Class 11 Maths · Original EZT practice
Sequences and Series (Class 11 Chapter 8): summary and practice
Sequences and Series turns lists of numbers into formulas: what a sequence is, how a series sums one, and the full toolkit for geometric progressions - the nth term a r^(n-1), the sum S_n, the geometric mean and the AM-GM inequality. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Sequences
A sequence is an ordered list built by a rule - each position n gets a term a_n. Fibonacci's 1, 1, 2, 3, 5, ... grows by adding the previous two terms; the point is that a pattern, not a guess, fixes every term.
2. Series
A series is what you get when you ADD the terms of a sequence: a_1 + a_2 + ... + a_n, written compactly with the sigma symbol. Knowing the sequence rule lets you sum a hundred terms without writing them out.
3. GP and its general term
A geometric progression multiplies by a fixed ratio r each step: a, ar, ar^2, ar^3, ... The nth term is a r^(n-1) - the exponent is one less than the position, because the first term carries no multiplication yet.
4. Sum to n terms
For r not equal to 1, S_n = a(r^n - 1)/(r - 1). When |r| < 1 the terms shrink toward zero and the infinite series settles at a/(1 - r); when |r| is 1 or more the infinite sum does not exist.
5. Geometric mean and AM >= GM
The geometric mean of two positive numbers a and b is sqrt(ab) - it sits between them the way a GP middle term does. Section 8.5 proves the arithmetic mean is always at least the geometric mean, with equality only when a = b.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Find a term deep into a GP
Choose the method: nth term = a r^(n-1), exponent one less than the position.
EZT practice question
A GP starts 3, 6, 12, 24, ... Find its 5th term.
Show the worked solution
- a = 3 and r = 6 / 3 = 2.
- 5th term = a r^4 = 3 x 2^4.
- = 3 x 16.
Answer: 48.
Check: Count the multiplications: 3 -> 6 -> 12 -> 24 -> 48 is four steps, so the exponent is 4, not 5.
Watch for this: Using r^n instead of r^(n-1) gives 96 - one whole multiplication too many.
Now try it yourself
Same GP - what is the 7th term?
Check your answer
192.
3 x 2^6 = 3 x 64 = 192.
Sum the first n terms of a GP
Choose the method: S_n = a(r^n - 1)/(r - 1) when r is not 1.
EZT practice question
Sum the first 8 terms of the GP 3, 6, 12, 24, ...
Show the worked solution
- a = 3, r = 2, n = 8.
- S_8 = 3(2^8 - 1)/(2 - 1).
- = 3(256 - 1)/1.
Answer: 765.
Check: The next term after the 8th is 3 x 2^8 = 768, and 765 = 768 - 3 - the sum always ends one step behind the next term here.
Watch for this: Subtract inside the bracket first - (2^8 - 1) is 255, not 256.
Now try it yourself
Sum the first 5 terms of the GP with a = 5, r = 3.
Check your answer
605.
5(3^5 - 1)/(3 - 1) = 5 x 242 / 2 = 605.
Sum an infinite GP when it converges
Choose the method: If |r| < 1, the infinite sum is a/(1 - r).
EZT practice question
Find the sum of the infinite GP 1 + 1/2 + 1/4 + 1/8 + ...
Show the worked solution
- a = 1 and r = 1/2, and |1/2| < 1 so the sum exists.
- S = a/(1 - r) = 1/(1 - 1/2).
- = 1/(1/2).
Answer: 2.
Check: Partial sums 1, 1.5, 1.75, 1.875 creep toward 2 from below - the formula matches the creep.
Watch for this: Never apply a/(1 - r) when |r| is 1 or more - the terms do not shrink and no finite sum exists.
Now try it yourself
Sum the infinite GP with a = 9, r = 1/3.
Check your answer
13.5.
9/(1 - 1/3) = 9/(2/3) = 13.5.
Insert the geometric mean
Choose the method: GM of positive a and b is sqrt(ab).
EZT practice question
Find the geometric mean of 4 and 9.
Show the worked solution
- GM = sqrt(4 x 9).
- = sqrt(36).
Answer: 6.
Check: 4, 6, 9 form a GP with ratio 1.5: 4 x 1.5 = 6 and 6 x 1.5 = 9 - the GM really is the middle term.
Watch for this: sqrt(ab) is not (a + b)/2 - the arithmetic mean of 4 and 9 is 6.5, a different number.
Now try it yourself
Geometric mean of 2 and 8?
Check your answer
4.
sqrt(16) = 4, and 2, 4, 8 has ratio 2.
Compare AM and GM
Choose the method: AM = (a + b)/2, GM = sqrt(ab); AM >= GM always.
EZT practice question
For 4 and 9, verify that the arithmetic mean beats the geometric mean.
Show the worked solution
- AM = (4 + 9)/2 = 6.5.
- GM = sqrt(36) = 6.
- 6.5 > 6.
Answer: AM 6.5 exceeds GM 6, exactly as AM >= GM predicts.
Check: Try 5 and 5: AM = 5 and GM = 5 - equality happens only when the two numbers are equal.
Watch for this: AM >= GM needs positive numbers - keep both a and b positive before comparing.
Now try it yourself
AM and GM of 2 and 8 - which is larger?
Check your answer
AM = 5 beats GM = 4.
(2 + 8)/2 = 5 and sqrt(16) = 4.
Recover the ratio from two terms
Choose the method: Divide a later term by an earlier one and match the exponent gap.
EZT practice question
In a GP, the 2nd term is 12 and the 4th term is 108. Find r (taking r positive).
Show the worked solution
- a r = 12 and a r^3 = 108.
- Divide: r^2 = 108 / 12 = 9.
- r = 3 (positive root).
Answer: r = 3.
Check: Then a = 4: the GP runs 4, 12, 36, 108 - the 4th term really is 108.
Watch for this: r^2 = 9 allows r = -3 too - say which root the question wants before dropping one.
Now try it yourself
3rd term 20, 5th term 80 - positive r?
Check your answer
r = 2.
r^2 = 80 / 20 = 4, so r = 2.
Try a mixed set
Choose a method or explanation before opening each answer.
1. GP nth term?
Show answer and reason
a r^(n-1).
First term has no multiplication yet.
2. GP sum to n terms?
Show answer and reason
a(r^n - 1)/(r - 1), r not 1.
Derived by subtracting rS_n from S_n.
3. Infinite GP sum?
Show answer and reason
a/(1 - r), only when |r| < 1.
Terms must shrink to zero.
4. AM vs GM?
Show answer and reason
AM >= GM, equal only if a = b.
(sqrt a - sqrt b)^2 >= 0.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- For a = 5, r = 3: find the 4th term and the sum of 5 terms.
- Sum the infinite GP with a = 9, r = 1/3.
- Find the GM of 2 and 8 and compare it with their AM.
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Questions about this practice page
- What makes a sequence geometric?
- Each term is the previous term times one fixed number - the common ratio r. Divide any term by the one before it; if every quotient is the same r, the sequence is geometric.
- When does an infinite GP have a finite sum?
- Only when |r| < 1, because then the terms shrink toward zero and the partial sums settle at a/(1 - r). With |r| equal to or above 1 the terms stay big and the series grows without bound.
- What does AM >= GM actually say?
- For any two positive numbers, the arithmetic mean (a + b)/2 is never smaller than the geometric mean sqrt(ab) - and the two are equal only when a = b. It follows from (sqrt a - sqrt b)^2 being zero or more.
- Does this chapter cover arithmetic progressions?
- In the current Reprint 2026-27 this chapter centres on geometric progressions - the nth term, the sum to n terms, the geometric mean and the AM-GM relationship. AP ideas appear only as background inside the sequences and series sections.
- Are these the NCERT exercise questions for Sequences and Series (Class 11)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.