JEEVisionary rocket logoJEEVisionaryIIT Mission Control
Join
Maths · Class 11 · Sequence and Series

Sequence and Series for JEE

Everything you need for this topic on one page: the ideas, the formulas, the traps, and free lectures that explain it well.

Last updated: 26 October 2026

Direct answer

Arithmetic and geometric progressions, means, AM–GM inequality, arithmetic–geometric series and sums of special series.

Arithmetic and geometric progressions, means, AM–GM inequality, arithmetic–geometric series and sums of special series.

Key ideas

Sequence and Series: what you must understand

  • AP: constant difference; GP: constant ratio
  • AM ≥ GM ≥ HM for positive numbers
  • Sum to infinity of GP exists if |r| < 1
  • Use method of differences for unusual series
Formula sheet

Must-know formulas for Sequence and Series

  • AP: aₙ = a + (n−1)d, Sₙ = n/2 [2a + (n−1)d]
  • GP: Sₙ = a(rⁿ − 1)/(r − 1), S∞ = a/(1 − r)
  • Σn = n(n+1)/2, Σn² = n(n+1)(2n+1)/6, Σn³ = [n(n+1)/2]²
  • AM ≥ GM: (a + b)/2 ≥ √(ab)
Print

One-page formula sheet

Printable Sequence and Series formula sheet: formulas, key ideas and traps on one page. Save it as a PDF for offline revision.

Avoid

Common mistakes students make

  • Using S∞ when |r| ≥ 1
  • Off-by-one errors in n
  • Applying AM–GM to negative numbers
For the exam

How this topic is asked in JEE

Appears in almost every JEE Main shift. High frequency. AM–GM min/max questions and special series sums are favourites.

NCERT map

Where Sequence and Series is in NCERT

Chapter names follow the current rationalised NCERT textbooks. Read these sections before solving PYQs.

ClassBookChapterSections JEE asks from
Class 11MathematicsSequences and SeriesArithmetic Progression; Geometric Progression; Relationship Between AM and GM; Sum to n Terms of Special Series; Arithmetic Mean and Geometric Mean
JEE Advanced level

A JEE Advanced-level example on Sequence and Series

Question. The sum of an infinite geometric progression is 2 and the sum of the GP formed by squaring each term of the original GP is 4/3. If the common ratio is positive, find the sum of the cubes of all the terms of the original GP.

Show solution
  1. Let first term = a and common ratio = r. Given: a/(1−r) = 2 and a²/(1−r²) = 4/3
  2. From first equation: a = 2(1−r). Substitute into second equation: 4(1−r)²/[(1−r)(1+r)] = 4/3
  3. Simplify: 4(1−r)/(1+r) = 4/3, which gives (1−r)/(1+r) = 1/3. Solving: 3−3r = 1+r, so r = 1/2
  4. Back substitute: a = 2(1−1/2) = 1. Verify: 1/(1−1/2) = 2 ✓ and 1/(1−1/4) = 4/3 ✓
  5. Sum of cubes: first term = a³ = 1, ratio = r³ = 1/8. Sum = 1/(1−1/8) = 8/7

Answer: 8/7

This example was solved twice independently and both solutions agree.

Free lectures

Best free lectures for Sequence and Series

Full-chapter free lectures picked from YouTube search results by views, from established JEE channels (mostly in Hinglish). Use the subtopic list to jump straight to the part you need (timestamps come from each video's own chapter list). Watch one, then solve PYQs the same day.

Practise

Where to practise for free

FAQs

Frequently asked questions

Is Sequence and Series important for JEE Main 2027?

Appears in almost every JEE Main shift. It is part of the official 'Sequence and Series' unit in the JEE Main 2027 syllabus.

What are the most important formulas in Sequence and Series?

AP: aₙ = a + (n−1)d, Sₙ = n/2 [2a + (n−1)d]; GP: Sₙ = a(rⁿ − 1)/(r − 1), S∞ = a/(1 − r); Σn = n(n+1)/2, Σn² = n(n+1)(2n+1)/6, Σn³ = [n(n+1)/2]².

How should I study Sequence and Series for JEE?

Understand the key ideas from NCERT or one full lecture, learn the formula sheet, then solve 25–30 past JEE questions on the topic. High frequency. AM–GM min/max questions and special series sums are favourites.

WA✎
Sitemap