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Sequences and Series

NCERT Class 11 · Mathematics Based on NCERT Class 11 Mathematics textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter formalizes the concepts of sequences (ordered lists of numbers following a rule) and series (the sum of the terms of a sequence). It builds on the arithmetic progression (A.P.) studied in earlier classes and introduces geometric progression (G.P.), including the general term, sum to n terms, geometric mean (G.M.), and the relationship between A.M. and G.M. The one big idea a student must walk away with is that sequences and series are functions on natural numbers, and that A.P. and G.P. are the two most important patterns with closed-form formulas for their nth term and sum, which are essential for solving real-world problems involving growth, depreciation, and compound interest.
Key Concepts (10)
Sequence
A sequence is an ordered list of numbers arranged according to a definite rule, where each number is called a term. It can be regarded as a function whose domain is the set of natural numbers or a subset of it.
e.g. The number of ancestors over 10 generations: 2, 4, 8, 16, 32, …, 1024 forms a sequence.
Term of a sequence
The various numbers occurring in a sequence are called its terms. The nth term is denoted by aₙ and is also called the general term of the sequence.
e.g. In the sequence 2, 4, 6, 8, …, a₁ = 2, a₂ = 4, a₃ = 6, and aₙ = 2n.
Finite sequence
A sequence containing a finite number of terms is called a finite sequence.
e.g. The sequence of ancestors (2, 4, 8, …, 1024) has exactly 10 terms, so it is a finite sequence.
Infinite sequence
A sequence is called infinite if it is not a finite sequence, i.e., it never ends.
e.g. The sequence of successive quotients obtained by dividing 10 by 3: 3, 3.3, 3.33, 3.333, … is an infinite sequence.
Series
If a₁, a₂, a₃, …, aₙ, … is a given sequence, then the expression a₁ + a₂ + a₃ + … + aₙ + … is called the series associated with the given sequence.
e.g. For the sequence 1, 3, 5, 7, the corresponding series is 1 + 3 + 5 + 7.
Sigma notation (∑)
The Greek letter ∑ (sigma) is used to indicate summation. The series a₁ + a₂ + a₃ + … + aₙ is abbreviated as ∑_{k=1}^{n} a_k.
e.g. ∑_{k=1}^{4} (2k) = 2 + 4 + 6 + 8 = 20.
Geometric Progression (G.P.)
A sequence a₁, a₂, a₃, …, aₙ, … is called a geometric progression if each term is non-zero and the ratio of any term to its preceding term is constant. This constant is called the common ratio (r).
e.g. 2, 4, 8, 16, … is a G.P. with first term a = 2 and common ratio r = 2.
Common ratio (r) of a G.P.
In a geometric progression, the constant ratio obtained by dividing any term (except the first) by its immediately preceding term is called the common ratio, denoted by r.
e.g. In the G.P. 1, -1/3, 1/9, -1/27, …, the common ratio r = -1/3.
Geometric Mean (G.M.)
The geometric mean of two positive numbers a and b is the number √(ab). If a, G, b are in G.P., then G is the geometric mean of a and b.
e.g. The geometric mean of 2 and 8 is √(2×8) = √16 = 4.
Fibonacci sequence
A sequence generated by the recurrence relation a₁ = a₂ = 1, and aₙ = aₙ₋₂ + aₙ₋₁ for n > 2.
e.g. 1, 1, 2, 3, 5, 8, 13, … is the Fibonacci sequence.
Formulas & Equations (9)
aₙ = arⁿ⁻¹
Reads as: a sub n equals a times r to the power n minus one
Variables: a = first term, r = common ratio, n = position of the term (natural number), aₙ = nth term
Use when: Finding the nth term of a geometric progression when the first term and common ratio are known.
⚠ Students often write aₙ = arⁿ instead of arⁿ⁻¹, forgetting that the first term corresponds to n = 1, not n = 0.
Sₙ = a(1 - rⁿ) / (1 - r), for r ≠ 1
Reads as: S sub n equals a times one minus r to the power n divided by one minus r
Variables: a = first term, r = common ratio, n = number of terms, Sₙ = sum of first n terms
Use when: Finding the sum of the first n terms of a G.P. when the common ratio is not equal to 1.
⚠ Students forget to check that r ≠ 1 before applying this formula. If r = 1, the sum is simply n × a.
Sₙ = a(rⁿ - 1) / (r - 1), for r > 1
Reads as: S sub n equals a times r to the power n minus one divided by r minus one
Variables: a = first term, r = common ratio, n = number of terms, Sₙ = sum of first n terms
Use when: An alternative form of the sum formula, convenient when r > 1 to avoid negative denominators.
⚠ Students mix up the two forms and use the wrong one, leading to sign errors.
G.M. of a and b = √(ab)
Reads as: Geometric mean of a and b equals the square root of a times b
Variables: a, b = two positive real numbers
Use when: Calculating the geometric mean of two positive numbers, or inserting a single geometric mean between them.
⚠ Students forget that G.M. is defined only for positive numbers and apply it to negative numbers without checking.
A.M. = (a + b) / 2
Reads as: Arithmetic mean equals a plus b divided by 2
Variables: a, b = two real numbers
Use when: Calculating the arithmetic mean of two numbers, often in problems comparing A.M. and G.M.
⚠ Students confuse A.M. with G.M. and use the wrong formula in comparison problems.
A ≥ G (for positive numbers)
Reads as: Arithmetic mean is greater than or equal to geometric mean
Variables: A = arithmetic mean, G = geometric mean
Use when: Proving inequalities or establishing relationships between two positive numbers.
⚠ Students forget that this holds only for positive numbers and apply it to all real numbers.
Sum of first n natural numbers: ∑k = n(n+1)/2
Reads as: Sum of k from 1 to n equals n times n plus 1 divided by 2
Variables: n = number of natural numbers
Use when: Finding the sum of the first n natural numbers, often in problems involving special series.
⚠ Students incorrectly write n(n-1)/2 instead of n(n+1)/2.
Sum of squares of first n natural numbers: ∑k² = n(n+1)(2n+1)/6
Reads as: Sum of k squared from 1 to n equals n times n plus 1 times 2n plus 1 divided by 6
Variables: n = number of natural numbers
Use when: Finding the sum of squares of the first n natural numbers, often in miscellaneous problems.
⚠ Students forget the factor 1/6 or misplace the (2n+1) term.
Sum of cubes of first n natural numbers: ∑k³ = [n(n+1)/2]²
Reads as: Sum of k cubed from 1 to n equals the square of n times n plus 1 divided by 2
Variables: n = number of natural numbers
Use when: Finding the sum of cubes of the first n natural numbers.
⚠ Students think ∑k³ = ∑k² × ∑k, which is incorrect; it is the square of ∑k.
How It Works
Finding the nth term of a G.P.
  1. Step 1: Identify the first term (a) and the common ratio (r) from the given sequence.
  2. Step 2: Use the formula aₙ = arⁿ⁻¹.
  3. Step 3: Substitute the value of n to find the required term.
💡 The exponent is (n-1), not n. The first term corresponds to n=1, giving a₁ = ar⁰ = a.
Finding the sum of first n terms of a G.P.
  1. Step 1: Identify a, r, and n.
  2. Step 2: Check if r = 1. If yes, Sₙ = n × a.
  3. Step 3: If r ≠ 1, use Sₙ = a(1 - rⁿ)/(1 - r) or Sₙ = a(rⁿ - 1)/(r - 1).
  4. Step 4: Simplify the expression to get the numerical sum.
💡 Always check the value of r first. If r = 1, the formula changes completely.
Inserting n geometric means between two numbers a and b
  1. Step 1: Let G₁, G₂, …, Gₙ be the n numbers to be inserted between a and b such that a, G₁, G₂, …, Gₙ, b are in G.P.
  2. Step 2: The total number of terms in the G.P. is (n+2). The last term b is the (n+2)th term.
  3. Step 3: Use b = arⁿ⁺¹ to find r = (b/a)^(1/(n+1)).
  4. Step 4: Then G₁ = ar, G₂ = ar², …, Gₙ = arⁿ.
💡 The number of terms in the G.P. after insertion is (n+2), so the exponent for r is (n+1), not n.
Finding terms of a G.P. when sum and product are given
  1. Step 1: Assume the three terms as a/r, a, ar (for three terms in G.P.).
  2. Step 2: Write equations for sum and product using these variables.
  3. Step 3: Solve the product equation first to find a.
  4. Step 4: Substitute a into the sum equation to form a quadratic in r.
  5. Step 5: Solve for r and then find the actual terms.
💡 For three terms in G.P., always assume them as a/r, a, ar. This simplifies the product to a³.
Converting a non-G.P. series (like 7, 77, 777) to a G.P. form
  1. Step 1: Factor out a common factor (e.g., 7/9) to rewrite each term as (10ⁿ - 1).
  2. Step 2: Write the sum as a combination of a geometric series (sum of powers of 10) and a constant series.
  3. Step 3: Apply the G.P. sum formula to the geometric part.
  4. Step 4: Simplify to get the final expression.
💡 The key trick is to express numbers like 77 as (7/9)(99) = (7/9)(10² - 1).
Exam Traps — Don't Make These Mistakes
⚠ Confusing the nth term formula exponent
✗ aₙ = arⁿ
→ The first term is a = ar⁰, so the nth term should have exponent (n-1). Writing arⁿ gives the (n+1)th term.
✓ aₙ = arⁿ⁻¹
⚠ Using the sum formula when r = 1
✗ Sₙ = a(1 - 1ⁿ)/(1 - 1) = 0/0 (undefined)
→ The formula Sₙ = a(1 - rⁿ)/(1 - r) is valid only for r ≠ 1. When r = 1, the series is a + a + a + … (n terms), so Sₙ = n × a.
✓ When r = 1, Sₙ = n × a.
⚠ Forgetting the sign when using Sₙ = a(rⁿ - 1)/(r - 1)
✗ Sₙ = a(rⁿ - 1)/(1 - r)
→ The formula Sₙ = a(rⁿ - 1)/(r - 1) is used when r > 1. If you write (1 - r) in the denominator, the numerator must be (1 - rⁿ) to keep the sign correct.
✓ Use Sₙ = a(rⁿ - 1)/(r - 1) for r > 1, and Sₙ = a(1 - rⁿ)/(1 - r) for r < 1.
⚠ Assuming G.M. is defined for negative numbers
✗ G.M. of -4 and -9 is √(36) = 6
→ The geometric mean is defined only for positive numbers in the NCERT context. For negative numbers, the square root of a positive product gives a positive number, but the sequence -4, 6, -9 is not a G.P. because the common ratio is not constant.
✓ G.M. is defined for positive numbers only. For a and b positive, G.M. = √(ab).
⚠ Mistaking the number of terms when inserting geometric means
✗ To insert 3 numbers between 1 and 256, the total terms are 3, so r⁴ = 256 gives r = 4.
→ The total number of terms in the G.P. after insertion is (3 + 2) = 5. The last term (256) is the 5th term, so the equation should be ar⁴ = 256, not r⁴ = 256 if a is not 1. Here a=1, so r⁴=256 is correct, but students often forget to add 2 to the number of inserted terms.
✓ If n numbers are inserted, total terms = n + 2. So b = arⁿ⁺¹.
Compare & Contrast
Definition of progression
Arithmetic Progression (A.P.)
Geometric Progression (G.P.)
In A.P., the difference between consecutive terms is constant (common difference d). In G.P., the ratio of consecutive terms is constant (common ratio r).
Formula and relationship
Arithmetic Mean (A.M.)
Geometric Mean (G.M.)
A.M. of two numbers a and b is (a+b)/2, while G.M. is √(ab). For positive numbers, A.M. ≥ G.M., with equality only when a = b.
Exam Focus
Marks pattern: Typically 1-2 MCQs (1 mark each), 1 short answer question (3 marks), and 1 long answer question (5 marks) from this chapter, totalling around 8-10 marks.
Most asked: Finding the nth term and sum to n terms of a G.P. · Insertion of geometric means between two numbers · Relationship between A.M. and G.M. (including proof that A ≥ G) · Word problems involving G.P. (e.g., population growth, compound interest, depreciation) · Sum of special series like 7, 77, 777, … and 0.6, 0.66, 0.666, …
Don't skip: Long answer (5 marks) on finding terms of a G.P. given sum and product, or inserting geometric means.; Proof-based question (3-5 marks) on A.M. ≥ G.M. or showing that a, b, c, d are in G.P. from a given condition.
One-Line Revision (read aloud before the exam)
  1. A sequence is an ordered list of numbers following a rule; a series is the sum of the terms of a sequence.
  2. The nth term of a G.P. is given by aₙ = arⁿ⁻¹, where a is the first term and r is the common ratio.
  3. The sum of the first n terms of a G.P. is Sₙ = a(1 - rⁿ)/(1 - r) for r ≠ 1, and Sₙ = n × a for r = 1.
  4. The geometric mean of two positive numbers a and b is √(ab).
  5. For any two positive numbers, A.M. ≥ G.M., with equality only when the numbers are equal.
  6. To insert n geometric means between a and b, use r = (b/a)^(1/(n+1)).
  7. The sum of the first n natural numbers is n(n+1)/2.
  8. The sum of squares of the first n natural numbers is n(n+1)(2n+1)/6.
  9. The sum of cubes of the first n natural numbers is [n(n+1)/2]².
  10. A series can be finite or infinite depending on whether the corresponding sequence is finite or infinite.
  11. Sigma notation ∑ is used to represent the sum of a series compactly.
  12. In a G.P., the product of three terms taken as a/r, a, ar simplifies to a³.
  13. The common ratio r can be negative, leading to alternating signs in the G.P.
  14. The Fibonacci sequence is defined by a₁ = a₂ = 1 and aₙ = aₙ₋₂ + aₙ₋₁ for n > 2.

MCQs — 10 Questions with Answers

Q1. The first three terms of the sequence defined by aₙ = 2n + 5 are:

  • A. 7, 9, 11 ✓
  • B. 5, 7, 9
  • C. 3, 5, 7
  • D. 2, 4, 6

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?

  • A. The set of all natural numbers
  • B. The 10 generations of a person's ancestors over 300 years ✓
  • C. The sequence 1, 1/2, 1/3, 1/4, ...
  • D. The sequence of prime numbers

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:

  • A. 7866
  • B. –7866 ✓
  • C. 6930
  • D. –6930

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:

  • A. 1
  • B. 2 ✓
  • C. 3
  • D. 4

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?

  • A. 1, –1/3, 1/9, –1/27, ...
  • B. 2, 4, 8, 16, ...
  • C. 1, 1, 2, 3, 5, 8, ... ✓
  • D. 0.1, 0.001, 0.00001, ...

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:

  • A. 125
  • B. –125 ✓
  • C. 625
  • D. –625

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:

  • A. Σₖ₌₁⁵ 2k
  • B. Σₖ₌₁⁵ (2k – 1) ✓
  • C. Σₖ₌₁⁵ k²
  • D. Σₖ₌₁⁵ (k + 1)

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:

  • A. 12
  • B. 18
  • C. 24 ✓
  • D. 27

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:

  • A. 5
  • B. 8 ✓
  • C. 13
  • D. 21

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:

  • A. 1, 3, 5, 7 and sum = 16 ✓
  • B. 1, 2, 3, 4 and sum = 10
  • C. 1, 3, 5, 7 and sum = 14
  • D. 0, 2, 4, 6 and sum = 12

Answer: A — a₁ = 1, a₂ = 1 + 2 = 3, a₃ = 3 + 2 = 5, a₄ = 5 + 2 = 7; sum = 1 + 3 + 5 + 7 = 16.

Flashcards

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, ...

Important Board Questions

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.

Assertion-Reason Questions for Sequences and Series

Q1. Assertion (A):

In a geometric progression with first term a and common ratio r, the nth term is given by an = ar^(n−1).

Reason (R):

The terms of a G.P. are obtained by multiplying the first term repeatedly by the common ratio, maintaining a constant multiplicative relationship between consecutive terms.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains why the formula works. The recurrence relation an = r·an−1 leads to the explicit formula an = ar^(n−1) through repeated multiplication.

Q2. Assertion (A):

A Fibonacci-like sequence defined by a1 = 1, a2 = 1, and an = an−1 + an−2 for n > 2 is an arithmetic progression.

Reason (R):

The sequence does not have a constant difference between consecutive terms.

Answer: D — A is false but R is true — The assertion is false: Fibonacci is not an A.P. (differences are 0, 1, 1, 2, 3, 5, ..., not constant). However, the reason is true—there is no constant difference. Since assertion is false and reason is true, the answer is C.

Q3. Assertion (A):

Every sequence can be expressed using an algebraic formula of the form an = f(n).

Reason (R):

Some sequences, like the sequence of prime numbers, cannot be described by a single algebraic formula, though they can be generated by other rules or verbal descriptions.

Answer: D — A is false but R is true — The assertion is false because not all sequences have algebraic formulas. The reason is also true but supports why the assertion is false. Since the assertion itself is false, and the reason (though true) doesn't explain a true assertion, the answer is D.

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Case-Based Questions (CBQ) for Sequences and Series

Case Study:

A pharmaceutical company is developing a vaccine and testing its effectiveness by measuring antibody levels in a patient's body over weeks. The antibody level follows a geometric progression model: after the first week, the level is 500 units; after the second week, it is 750 units; after the third week, it is 1125 units. The company wants to predict antibody levels for weeks 6 and 10 to assess long-term immunity.

Q1. What is the common ratio of this geometric progression?

  • A. 1.5 ✓
  • B. 0.5
  • C. 250
  • D. 2.25

Answer: A

Q2. Using the G.P. formula, what will the antibody level be after 6 weeks?

  • A. 3796.875 units
  • B. 5062.5 units
  • C. 7593.75 units ✓
  • D. 2531.25 units

Answer: C

Q3. Which week marks the point where the antibody level exceeds 5000 units?

  • A. Week 5
  • B. Week 6 ✓
  • C. Week 7
  • D. Week 4

Answer: B

Q4. If the immune system requires a minimum of 10000 units for sustained immunity, in which week will this threshold be reached?

  • A. Week 8
  • B. Week 9
  • C. Week 10 ✓
  • D. Week 11

Answer: C

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