Sequences and Series Class 11 Notes: Formulas and Worked Examples

If you are revising the night before a test, these sequences and series class 11 notes give you the whole chapter in one place: definitions, geometric progression formulas, the geometric mean, worked examples and exam pointers. The page follows the rationalised NCERT text for the 2026-27 session, so every formula matches your textbook and the chapter’s summary (NCERT, p. 15).

Read the concept sections first, then work the examples and finish with the mistakes table.

The chapter builds one idea on another: a sequence is an ordered list, a series is the sum of its terms, and a geometric progression is the one family of sequences that has formulas for any term and any sum. Once you see that structure, the formulas stop being memorisation.

These notes sit inside our Class 11 Mathematics notes, part of the wider Class 11 notes and the full CBSE notes index.

Sequences: Terms, General Term and the Fibonacci Pattern

In everyday language a sequence is an ordered collection: it has a first member, a second member, a third member and so on. Mathematics uses the same idea, and the order is part of the definition (NCERT, p. 1).

The chapter opens with two examples that carry the whole section:

• Ancestors over 300 years: with a 30-year generation gap, that is 10 generations. The number of ancestors in each generation runs 2, 4, 8, 16, …, 1024. Ten terms, so it is a finite sequence.
• Successive quotients of 10 ÷ 3: the division steps keep producing 3, 3.3, 3.33, 3.333, … without end. That makes it an infinite sequence.

The numbers in a sequence are its terms, written \( a_1, a_2, a_3, \dots, a_n, \dots \) where the subscript shows the position. The \( n^{th} \) term \( a_n \) — the number at the \( n^{th} \) position — is also called the general term (NCERT, p. 1).

Often a formula produces every term. The even numbers follow \( a_n = 2n \), and the odd numbers follow \( a_n = 2n – 1 \). Some sequences hide their rule: the Fibonacci sequence 1, 1, 2, 3, 5, 8, … is built by the recurrence \( a_1 = a_2 = 1 \) and \( a_n = a_{n-2} + a_{n-1} \) for \( n \gt 2 \).

The figure above is the chapter’s headline idea: sequences that follow a specific pattern are called progressions. Every list in this section — ancestors, quotients, even numbers — is a progression in that sense.

Because a sequence is just a rule generating term after term, it can be viewed as a function whose domain is the natural numbers or a subset of them (NCERT, p. 1).

Series and Sigma Notation: From a List of Terms to Their Sum

Take a sequence \( a_1, a_2, a_3, \dots \) and write its terms with plus signs between them. The expression \( a_1 + a_2 + a_3 + \dots \) is the series associated with that sequence (NCERT, p. 3). A series is finite or infinite exactly as the underlying sequence is.

Sigma notation compresses the writing. The Greek letter \( \Sigma \) means “sum”, and

\[ \sum_{k=1}^{n} a_k = a_1 + a_2 + a_3 + \dots + a_n \]

reads “the sum of \( a_k \) as \( k \) runs from 1 to \( n \)”.

Misconception autopsy: sequence versus series. The same numbers appear in both, which is why students mix them up. The sequence is the list — 2, 4, 6. The series is the indicated sum — 2 + 4 + 6. And the sum of the series is the number you get after adding, 12.

The textbook’s remark is worth keeping straight: a name like 1 + 3 + 5 + 7 labels an expression, and the sum of that series is 16 (NCERT, p. 3).

Before summing you need the terms. If \( a_n = 3n – 2 \), then \( a_1 = 3(1) – 2 = 1 \), \( a_2 = 3(2) – 2 = 4 \) and \( a_3 = 3(3) – 2 = 7 \), giving the series \( 1 + 4 + 7 + \dots \).

Geometric Progression: General Term and Sum to n Terms

A sequence is a geometric progression (G.P.) when every term is non-zero and each term divided by the previous one gives the same constant \( r \):

\[ \frac{a_{k+1}}{a_k} = r, \quad k \geq 1 \]

The first term is written \( a \), and the constant \( r \) is the common ratio. The G.P. then reads \( a, ar, ar^2, ar^3, \dots \) (NCERT, p. 5). In the chapter’s examples, 2, 4, 8, 16, … has \( r = 2 \); \( \frac{1}{9}, -\frac{1}{27}, \frac{1}{81}, \dots \) has \( r = -\frac{1}{3} \); and 0.01, 0.0001, … has \( r = 0.01 \).

A useful picture: a G.P. is like a chain of photocopies made at one fixed reduction. Every copy is the previous one multiplied by the same factor \( r \) — at 80% size the lengths run 100, 80, 64, 51.2, …, a G.P. with \( r = 0.8 \).

General term

Write the opening terms with their powers of \( r \):

\[ a_1 = a = ar^0, \quad a_2 = ar = ar^1, \quad a_3 = ar^2, \quad a_4 = ar^3 \]

The pattern settles the \( n^{th} \) term (NCERT, p. 5):

\[ a_n = ar^{n-1} \]

Why \( n – 1 \) and not \( n \): the first term already contains \( r^0 \). Moving from the first to the \( n^{th} \) term takes \( n – 1 \) multiplications by \( r \), so the exponent counts the steps after the first term.

Sum to n terms

Let \( S_n = a + ar + ar^2 + \dots + ar^{n-1} \). If \( r = 1 \), all \( n \) terms equal \( a \), so \( S_n = na \). For \( r \neq 1 \), multiply the whole sum by \( r \) and subtract the two lines:

\[ S_n – rS_n = (a + ar + \dots + ar^{n-1}) – (ar + ar^2 + \dots + ar^n) = a – ar^n \]
\[ (1 – r)S_n = a(1 – r^n) \]

Dividing by \( 1 – r \) gives the two equivalent forms (NCERT, p. 15):

\[ S_n = \frac{a(r^n – 1)}{r – 1} = \frac{a(1 – r^n)}{1 – r}, \quad r \neq 1 \]

They are identical — multiply numerator and denominator of one by −1 to reach the other. The case \( r = 1 \) is separate because the fraction becomes \( \frac{0}{0} \).

Where G.P.s appear in real life

• Bacteria growth: a culture starting with 30 bacteria that doubles every hour gives 30, 60, 120, … — a G.P. with \( r = 2 \) (Exercise 8.2, Q30).
• Compound interest: Rs 500 at 10% compounded annually grows as 500, 550, 605, … — a G.P. with \( r = 1.1 \) (Exercise 8.2, Q31).
• Depreciation: a machine losing 20% of its value yearly multiplies by 0.8 each year, again a G.P. (Miscellaneous Exercise, Q17).

Geometric Mean: How to Insert Numbers Between Two Terms

The geometric mean (G.M.) of two positive numbers \( a \) and \( b \) is \( \sqrt{ab} \) (NCERT, p. 9). For 2 and 8 the G.M. is \( \sqrt{16} = 4 \), and 2, 4, 8 are consecutive terms of a G.P. — that is the defining property of a geometric mean.

The idea generalises to inserting any number of means. If \( G_1, G_2, \dots, G_n \) are \( n \) means between positive \( a \) and \( b \), then \( a, G_1, G_2, \dots, G_n, b \) is a G.P. with \( n + 2 \) terms, so \( b \) is the \( (n + 2)^{th} \) term (NCERT, p. 9):

\[ b = ar^{n+1} \quad \Rightarrow \quad r = \left( \frac{b}{a} \right)^{\frac{1}{n+1}} \]

Each inserted mean is \( G_k = ar^k \), and the whole inserted string is itself a G.P. with that same ratio \( r \).

To place three numbers between 1 and 256, solve \( 256 = r^4 \), giving \( r = \pm 4 \) (real roots only). For \( r = 4 \) the means are 4, 16, 64; for \( r = -4 \) they are −4, 16, −64 — the chapter’s example works both branches.

The A.M.-G.M. Relationship

For two positive numbers \( a \) and \( b \), set \( A = \frac{a + b}{2} \) and \( G = \sqrt{ab} \). The chapter’s key relationship is \( A \geq G \), with equality exactly when \( a = b \) (NCERT, p. 10). The proof is a single chain:

\[ A – G = \frac{a + b}{2} – \sqrt{ab} = \frac{a + b – 2\sqrt{ab}}{2} = \frac{(\sqrt{a} – \sqrt{b})^2}{2} \geq 0 \]

A square is never negative, so \( A – G \geq 0 \). Equality holds precisely when \( \sqrt{a} = \sqrt{b} \), which means \( a = b \).

Method demonstration: suppose the A.M. and G.M. of two positive numbers are 10 and 8. Then \( a + b = 20 \) and \( ab = 64 \). The identity \( (a-b)^2 = (a+b)^2 – 4ab \) gives \( (a-b)^2 = 400 – 256 = 144 \), so \( a – b = \pm 12 \). Solving with \( a + b = 20 \) yields the pair 16 and 4.

The two means answer different questions, which explains why they differ:

Key Definitions at a Glance

Every term you must be able to name in a one-mark question, with a concrete example attached:

Sequences and Series Class 11 Notes: Formula Sheet

Quick-scan table for the night before the exam. Throughout, \( a \) is the first term, \( r \) the common ratio, \( n \) the number of terms and \( S_n \) the sum of the first \( n \) terms.

Memory device — keeping the two sum forms apart. Put the bigger quantity on top. If \( r \gt 1 \), then \( r^n \) beats 1, so write \( r^n – 1 \) over \( r – 1 \). If \( r \lt 1 \), then 1 beats \( r^n \), so write \( 1 – r^n \) over \( 1 – r \). Either form returns the same number; the rule only stops you carrying a negative fraction and making sign slips.

Worked Examples: Solved Step by Step

Three fresh problems, one for each standard question type. Copy the working style — naming the formula first is what earns the setup mark.

Example 1: General term of a G.P.

Example 2: Sum of the first 5 terms of a G.P.

Example 3: Inserting two geometric means

Common Mistakes Students Make

Each error is paired with the correction and a quick way to catch it in your own working:

Why each correction holds: the exponent rule follows from the very form \( a, ar, ar^2, \dots \); the \( r = 1 \) case exists because the derivation divides by \( 1 – r \); and sign mistakes evaporate when the larger quantity sits on top of the fraction.

Exam Notes: Where the Marks Are

• The \( A \geq G \) proof: the marks sit in the chain \( A – G = \frac{a+b-2\sqrt{ab}}{2} = \frac{(\sqrt{a}-\sqrt{b})^2}{2} \geq 0 \). Write every equality; a bare “clearly \( A \geq G \)” gains nothing.
• “Which term is X” questions: state \( a_n = ar^{n-1} \), substitute the given \( a \) and \( r \), equate to X and solve for \( n \). Showing the substitution earns partial credit even if the final \( n \) is wrong.
• Sum questions: write the chosen formula with numbers substituted before any arithmetic. That single line is the setup mark.
• The three-term trick: when a G.P.’s three consecutive terms are written \( \frac{a}{r}, a, ar \), their product collapses to \( a^3 \). The chapter uses this in its worked examples, and it is the fastest route in sum-and-product problems.

The exercise set splits into the same three groups the examples above model: Exercise 8.2 Q1–Q6 test the general term and the G.P. property; Q7–Q15 are sums, “how many terms” problems and sigma evaluation; Q26–Q32 apply geometric means and the A.M.-G.M. relationship, including real-life settings such as bacteria growth and compound interest.

Practise one question from each group. You can also verify every formula against the official NCERT Class 11 Mathematics textbook (Chapter 8).

If you are revising the full syllabus, the binomial theorem notes use the same summation instincts, and the next chapter, straight lines, is pure coordinate geometry — both have complete notes on this site.

Fast Revision: The Chapter in Seven Points

1. A sequence is an ordered arrangement of numbers following a rule, viewed as a function on the natural numbers (NCERT, p. 15).
2. A finite sequence has a fixed number of terms; an infinite one does not end.
3. A series is the indicated sum \( a_1 + a_2 + a_3 + \dots \), written \( \sum_{k=1}^{n} a_k \) in compact form.
4. A geometric progression keeps a constant ratio \( r \) between consecutive terms, with first term \( a \).
5. The \( n^{th} \) term of a G.P. is \( a_n = ar^{n-1} \).
6. The sum is \( S_n = \frac{a(r^n – 1)}{r – 1} = \frac{a(1 – r^n)}{1 – r} \) when \( r \neq 1 \), and \( S_n = na \) when \( r = 1 \).
7. The geometric mean of two positive numbers is \( \sqrt{ab} \), and \( A \geq G \) with equality only when \( a = b \).

Frequently Asked Questions

What is the difference between a sequence and a series in Class 11?

A sequence is the ordered list of terms, like 2, 4, 6. A series is the indicated sum of those terms, like 2 + 4 + 6. The sum of the series is the number you get after adding — for 2 + 4 + 6 that is 12.

What is the nth term formula of a geometric progression and why is the exponent n-1?

The formula is \( a_n = ar^{n-1} \). The exponent is \( n – 1 \) because the first term is \( a = ar^0 \) — reaching the \( n^{th} \) term takes \( n – 1 \) multiplications by \( r \).

Which G.P. sum formula should I use when r is greater than 1?

Use \( S_n = \frac{a(r^n – 1)}{r – 1} \) when \( r \gt 1 \) and \( S_n = \frac{a(1 – r^n)}{1 – r} \) when \( r \lt 1 \). Both give the same value for every \( r \neq 1 \); the choice only keeps numerator and denominator positive.

How do I insert geometric means between two given numbers?

Treat the second number as the \( (n+2)^{th} \) term of a G.P.: set \( b = ar^{n+1} \), solve for \( r = \left( \frac{b}{a} \right)^{\frac{1}{n+1}} \), then multiply term by term to get each mean \( G_k = ar^k \).

Is the arithmetic mean always greater than or equal to the geometric mean?

For two positive numbers, yes. The difference \( A – G = \frac{(\sqrt{a} – \sqrt{b})^2}{2} \) is a square divided by 2, never negative, so \( A \geq G \). Equality happens exactly when \( a = b \).

Reference: NCERT Class 11 Mathematics textbook, chapter 8, Sequences and Series.