A **sequence** is an ordered collection of numbers following a specific pattern or rule, where each number has a definite position (first, second, third, etc.).
**Key Terminology:**
**Definition:** A sequence is essentially a function whose domain is the set of natural numbers or a subset thereof. We can write aₙ or a(n) to represent the nth term.
**Common Representations:**
1. **Explicit Formula (Algebraic):** The general term is given directly as a formula in terms of n
2. **Recurrence Relation:** Each term is defined in relation to one or more previous terms
3. **Verbal Description:** When no formula exists (e.g., sequence of prime numbers: 2, 3, 5, 7, 11, ...)
**Worked Example 1:**
Write the first three terms of the sequence defined by aₙ = 2n + 5.
**Worked Example 2:**
Find the 20th term of aₙ = (n - 1)(2 - n)(3 + n).
**Exam Important Points:**
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A **series** is the sum of the terms of a sequence, expressed as:
a₁ + a₂ + a₃ + ... + aₙ + ...
**Key Distinction:**
**Sigma Notation (Summation Notation):**
The series a₁ + a₂ + a₃ + ... + aₙ is written compactly as:
$$\sum_{k=1}^{n} a_k$$
where:
**Example:** The series 1 + 3 + 5 + 7 + 9 is written as $$\sum_{k=1}^{5} (2k-1)$$
**Important Remark:** When we write a series, we indicate the sum without calculating it. When we ask for the "sum of the series," we mean the actual numerical result.
**Worked Example 3:**
A sequence is defined by a₁ = 1, aₙ = aₙ₋₁ + 2 for n ≥ 2. Find the first five terms and write the corresponding series.
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A **geometric progression** is a sequence where each term (except the first) has a constant ratio to the term immediately preceding it.
**Definition:** A sequence a₁, a₂, a₃, ..., aₙ, ... is a G.P. if each term is non-zero and:
$$\frac{a_{k+1}}{a_k} = r \text{ (constant, for all } k \geq 1\text{)}$$
where r is called the **common ratio**.
**Standard Form:** If the first term is 'a' and common ratio is 'r', a G.P. can be written as:
a, ar, ar², ar³, ..., arⁿ⁻¹, ...
**Examples of G.P.:**
1. 2, 4, 8, 16, ... (a = 2, r = 2)
2. 1, -1/3, 1/9, -1/27, ... (a = 1, r = -1/3)
3. 0.01, 0.0001, 0.000001, ... (a = 0.01, r = 0.01)
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The **nth term** of a geometric progression is given by:
$$a_n = ar^{n-1}$$
where:
**Derivation:**
**Worked Example 4:**
Find the 10th and nth terms of the G.P. 5, 25, 125, ...
**Worked Example 5:**
Which term of the G.P. 2, 8, 32, ... is 131072?
**Exam Important Points:**
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The **sum of first n terms** of a G.P. is denoted by Sₙ and is given by different formulas depending on the common ratio.
**Formula (when r ≠ 1):**
$$S_n = \frac{a(1-r^n)}{1-r} \quad \text{or} \quad S_n = \frac{a(r^n-1)}{r-1}$$
**Formula (when r = 1):**
$$S_n = na$$
**Derivation for r ≠ 1:**
Given: Sₙ = a + ar + ar² + ... + arⁿ⁻¹ ... (1)
Multiply equation (1) by r:
rSₙ = ar + ar² + ar³ + ... + arⁿ ... (2)
Subtract (2) from (1):
Sₙ - rSₙ = a - arⁿ
(1 - r)Sₙ = a(1 - rⁿ)
Therefore: $$S_n = \frac{a(1-r^n)}{1-r}$$
**Alternative form using r > 1:**
When r > 1, it's convenient to use: $$S_n = \frac{a(r^n-1)}{r-1}$$
**Worked Example 6:**
Find the sum of first n terms and the sum of first 5 terms of the G.P. 1/3, 2/9, 4/27, ...
**Worked Example 7:**
How many terms of the G.P. 3, (3√3)/2, (3·3)/4, ... are needed to give sum 3069/512?
**Worked Example 8:**
The sum of first three terms of a G.P. is 13/12 and their product is -1. Find the common ratio and the three terms.
Let the three terms be a/r, a, ar.
**Given conditions:**
**From equation (2):**
a³ = -1
a = -1 (taking real root)
**Substitute a = -1 in equation (1):**
-1/r - 1 - r = 13/12
Multiply by -12r:
12 + 12r + 12r² = -13r
12r² + 25r + 12 = 0
**Using quadratic formula:**
$$r = \frac{-25 \pm \sqrt{625-576}}{24} = \frac{-25 \pm 7}{24}$$
r = -3/4 or r = -4/3
**When r = -3/4:**
Three terms: 4/3, -1, -3/4
**When r = -4/3:**
Three terms: 3/4, -1, -4/3
**Exam Important Points:**
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**Series Not Following G.P.:**
Some series can be related to G.P. by algebraic manipulation.
**Worked Example 9:**
Find the sum of the sequence 7, 77, 777, 7777, ... to n terms.
Sₙ = 7 + 77 + 777 + 7777 + ... (to n terms)
**Rewrite each term:**
= 7(1 + 11 + 111 + 1111 + ... to n terms)
= (7/9)(9 + 99 + 999 + 9999 + ... to n terms)
= (7/9)[(10 - 1) + (10² - 1) + (10³ - 1) + ... to n terms]
= (7/9)[(10 + 10² + 10³ + ...) - n]
The sum 10 + 10² + 10³ + ... (n terms) is a G.P. with a = 10, r = 10:
= 10(10ⁿ - 1)/(10 - 1) = 10(10ⁿ - 1)/9
Therefore:
$$S_n = \frac{7}{9}\left[\frac{10(10^n-1)}{9} - n\right] = \frac{7[10(10^n-1) - 9n]}{81}$$
**Worked Example 10:**
A person has 2 parents, 4 grandparents, 8 great-grandparents, and so on. Find the total number of ancestors over 10 generations.
This is a G.P. with a = 2, r = 2, n = 10.
$$S_n = \frac{a(r^n-1)}{r-1} = \frac{2(2^{10}-1)}{2-1} = 2(1024-1) = 2046$$
Therefore, the total number of ancestors is **2046**.
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The **geometric mean** of two positive numbers a and b is defined as:
$$G.M. = \sqrt{ab}$$
**Example:** G.M. of 2 and 8 is √(2 × 8) = √16 = 4
**Observation:** The numbers 2, 4, 8 are in G.P., where 4 is the G.M. of 2 and 8.
**Insertion of Geometric Means:**
If n numbers G₁, G₂, ..., Gₙ are inserted between two positive numbers a and b such that a, G₁, G₂, ..., Gₙ, b form a G.P., then:
**The geometric means are:**
$$G_k = a \cdot r^k = a \cdot \left(\frac{b}{a}\right)^{\frac{k}{n+1}} \quad \text{for } k = 1, 2, ..., n$$
**Worked Example 11:**
Insert three numbers between 1 and 256 so that the resulting sequence is a G.P.
Let G₁, G₂, G₃ be the three numbers.
The sequence is: 1, G₁, G₂, G₃, 256
This is a G.P. with a = 1, n + 2 = 5 terms, so n + 1 = 4.
**Find r:**
$$256 = 1 \cdot r^4$$
r⁴ = 256
r = ±4
**For r = 4:**
**For r = -4:**
**Answer:** Insert 4, 16, 64 (taking positive values for the conventional answer).
**Exam Important Points:**
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For two positive real numbers a and b:
**Arithmetic Mean (A.M.):**
$$A = \frac{a+b}{2}$$
**Geometric Mean (G.M.):**
$$G = \sqrt{ab}$$
**Fundamental Inequality:**
$$A \geq G$$
with equality if and only if a = b.
**Proof:**
$$A - G = \frac{a+b}{2} - \sqrt{ab}$$
$$= \frac{a + b - 2\sqrt{ab}}{2} = \frac{(\sqrt{a} - \sqrt{b})^2}{2} \geq 0$$
Since the square of any real number is non-negative, A - G ≥ 0.
Therefore: **A ≥ G**
Equality holds when (√a - √b)² = 0, which means a = b.
**Worked Example 12:**
If A.M. and G.M. of two positive numbers a and b are 10 and 8 respectively, find the numbers.
**Given:**
**Find a - b:**
(a - b)² = (a + b)² - 4ab = 400 - 256 = 144
a - b = ±12 ... (3)
**Case 1: a - b = 12**
From (1) and a - b = 12:
**Case 2: a - b = -12**
From (1) and a - b = -12:
**Answer:** The numbers are 4 and 16.
**Exam Important Points:**
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**Finding Missing Terms:**
When certain terms of a G.P. are given, we use aₙ = arⁿ⁻¹ to form equations.
**Worked Example 13:**
In a G.P., the 3rd term is 24 and the 6th term is 192. Find the 10th term.
**Given:**
**Divide (2) by (1):**
$$\frac{ar^5}{ar^2} = \frac{192}{24}$$
r³ = 8
r = 2
**Substitute r = 2 in (1):**
a(2)² = 24
4a = 24
a = 6
**Find a₁₀:**
a₁₀ = ar⁹ = 6(2)⁹ = 6 × 512 = 3072
**Product of Terms in G.P.:**
If a G.P. has n terms with first term a and last term l, the product P of all terms is:
$$P = (al)^{n/2} \text{ or } P = a^n \cdot r^{n(n-1)/2}$$
This is because for terms equidistant from the ends, their product equals al.
**Worked Example 14:**
If P is the product of n terms of a G.P. with first term a and last term b, prove that P² = (ab)ⁿ.
**Proof:**
The n terms are: a, ar, ar², ..., arⁿ⁻¹
Product P = a · ar · ar² · ... · arⁿ⁻¹
= aⁿ · r⁰⁺¹⁺²⁺...⁺⁽ⁿ⁻¹⁾
= aⁿ · r^[n(n-1)/2]
Last term b = arⁿ⁻¹
Now: (ab)ⁿ = aⁿ · (arⁿ⁻¹)ⁿ = aⁿ · aⁿ · rⁿ⁽ⁿ⁻¹⁾
= a²ⁿ · r^[n(n-1)]
And: P² = (aⁿ · r^[n(n-1)/2])² = a²ⁿ · r^[n(n-1)]
Therefore: **P² = (ab)ⁿ**
**Practical Applications:**
**Problem:** What will Rs 500 amount to in 10 years if deposited in a bank paying 10% annual interest compounded annually?
**Solution:**
Initial amount = 500
Rate = 10% per annum
Time = 10 years
Amount forms a G.P.:
500, 500(1.1), 500(1.1)², ..., 500(1.1)¹⁰
After 10 years:
A = 500(1.1)¹⁰ ≈ Rs 1,296.87
**Problem:** A bacterial culture doubles every hour. If 30 bacteria are present initially, find the number at the end of 2nd hour, 4th hour, and nth hour.
**Solution:**
Initial: a₀ = 30
After 1 hour: a₁ = 30 × 2 = 60
After 2 hours: a₂ = 30 × 2² = 120
After 4 hours: a₄ = 30 × 2⁴ = 480
After n hours: aₙ = 30 × 2ⁿ
**Exam Important Points:**
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1. **Forgetting the condition r ≠ 1:** When r = 1, the sum formula changes to Sₙ = na, not the general formula.
2. **Confusing a and aₙ:** a represents the first term a₁, not the nth term.
3. **Sign errors with r < 0:** When r is negative, terms alternate in sign. Care must be taken with powers of r.
4. **Misidentifying the sequence type:** Always check if it's truly a G.P. by calculating r from consecutive terms consistently.
5. **Rounding prematurely:** In problems requiring finding n (which must be a natural number), avoid rounding until the final answer.
6. **Incorrect insertion formula:** When inserting k numbers between a and b to form a G.P., the total terms = k + 2, so $$r = (b/a)^{1/(k+1)}$$.
7. **A.M. - G.M. inequality direction:** Always remember A ≥ G, not A ≤ G.
8. **Applying sum formula for infinite series:** The formula Sₙ = a(1 - rⁿ)/(1 - r) applies only to finite series. For infinite series with |r| < 1, S∞ = a/(1 - r).
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| Concept | Formula | Condition |
|---------|---------|-----------|
| General term of G.P. | aₙ = arⁿ⁻¹ | a = first term, r = common ratio |
| Sum (finite G.P.) | Sₙ = a(rⁿ - 1)/(r - 1) | r ≠ 1 |
| Sum (when r = 1) | Sₙ = na | r = 1 |
| Geometric Mean | G = √(ab) | a, b > 0 |
| A.M. - G.M. relation | A ≥ G | a, b > 0; equality when a = b |
| Geometric means insertion | Gₖ = a(b/a)^[k/(n+1)] | k = 1,2,...,n |
| Product of n terms | P² = (ab)ⁿ | a = 1st term, b = last term |
Q1. The first three terms of the sequence defined by aₙ = 2n + 5 are:
Answer: A — Substituting n = 1, 2, 3: a₁ = 2(1) + 5 = 7; a₂ = 2(2) + 5 = 9; a₃ = 2(3) + 5 = 11.
Q2. Which of the following is a finite sequence?
Answer: B — A finite sequence has a fixed, countable number of terms; the ancestor sequence has exactly 10 generations, while the others are infinite.
Q3. The 20th term of the sequence aₙ = (n – 1)(2 – n)(3 + n) is:
Answer: B — a₂₀ = (20 – 1)(2 – 20)(3 + 20) = 19 × (–18) × 23 = –7866.
Q4. In the geometric progression 2, 4, 8, 16, ..., the common ratio r is:
Answer: B — The ratio between consecutive terms: 4/2 = 2, 8/4 = 2, 16/8 = 2, so r = 2.
Q5. Which sequence is NOT a geometric progression?
Answer: C — The Fibonacci sequence 1, 1, 2, 3, 5, 8 has no constant ratio; the ratio varies (1/1 = 1, 2/1 = 2, 3/2 = 1.5), so it is not a G.P.
Q6. If aₙ = (–1)ⁿ⁻¹ · 5ⁿ⁺¹, then a₂ equals:
Answer: B — a₂ = (–1)^(2–1) · 5^(2+1) = (–1)¹ · 5³ = –1 × 125 = –125.
Q7. The series corresponding to the sequence 1, 3, 5, 7, 9 is written in sigma notation as:
Answer: B — The sequence 1, 3, 5, 7, 9 are odd numbers with formula aₖ = 2k – 1; the series is Σₖ₌₁⁵ (2k – 1) = 1 + 3 + 5 + 7 + 9.
Q8. In a geometric progression with first term a = 3 and common ratio r = 2, the 4th term is:
Answer: C — In G.P., the nth term is ar^(n–1); the 4th term = 3 · 2³ = 3 · 8 = 24.
Q9. For the Fibonacci sequence where a₁ = a₂ = 1 and aₙ = aₙ₋₁ + aₙ₋₂ (n > 2), the 6th term is:
Answer: B — a₁ = 1, a₂ = 1, a₃ = 1 + 1 = 2, a₄ = 1 + 2 = 3, a₅ = 2 + 3 = 5, a₆ = 3 + 5 = 8.
Q10. A sequence is defined by a₁ = 1 and aₙ = aₙ₋₁ + 2 for all n > 1. The first four terms and their sum are respectively:
Answer: A — a₁ = 1, a₂ = 1 + 2 = 3, a₃ = 3 + 2 = 5, a₄ = 5 + 2 = 7; sum = 1 + 3 + 5 + 7 = 16.
What is a sequence?
An ordered collection of numbers a₁, a₂, a₃, ... where each position n has an associated term aₙ following a specific rule or pattern.
Define the nth term (general term) of a sequence.
The formula aₙ that allows you to calculate the value at position n without listing all previous terms.
What is the difference between a sequence and a series?
A sequence is an ordered list of numbers; a series is the sum of the terms in that sequence, written as a₁ + a₂ + a₃ + ... or Σaₖ.
What is a finite sequence?
A sequence that contains a fixed, countable number of terms, such as the 10 generations of ancestors over 300 years.
What is the Fibonacci sequence and its recurrence relation?
Sequence 1, 1, 2, 3, 5, 8, ... defined by a₁ = a₂ = 1 and aₙ = aₙ₋₂ + aₙ₋₁ for n > 2, where each term is the sum of the two preceding terms.
What is sigma notation (Σ)?
Compact mathematical notation for series: Σₖ₌₁ⁿ aₖ represents the sum a₁ + a₂ + a₃ + ... + aₙ.
Define a geometric progression (G.P.).
A sequence where each non-zero term bears a constant ratio r to the preceding term, so consecutive terms are a, ar, ar², ar³, ... for first term a and common ratio r.
What is the condition for a geometric progression?
The ratio aₖ₊₁/aₖ = r (constant) for all k ≥ 1, meaning every term divided by the previous term gives the same value r.
How do you identify the pattern in an unfamiliar sequence?
Check if it follows an explicit formula aₙ = f(n), a recurrence relation linking terms, or describe verbally if no pattern exists (like prime numbers).
What is the domain of a sequence when viewed as a function?
The set of natural numbers ℕ or a subset of natural numbers, since we index terms by position numbers 1, 2, 3, ...
Define a sequence and state whether the sequence 2, 4, 8, 16, ... is finite or infinite. What is its nth term? [2 marks]
A sequence is an ordered list of numbers aₙ indexed by position n. This sequence is infinite (never ends). Identify the pattern: each term is double the previous; the nth term is aₙ = 2ⁿ.
For the sequence defined by a₁ = 3 and aₙ = 3aₙ₋₁ + 2 for all n > 1, find the first five terms and write the corresponding series using sigma notation. [5 marks]
Use the recurrence relation step by step: a₂ = 3(3) + 2 = 11, a₃ = 3(11) + 2 = 35, and so on. After finding all five terms, write the series as a₁ + a₂ + a₃ + a₄ + a₅ and express it in sigma form Σₖ₌₁⁵ aₖ.
Explain with examples the difference between an explicit formula and a recurrence relation for defining a sequence. Show that the sequence aₙ = 2n (explicit) and the sequence a₁ = 2, aₙ = aₙ₋₁ + 2 (recurrence) generate the same terms. Why might we prefer one over the other? [6 marks]
Explicit formula directly calculates aₙ from n (e.g., a₅ = 2(5) = 10); recurrence relation defines aₙ using previous terms (a₁ = 2, then a₂ = 2 + 2 = 4, a₃ = 4 + 2 = 6, ...). Show both give 2, 4, 6, 8, 10, ... Explicit is faster for large n; recurrence shows how the sequence builds logically.
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