰ (3k + 1)
As a check, the sum contains ten terms and equals:
10(4 + 31) ÷ 2 = 175
Can the summation index be changed?
Yes, provided the expression and limits are adjusted consistently.
For example:
Σₖ₌₁⁵ (2k + 1)
is equivalent to:
Σⱼ₌₀⁴ (2j + 3)
Both expand to:
3 + 5 + 7 + 9 + 11
Changing the index does not change the sum when the re-indexing is performed correctly.
10. Arithmetic versus geometric sequences: a quick comparison
| Feature | Arithmetic | Geometric | ||
|---|---|---|---|---|
| Pattern | Add the same difference | Multiply by the same ratio | ||
| Key quantity | Common difference d | Common ratio r | ||
| nth term | a₁ + (n − 1)d | a₁ × rⁿ⁻¹ | ||
| Recursive rule | aₙ = aₙ₋₁ + d | aₙ = raₙ₋₁ | ||
| Finite sum | n(a₁ + aₙ) ÷ 2 | a₁(1 − rⁿ) ÷ (1 − r), when r ≠ 1 | ||
| Typical model | Fixed annual increase | Fixed percentage change | ||
| Infinite sum | Not generally a finite total for an infinite arithmetic sequence | Exists when | r | < 1, apart from the trivial all-zero case |
A constant sequence can be both arithmetic and geometric when its common value is nonzero. The all-zero sequence is arithmetic, but its common ratio cannot be determined by dividing consecutive terms.
11. Real-world applications of sequences and series
Monthly savings with compound interest
Suppose you deposit $200 at the end of each month into an account that earns a constant 0.5% interest per month. You make 12 deposits and assume the rate remains unchanged.
The first deposit earns interest for 11 months. The second earns interest for ten months, and the final deposit earns no interest before the calculation date.
The accumulated value is:
200 × 1.005¹¹ + 200 × 1.005¹⁰ + … + 200
This is a finite geometric series.
S = 200(1.005¹² − 1) ÷ 0.005
S ≈ $2,467.11
The deposits total $2,400, so approximately $67.11 comes from interest.
This is a simplified mathematical illustration. Actual account terms, compounding conventions and fees can change the result.
Seating arrangements and arithmetic growth
A theatre has 20 seats in the first row. Each subsequent row has two more seats than the previous row. How many seats are there in 25 rows?
The final row contains:
a₂₅ = 20 + 24 × 2 = 68 seats
The total number of seats is:
S₂₅ = 25(20 + 68) ÷ 2
Answer: 1,100 seats.
Doubling patterns and geometric growth
Suppose one person sends a message to two new people in each round, and every person reached in one round passes it to two new people in the next round. Assume no one receives the message twice.
The number of newly reached people in each round follows:
2, 4, 8, 16, …
After ten rounds, the number reached in the tenth round is:
2¹⁰ = 1,024
If the original sender is included, the total number of people reached through the ten rounds is:
1 + 2 + 4 + … + 2¹⁰
Using the geometric series formula:
Total = 2¹¹ − 1 = 2,047
This is an idealised model. Real messages rarely spread with perfect doubling because of overlapping contacts and people who do not forward them.
12. Eight common mistakes in Precalculus sequence and series problems
Mistake 1: Using nd instead of (n − 1)d
The first term requires zero steps from the starting position. The correct arithmetic nth-term formula is a₁ + (n − 1)d.
Mistake 2: Counting sigma terms incorrectly
For inclusive integer limits, subtract the lower limit from the upper limit and add one.
Mistake 3: Using the infinite geometric series formula without checking the ratio
First check that |r| < 1. If the condition fails, the ordinary sum-to-infinity formula does not apply.
Mistake 4: Confusing the first term with the zeroth term
If indexing begins at zero, use the formula with the correct starting index. For example, aₙ = a₀rⁿ for a geometric sequence beginning at a₀.
Mistake 5: Misreading the exponent
In aₙ = a₁ × rⁿ⁻¹, the exponent n − 1 applies to r, not to the entire product.
Mistake 6: Forgetting the upward and downward journey
In a bouncing-ball problem, the total distance after the first fall includes two journeys for each rebound height.
Mistake 7: Losing negative signs
When r is negative, powers alternate in sign. Use brackets when evaluating powers such as (−2)³ and (−2)⁴.
Mistake 8: Giving an incomplete recursive rule
A recurrence normally needs the initial value or values required to generate the sequence.
13. How sequences and series appear in US and UK mathematics courses
The exact syllabus depends on the course, exam board and specification.
US students: Arithmetic and geometric patterns connect naturally to linear and exponential functions in Algebra 2 and Precalculus. They also provide useful practice for interpreting growth models and solving quantitative word problems. Students preparing for AP Precalculus should follow the current course framework for the precise required content. Do not assume that every sigma identity or infinite-series technique is tested in every course or exam.
UK students: Arithmetic sequences and nth-term rules are common at GCSE level. Geometric sequences, arithmetic and geometric series, recurrence relations, sigma notation and sums to infinity may appear in AS or A-level Mathematics, depending on the specification and course.
The notation can differ between textbooks. A formula written with aₙ in one book might use uₙ or Tₙ in another.
What do common exam instructions mean?
Find: Calculate the requested value.
Show that: Present sufficient mathematical working to establish the result.
Hence: Use a previous result to solve the next part.
Find the sum to infinity: Check convergence before applying the formula.
Write in sigma notation: Identify the summation expression and its correct limits.
Practise writing each step clearly. Correct reasoning is easier to check than a final answer without supporting work.
14. Practice questions on sequences and series
Try these problems before checking the worked answers.
Find the 25th term of the arithmetic sequence 12, 9, 6, 3, …
An arithmetic sequence has a₄ = 22 and a₁₀ = 52. Find a₁ and d.
Find the sum of the first 30 terms of 5, 9, 13, …
Find a₇ for the geometric sequence 2, 6, 18, …
A geometric sequence has a₃ = 20 and a₆ = 160. Find a₁ and r, assuming real terms.
Evaluate Σₖ₌₁⁶ (3k − 1).
Evaluate Σₖ₌₀⁵ 4(1/2)ᵏ.
Find the sum to infinity of 18 + 6 + 2 + …
Write 5 + 8 + 11 + … + 50 in sigma notation.
A sequence is defined by a₁ = 4 and aₙ = 3aₙ₋₁ − 2. Find the first five terms and an explicit formula.
Express 0.272727… as a fraction using a geometric series.
A ball is dropped from 20 feet and rebounds to 75% of its previous height. Assuming this continues indefinitely, find the total distance travelled.
15. Worked answers to the practice questions
1. The 25th arithmetic term
The common difference is −3.
a₂₅ = 12 + 24(−3)
Answer: −60.
2. Find the first term and common difference
There are six steps between positions 4 and 10.
d = (52 − 22) ÷ 6 = 5
a₁ = 22 − 3 × 5
Answer: a₁ = 7 and d = 5.
3. Sum the first 30 terms
a₃₀ = 5 + 29 × 4 = 121
S₃₀ = 30(5 + 121) ÷ 2
Answer: 1,890.
4. Find the seventh geometric term
a₇ = 2 × 3⁶
Answer: 1,458.
5. Find the first term and ratio
a₆ ÷ a₃ = r³ = 160 ÷ 20 = 8
Therefore r = 2.
a₁ = 20 ÷ 2² = 5
Answer: a₁ = 5 and r = 2.
6. Evaluate the sigma sum
The terms are 2, 5, 8, 11, 14 and 17.
S = 6(2 + 17) ÷ 2
Answer: 57.
7. Evaluate the geometric sum
The terms are 4, 2, 1, 0.5, 0.25 and 0.125.
Adding them gives:
Answer: 7.875.
8. Find the sum to infinity
The first term is 18 and the common ratio is 1/3.
S∞ = 18 ÷ (1 − 1/3)
Answer: 27.
9. Write the sum in sigma notation
The nth term is:
aₖ = 5 + 3(k − 1) = 3k + 2
The last term is 50, so:
3k + 2 = 50
k = 16
Answer: Σₖ₌₁¹⁶ (3k + 2).
10. Find an explicit formula
The first five terms are:
4, 10, 28, 82, 244
The differences are 6, 18, 54 and 162.
The pattern suggests aₙ = 3ⁿ + 1.
This can be verified using the recurrence:
3(3ⁿ⁻¹ + 1) − 2 = 3ⁿ + 1
Answer: aₙ = 3ⁿ + 1.
11. Convert 0.272727… into a fraction
Expand the repeating decimal:
0.27 + 0.0027 + 0.000027 + …
The first term is 0.27 and the common ratio is 0.01.
S∞ = 0.27 ÷ (1 − 0.01)
S∞ = 0.27 ÷ 0.99
Answer: 3/11.
12. Find the total bouncing-ball distance
The initial fall is 20 feet. The rebound heights are 15, 11.25, … with ratio 0.75.
S∞ = 15 ÷ (1 − 0.75) = 60 feet
The ball travels up and down for each rebound height.
Total distance = 20 + 2 × 60
Answer: 140 feet.
16. Frequently asked questions about sequences and series in Precalculus
What is the difference between a sequence and a series?
A sequence is an ordered list of terms. A series is the sum of those terms. For example, 2, 4, 6 is a sequence, while 2 + 4 + 6 = 12 is its three-term series.
How do I know whether a sequence is arithmetic or geometric?
Subtract consecutive terms. If the difference is constant, the sequence is arithmetic. If it is not, check the ratios of consecutive nonzero terms. A constant ratio indicates a geometric sequence.
What is the nth-term formula for an arithmetic sequence?
The formula is aₙ = a₁ + (n − 1)d, where a₁ is the first term and d is the common difference.
