MathsCBSE Class 11 · Chapter 8

Sequences and Series Class 11: G.P., A.M., G.M. and Solved Questions

Q: Is arithmetic progression (A.P.) in the 2026-27 Class 11 syllabus?

The full A.P. section was removed from the Class 11 book because A.P.s are covered in Class 10. The arithmetic mean is still in the syllabus, mainly for the A.M.-G.M. relation.

Q: Are the sums $\sum k$, $\sum k^2$ and $\sum k^3$ in the exam?

No. CBSE lists these special sums as assessed only formatively for 2026-27, and the section on special series was removed from the NCERT book.

Q: Is the infinite G.P. in the syllabus if the NCERT chapter has no section on it?

Yes. The CBSE syllabus names 'infinite G.P. and its sum', so learn S_\infty = \frac{a}{1 - r} for |r| < 1 and its uses with recurring decimals and repeated processes.

Q: Why must the numbers be positive for the G.M.?

\sqrt{ab} is real only if ab is non-negative, and the A.M.-G.M. proof uses \sqrt{a} and \sqrt{b}, which need a, b \ge 0.

Q: How is the Class 11 Maths exam set?

It is a school annual exam of 80 marks (3 hours) plus 20 marks of internal assessment; Unit II, Algebra, which includes this chapter, carries 25 marks. CBSE gives no Class 11 sample paper, so schools usually mix MCQs, assertion-reason, short and long answers and case-based questions.

Updated 14 October 2026 · Fact-checked

Sequences and Series is Chapter 8 of CBSE Class 11 Maths. For 2026-27 it covers sequences and series, the arithmetic mean, geometric progressions with the general term and sum of n terms, the infinite G.P. and its sum, the geometric mean, and the relation A.M. ≥ G.M. between two positive numbers.

What Sequences and Series is about

A sequence is an ordered list of numbers following a rule: 3, 6, 12, 24, … Each number is a term, written $a_{1}, a_{2}, a_{3}, \dots$ ; the n-th term $a_{n}$ is often given by a formula or by a rule linking each term to earlier ones. Adding the terms gives a series: $3 + 6 + 12 + 24 + \dots$

The main sequence in this chapter is the geometric progression (G.P.), where each term is the previous one multiplied by a fixed common ratio r: $a, a r, a r^{2}, \dots$ Its n-th term is $a_{n} = a r^{n - 1}$ and the sum of its first n terms is $S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$ for $r \neq = 1$ . When $∣ r ∣ < 1$ the terms shrink so fast that the infinite sum settles to $\frac{a}{1 - r}$ .

For two positive numbers a and b, the arithmetic mean is $A = \frac{a + b}{2}$ and the geometric mean is $G = ab$ . A neat one-line proof shows $A \geq G$ always, with equality only when a = b. G.P.s model compound interest, population growth, bouncing balls and chain messages. Arithmetic progressions were studied in Class 10 and are not repeated here; the special sums $\sum k$ , $\sum k^{2}$ , $\sum k^{3}$ are assessed only formatively in 2026-27.

Weightage: This chapter is part of Unit II, Algebra, which the syllabus gives 25 of the 80 theory marks in the school annual exam (CBSE adds that there is no fixed chapter-wise weightage).Source: CBSE Mathematics (041) syllabus 2026-27, Class XI course structure, Unit II Algebra

Key concepts

Sequences

NCERT section 8.2

A sequence is a function whose domain is the natural numbers (or the first n natural numbers): it assigns a number $a_{n}$ to each position n.

Finite sequence: a fixed number of terms (e.g. the marks of 40 students). Infinite sequence: goes on for ever (e.g. 1, 2, 4, 8, …).
General term $a_{n}$ : a formula giving any term from its position, e.g. $a_{n} = n^{2} - 1$ .
Recursive rule: each term from earlier ones, e.g. $a_{1} = 2$ , $a_{n} = 2 a_{n - 1} - 1$ . The Fibonacci sequence $1, 1, 2, 3, 5, 8, \dots$ uses $a_{n} = a_{n - 1} + a_{n - 2}$ .

Not every sequence has a simple formula: the primes 2, 3, 5, 7, 11, … are a sequence, but no simple formula gives the n-th prime.

Worked example

Write the first five terms of (a) $a_{n} = \frac{2 n - 1}{n + 1}$ (b) $a_{1} = 2$ , $a_{n} = 2 a_{n - 1} - 1$ for $n > 1$ , and the series for (b). [2 marks]

Try it first, then show the solutionSolution

(a) $\frac{1}{2}, 1, \frac{5}{4}, \frac{7}{5}, \frac{3}{2}$
(b) 2, 3, 5, 9, 17; series $2 + 3 + 5 + 9 + 17 + \dots$

Answer: As above.

Series and sigma notation

NCERT section 8.3

If $a_{1}, a_{2}, a_{3}, \dots$ is a sequence, the expression $a_{1} + a_{2} + a_{3} + \dots$ is the associated series. It is finite or infinite as the sequence is.

Sigma notation: $\sum_{k = 1}^{n} a_{k} = a_{1} + a_{2} + \dots + a_{n}$ . The letter k is a dummy counter: $\sum_{k = 1}^{4} k^{2} = 1 + 4 + 9 + 16 = 30$ .

Useful rules:

$\sum (a_{k} + b_{k}) = \sum a_{k} + \sum b_{k}$
$\sum c a_{k} = c \sum a_{k}$ ; $\sum_{k = 1}^{n} c = n c$

A series means the sum, not the list: 'the sum of the series' is a single number (or an expression in n).

Worked example

Evaluate $\sum_{k = 1}^{8} (3 + 2^{k})$ . [2 marks]

Try it first, then show the solutionSolution

$\sum_{k = 1}^{8} 3 = 24$ ; $\sum_{k = 1}^{8} 2^{k} = 2 + 4 + \dots + 256 = 2 (2^{8} - 1) = 510$
Total $= 24 + 510 = 534$

Answer: 534

Arithmetic mean

NCERT section Syllabus topic (A.P. itself was done in Class 10)

The arithmetic mean (A.M.) of two numbers a and b is
$A = \frac{a + b}{2}$

It is the number that makes a, A, b an arithmetic progression (equal steps): $A - a = b - A$ .

More generally, inserting n numbers $A_{1}, \dots, A_{n}$ between a and b so that $a, A_{1}, \dots, A_{n}, b$ is an A.P. gives a common difference $d = \frac{b - a}{n + 1}$ .

The A.M. matters here because it is compared with the geometric mean in the A.M.-G.M. inequality.

Worked example

Find the A.M. of 7 and 19, and insert three numbers between 2 and 18 so that the five numbers form an A.P. [2 marks]

Try it first, then show the solutionSolution

$A = \frac{7 + 19}{2} = 13$
$d = \frac{18 - 2}{4} = 4$ : the numbers are 6, 10, 14

Answer: 13; 6, 10, 14

Geometric progression and its n-th term

NCERT section 8.4 and 8.4.1

A sequence of non-zero numbers is a G.P. if the ratio of each term to the one before it is constant:
$\frac{a _{k + 1}}{a _{k}} = r$ (the common ratio) for every k.

With first term a: $a, a r, a r^{2}, a r^{3}, \dots$ , so
$a_{n} = a r^{n - 1}$

r can be negative (signs alternate) or a fraction (terms shrink).
Three non-zero numbers a, b, c are in G.P. exactly when $b^{2} = a c$ .
To find a and r from two given terms, divide one equation by the other to remove a: if $a_{3} = 12$ and $a_{6} = 96$ , then $r^{3} = 8$ .
Handy choices of unknown terms: three terms in G.P. as $\frac{a}{r}, a, a r$ (their product is $a^{3}$ ).

Worked example

In a G.P. the 3rd term is 12 and the 6th term is 96. Find the 10th term. [2 marks]

Try it first, then show the solutionSolution

$a r^{2} = 12$ , $a r^{5} = 96$ ; dividing, $r^{3} = 8$ , so r = 2 and $a = \frac{12}{4} = 3$
$a_{10} = 3 \times 2^{9} = 1536$

Answer: 1536

Sum of n terms of a G.P.

NCERT section 8.4.2

Let $S_{n} = a + a r + a r^{2} + \dots + a r^{n - 1}$ . Multiply by r:
$r S_{n} = a r + a r^{2} + \dots + a r^{n}$

Subtract: almost every term cancels, leaving $S_{n} - r S_{n} = a - a r^{n}$ . So for $r \neq = 1$ :
$S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r} = \frac{a ( r ^{n} - 1 )}{r - 1}$

Use the first form when $∣ r ∣ < 1$ and the second when $r > 1$ , to keep the numbers positive. If r = 1, every term is a, so $S_{n} = na$ .

Type 'how many terms': set $S_{n}$ equal to the given sum and solve for $r^{n}$ , then for n.

Series like 7 + 77 + 777 + …: write each term as $\frac{7}{9} (1 0^{k} - 1)$ and sum a G.P.

Worked example

How many terms of the G.P. 2, 4, 8, … are needed to give the sum 510? [2 marks]

Try it first, then show the solutionSolution

$S_{n} = \frac{2 ( 2 ^{n} - 1 )}{2 - 1} = 510 \Rightarrow 2^{n} - 1 = 255$
$2^{n} = 256 = 2^{8}$ , so n = 8

Answer: 8 terms

Infinite G.P. and its sum

NCERT section Syllabus topic (infinite G.P.)

If $∣ r ∣ < 1$ , then $r^{n}$ gets closer and closer to 0 as n grows. In $S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}$ , the $r^{n}$ part fades away, so the sums approach a fixed value, the sum to infinity:
$S_{\infty} = \frac{a}{1 - r}, ∣ r ∣ < 1$

If $∣ r ∣ \geq 1$ the terms do not shrink and there is no finite sum.

Recurring decimals: $0. \overline{45} = 0.45 + 0.0045 + \dots = \frac{0.45}{1 - 0.01} = \frac{45}{99} = \frac{5}{11}$ .

Bouncing ball: a ball dropped from height h that rebounds to a fraction k of its height each time travels $h + 2 (k h + k^{2} h + \dots) = h + \frac{2 k h}{1 - k}$ in total.

Worked example

Find the sum to infinity of $8 + 4 + 2 + 1 + \dots$ and write $0.4 \overline{7}$ as a fraction. [2 marks]

Try it first, then show the solutionSolution

a = 8, $r = \frac{1}{2}$ : $S_{\infty} = \frac{8}{1 - \frac{1}{2}} = 16$
$0.4 \overline{7} = 0.4 + (0.07 + 0.007 + \dots) = \frac{4}{10} + \frac{0.07}{0.9} = \frac{4}{10} + \frac{7}{90} = \frac{43}{90}$

Answer: 16; $\frac{43}{90}$

Geometric mean and A.M. ≥ G.M.

NCERT section 8.4.3 and 8.5

The geometric mean (G.M.) of two positive numbers a and b is $G = ab$ ; then a, G, b is a G.P.

Inserting n G.M.s between a and b: the sequence $a, G_{1}, \dots, G_{n}, b$ has n + 2 terms, so $b = a r^{n + 1}$ , $r = (\frac{b}{a})^{\frac{1}{n + 1}}$ , and $G_{k} = a r^{k}$ .

Theorem (A.M. ≥ G.M.): for positive a, b with $A = \frac{a + b}{2}$ and $G = ab$ ,
$A - G = \frac{a + b - 2 ab}{2} = \frac{( a - b ) ^{2}}{2} \geq 0$
so $A \geq G$ , with equality exactly when a = b.

Uses: finding two numbers from their A.M. and G.M. ( $a + b = 2 A$ , $ab = G^{2}$ ); proving inequalities such as $x + \frac{1}{x} \geq 2$ for $x > 0$ ; finding least values such as the minimum of $x + \frac{9}{x}$ .

Worked example

The A.M. and G.M. of two positive numbers are 13 and 12. Find the numbers. [3 marks]

Try it first, then show the solutionSolution

$a + b = 26$ and $ab = 144$
$(a - b)^{2} = (a + b)^{2} - 4 ab = 676 - 576 = 100$ , so $a - b = \pm 10$
Solving with a + b = 26: the numbers are 18 and 8

Answer: 18 and 8

Maths in real life: Sequences and Series

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Exponential growth

Suppose a culture starts with 50 bacteria and the number doubles every hour.

How many bacteria are there after 5 hours, and after n hours?

Try it first, then show the solutionSolution

The counts at 0, 1, 2, … hours form a G.P. with a = 50, r = 2
After 5 hours (the 6th term): $50 \times 2^{5} = 1600$ ; after n hours: $50 \times 2^{n}$

Answer: 1600; $50 \times 2^{n}$

Memory hook: 'Doubles every hour' means r = 2.

Infinite G.P.

Imagine a ball dropped from a height of 10 m that always bounces back to $\frac{3}{5}$ of the height it fell from.

Find the total distance it travels before coming to rest.

Try it first, then show the solutionSolution

Down 10 m, then up and down 6 m, 3.6 m, … each time
Total $= 10 + 2 (6 + 3.6 + \dots) = 10 + 2 \times \frac{6}{1 - 0.6} = 10 + 30$

Answer: 40 m

Memory hook: Every bounce counts twice: up and down.

Sum of a G.P.

Suppose a person sends a message to 3 friends, each of whom forwards it to 3 new people, and so on, with each message costing ₹1.

Find the total cost when the 6th set of messages has been sent.

Try it first, then show the solutionSolution

Messages in each set: 3, 9, 27, …: a G.P. with a = 3, r = 3
$S_{6} = \frac{3 ( 3 ^{6} - 1 )}{3 - 1} = \frac{3 \times 728}{2} = 1092$

Answer: ₹1092

Memory hook: Total so far = sum of the G.P., not just the last term.

Topics in this chapter

Sequences and Series formula sheet

n-th term of a G.P.: $a_{n} = a r^{n - 1}$
Sum of n terms: $S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1} = \frac{a ( 1 - r ^{n} )}{1 - r} (r \neq = 1), S_{n} = na (r = 1)$
Sum to infinity: $S_{\infty} = \frac{a}{1 - r}, ∣ r ∣ < 1$
Means of a and b: $A = \frac{a + b}{2}, G = ab (a, b > 0)$
A.M.-G.M. inequality: $\frac{a + b}{2} \geq ab, equality iff a = b$
Three terms in G.P.: $a, b, c in G.P. ⟺ b^{2} = a c$
Inserting n G.M.s between a and b: $r = (\frac{b}{a})^{\frac{1}{n + 1}}, G_{k} = a r^{k}$

How the school exam asks Sequences and Series

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Terms of a sequence	Write terms from a formula or a recursive rule; find a given term; write the series (Exercise 8.1).	Substitute n = 1, 2, 3, …; for recursive rules compute in order.
G.P. terms	Find $a_{n}$ ; which term equals a value; find a and r from two terms; prove three terms are in G.P.	Use $a_{n} = a r^{n - 1}$ ; divide equations to remove a; use $b^{2} = a c$ .
Sums of G.P.s	Sum to n terms; number of terms for a given sum; sums like 8 + 88 + 888 + …; sigma sums (Exercise 8.2).	Pick the convenient form of $S_{n}$ ; rewrite repeated-digit terms using $1 0^{k} - 1$ .
Infinite G.P.	Sum to infinity; recurring decimals; bouncing balls.	Check $∣ r ∣ < 1$ , then use $\frac{a}{1 - r}$ .
Means and A.M.-G.M.	Insert G.M.s; find numbers from their A.M. and G.M.; prove or use $A \geq G$ .	Use $a + b = 2 A$ , $ab = G^{2}$ and $(a - b)^{2} = (a + b)^{2} - 4 ab$ .
Applications	Ancestors, chain letters, bacteria, compound interest, instalments.	Identify a and r from the story, decide whether a term or a sum is wanted.

Tricks to remember Sequences and Series

Short memory aids. Say each one aloud once, then use it on a question the same day.

Power is one behind
In $a_{n} = a r^{n - 1}$ , the power of r is one less than the term number: the 10th term has $r^{9}$ .
Use it when: n-th term.
Big r, big first
If r > 1 use $\frac{a ( r ^{n} - 1 )}{r - 1}$ ; if r < 1 use $\frac{a ( 1 - r ^{n} )}{1 - r}$ : both tops stay positive.
Use it when: Choosing the sum formula.
Shrinking ratio, finite total
Infinite sum exists only when $- 1 < r < 1$ ; then drop $r^{n}$ .
Use it when: Sum to infinity.
Middle squared = ends multiplied
a, b, c in G.P. ⇔ $b^{2} = a c$ .
Use it when: Proving terms are in G.P.
A beats G
Average ≥ square root of product, because $(a - b)^{2} \geq 0$ .
Use it when: A.M.-G.M. questions.

Sequences and Series: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Recognising a G.P. · 1 mark · MCQExam level

Which of the following is a G.P.?
(a) 2, 4, 6, 8
(b) 1, −2, 4, −8
(c) 1, 4, 9, 16
(d) 3, 6, 10, 15

Try it first, then show the solutionSolution

In (b) each term is −2 times the previous one.

Answer: (b)

Example 2: Sum to infinity · 1 mark · Assertion-reasonExam level

Assertion (A): $1 - \frac{1}{3} + \frac{1}{9} - \dots$ to infinity equals $\frac{3}{4}$ .
Reason (R): The sum to infinity of a G.P. with $∣ r ∣ < 1$ is $\frac{a}{1 - r}$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

a = 1, $r = - \frac{1}{3}$ : $\frac{1}{1 + \frac{1}{3}} = \frac{3}{4}$ .

Answer: (a)

Example 3: Sum of n terms with a fraction ratio · 3 marks · Short answerHarder exam level

Find the sum of the first n terms of $1 + \frac{1}{2} + \frac{1}{4} + \dots$ and the least n for which the sum exceeds 1.99. [3 marks]

Try it first, then show the solutionSolution

$S_{n} = \frac{1 - ( \frac{1}{2} ) ^{n}}{1 - \frac{1}{2}} = 2 (1 - \frac{1}{2 ^{n}})$
$2 - \frac{2}{2 ^{n}} > 1.99 \Rightarrow \frac{2}{2 ^{n}} < 0.01 \Rightarrow 2^{n} > 200$
$2^{7} = 128$ , $2^{8} = 256$ : least n = 8

Answer: $S_{n} = 2 (1 - 2^{- n})$ ; n = 8

Example 4: First and next three terms · 3 marks · Short answerHarder exam level

The sum of the first three terms of a G.P. is 7 and the sum of the next three terms is 56. Find a, r and $S_{n}$ . [3 marks]

Try it first, then show the solutionSolution

$a (1 + r + r^{2}) = 7$ and $a r^{3} (1 + r + r^{2}) = 56$
Dividing: $r^{3} = 8$ , r = 2; then $7 a = 7$ , a = 1
$S_{n} = \frac{2 ^{n} - 1}{2 - 1} = 2^{n} - 1$

Answer: a = 1, r = 2, $S_{n} = 2^{n} - 1$

Example 5: Recurring decimal · 2 marks · Very short answerHarder exam level

Express $0. \overline{45}$ as a fraction in lowest terms using an infinite G.P. [2 marks]

Try it first, then show the solutionSolution

$0.454545 \dots = 0.45 + 0.0045 + 0.000045 + \dots$ : a = 0.45, r = 0.01
$S_{\infty} = \frac{0.45}{0.99} = \frac{45}{99} = \frac{5}{11}$

Answer: $\frac{5}{11}$

Example 6: Terms in G.P. from positions · 2 marks · Very short answerHarder exam level

In a G.P., let p, q and s be the 3rd, 7th and 11th terms. Show that p, q, s are themselves in G.P. [2 marks]

Try it first, then show the solutionSolution

$p = a r^{2}$ , $q = a r^{6}$ , $s = a r^{10}$
$q^{2} = a^{2} r^{12} = (a r^{2}) (a r^{10}) = p s$ , so p, q, s are in G.P. (with ratio $r^{4}$ )

Answer: Proved.

Example 7: Least value with A.M.-G.M. · 3 marks · Short answerHOTS / competency

For $x > 0$ , find the least value of $x + \frac{9}{x}$ and the value of x where it occurs. [3 marks]

Try it first, then show the solutionSolution

Apply A.M. ≥ G.M. to the positive numbers x and $\frac{9}{x}$ : $\frac{x + \frac{9}{x}}{2} \geq x \cdot \frac{9}{x} = 3$
So $x + \frac{9}{x} \geq 6$
Equality when $x = \frac{9}{x}$ , i.e. x = 3, where the value is 6

Answer: Least value 6, at x = 3

Example 8: Sum of decimals · 5 marks · Long answerHOTS / competency

Find the sum to n terms of $0.4 + 0.44 + 0.444 + \dots$ [5 marks]

Try it first, then show the solutionSolution

$S_{n} = 4 (0.1 + 0.11 + 0.111 + \dots) = \frac{4}{9} (0.9 + 0.99 + 0.999 + \dots)$
$= \frac{4}{9} [(1 - 0.1) + (1 - 0.01) + \dots + (1 - 1 0^{- n})]$
$= \frac{4}{9} [n - (0.1 + 0.01 + \dots + 1 0^{- n})]$
G.P. sum: $0.1 + \dots + 1 0^{- n} = \frac{0.1 ( 1 - 1 0 ^{- n} )}{0.9} = \frac{1}{9} (1 - 1 0^{- n})$
$S_{n} = \frac{4}{9} [n - \frac{1}{9} (1 - 1 0^{- n})]$ ; check n = 2: $\frac{4}{9} (2 - 0.11) = 0.84 = 0.4 + 0.44$ ✓

Answer: $\frac{4}{9} [n - \frac{1}{9} (1 - \frac{1}{1 0 ^{n}})]$

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Family tree

Suppose a student draws her family tree. She has 2 parents, 4 grandparents, 8 great-grandparents, and so on, assuming no person appears twice.

(i) · 1 mark
How many ancestors are in the 6th generation back?
Answer
Answer: 64
$2 \times 2^{5} = 2^{6}$
(ii) · 1 mark
How many ancestors in total in the 8 generations before her?
Answer
Answer: 510
$S_{8} = \frac{2 ( 2 ^{8} - 1 )}{1} = 510$
(iii) · 1 mark
What kind of sequence is this, and what is r?
Answer
Answer: A G.P. with r = 2
Each generation doubles.
(iv) · 1 mark
Why can't the sum to infinity be used here?
Answer
Answer: r = 2 is not less than 1
The terms grow, so the sums grow without limit.

Comparing two averages

Suppose a shop's sales grew from ₹8 lakh in one year to ₹18 lakh two years later. An analyst wants a 'typical' middle-year value.

(i) · 1 mark
Find the A.M. of 8 and 18.
Answer
Answer: 13
$\frac{8 + 18}{2}$
(ii) · 1 mark
Find the G.M. of 8 and 18.
Answer
Answer: 12
$144$
(iii) · 1 mark
If sales grew by the same percentage each year, what was the middle-year value?
Answer
Answer: ₹12 lakh
Equal percentage growth means a G.P.: 8, 12, 18 with r = 1.5.
(iv) · 1 mark
Which is larger, and is that always true?
Answer
Answer: A.M. > G.M.; always A.M. ≥ G.M.
Equality only if the two values are equal.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Which term of the G.P. 3, 9, 27, … is 6561? [2 marks] (2 marks)

Step	Marks
a = 3, r = 3: $a_{n} = 3 \times 3^{n - 1} = 3^{n}$	1
$3^{n} = 6561 = 3^{8} \Rightarrow n = 8$ : the 8th term	1

Three numbers in G.P. have sum 21 and product 216. Find them. [3 marks] (3 marks)

Step	Marks
Take the numbers as $\frac{a}{r}, a, a r$ ; product $a^{3} = 216 \Rightarrow a = 6$	1
$\frac{6}{r} + 6 + 6 r = 21 \Rightarrow 6 r^{2} - 15 r + 6 = 0 \Rightarrow 2 r^{2} - 5 r + 2 = 0$	1
$r = 2$ or $\frac{1}{2}$ ; either way the numbers are 3, 6, 12 (or 12, 6, 3)	1

Find the sum to n terms of $7 + 77 + 777 + \dots$ and check your formula for n = 3. [5 marks] (5 marks)

Step	Marks
$S_{n} = 7 (1 + 11 + 111 + \dots) = \frac{7}{9} (9 + 99 + 999 + \dots)$	1
$= \frac{7}{9} [(10 - 1) + (1 0^{2} - 1) + \dots + (1 0^{n} - 1)] = \frac{7}{9} [(10 + 1 0^{2} + \dots + 1 0^{n}) - n]$	1
G.P. sum: $10 + \dots + 1 0^{n} = \frac{10 ( 1 0 ^{n} - 1 )}{9}$	1
$S_{n} = \frac{7}{9} [\frac{10 ( 1 0 ^{n} - 1 )}{9} - n] = \frac{7}{81} (1 0^{n + 1} - 10 - 9 n)$	1
n = 3: $\frac{7}{81} (10000 - 10 - 27) = \frac{7 \times 9963}{81} = 861 = 7 + 77 + 777$ ✓	1

Where students lose marks in Sequences and Series

The slips examiners see most often, and what to write instead.

Mistake: Using $r^{n}$ instead of $r^{n - 1}$ for the n-th term
Mixing up the term formula with the sum formula.
Wrong working
$a_{10} = 3 \times 2^{10}$
Right working
$a_{10} = 3 \times 2^{9} = 1536$
Do this: The first term has $r^{0}$ .
Usually costs: 1 mark.
Mistake: Applying $\frac{a}{1 - r}$ when $∣ r ∣ \geq 1$
Not checking the condition.
Do this: State $∣ r ∣ < 1$ before using the formula.
Usually costs: 1 mark.
Mistake: Ignoring the negative value of r
Taking only the positive root of $r^{2} = 4$ .
Do this: Check both r = 2 and r = −2 unless the question rules one out.
Usually costs: 1 mark.
Mistake: Counting the wrong number of terms when inserting means
Forgetting a and b are terms too.
Do this: Inserting n means gives n + 2 terms, so $b = a r^{n + 1}$ .
Usually costs: 1-2 marks.
Mistake: Using the term instead of the sum in word problems
Not reading 'total' carefully.
Do this: 'In the 6th generation' → term; 'up to the 6th generation' → sum.
Usually costs: 1-2 marks.
Mistake: Taking G.M. of negative numbers
Forgetting the condition.
Do this: G.M. and the A.M.-G.M. inequality are for positive numbers.
Usually costs: 1 mark.

NCERT exercise guide: Sequences and Series

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 8.1 (sequences and series)

What it tests: Terms from a general formula, particular terms, recursive sequences and their series, ratios of consecutive Fibonacci-type terms.

How to approach it: Substitute or apply the rule term by term.

Same kind as Exercise 8.1

A sequence is defined by $a_{1} = 1$ , $a_{2} = 2$ and $a_{n} = a_{n - 1} + a_{n - 2}$ for $n > 2$ . Write its first six terms and find $\frac{a _{n + 1}}{a _{n}}$ for n = 1 to 5.

Show the solutionSolution

Terms: 1, 2, 3, 5, 8, 13
Ratios: $\frac{2}{1} = 2$ , $\frac{3}{2}$ , $\frac{5}{3}$ , $\frac{8}{5}$ , $\frac{13}{8}$

Answer: 2, $\frac{3}{2}$ , $\frac{5}{3}$ , $\frac{8}{5}$ , $\frac{13}{8}$

Exercise 8.2 (G.P., G.M., A.M.-G.M.)

What it tests: n-th terms, which term, a and r from conditions, sums to n terms, number of terms, sums like 8 + 88 + …, proofs that terms are in G.P., inserting G.M.s, problems using A.M. and G.M.

How to approach it: Use $a_{n}$ , $S_{n}$ , $b^{2} = a c$ and the mean formulae.

Same kind as Exercise 8.2

Insert two numbers between 2 and 54 so that the resulting sequence is a G.P.

Show the solutionSolution

Four terms: $54 = 2 r^{3} \Rightarrow r^{3} = 27 \Rightarrow r = 3$
The numbers are $2 \times 3 = 6$ and $2 \times 9 = 18$

Answer: 6 and 18

Miscellaneous Exercise on Chapter 8

What it tests: Mixed G.P. problems: last term and number of terms, functional equations, chain letters, instalments, ratio of A.M. to G.M., sums like 0.6 + 0.66 + ….

How to approach it: Model each story as a G.P.; use the A.M.-G.M. relations.

Same kind as Miscellaneous Exercise

The ratio of the A.M. to the G.M. of two positive numbers is 5 : 4. Find the ratio of the numbers.

Show the solutionSolution

$\frac{a + b}{2 ab} = \frac{5}{4}$ . Let $t = \frac{a}{b}$ : $t + \frac{1}{t} = \frac{5}{2} \Rightarrow 2 t^{2} - 5 t + 2 = 0$
$t = 2$ or $\frac{1}{2}$ , so $\frac{a}{b} = 4$ or $\frac{1}{4}$

Answer: 4 : 1 (or 1 : 4)

Practice questions on Sequences and Series

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The 10th term of the G.P. 3, 6, 12, … is:
(a) 3072
(b) 1536
(c) 768
(d) 1024
Hint
$3 \times 2^{9}$ .
Answer
(b) 1536
Q2 · 1 mark · MCQExam level
The G.M. of 4 and 25 is:
(a) 14.5
(b) 100
(c) 10
(d) 29
Hint
$ab$ .
Answer
(c) 10
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): 2, x, 18 are in G.P. if x = 6 or x = −6.
Reason (R): a, b, c are in G.P. if $b^{2} = a c$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
$x^{2} = 36$ .
Answer
(a)
Q4 · 2 marks · Very short answerExam level
Find the sum of the first 5 terms of the G.P. 2, 6, 18, …
Hint
r = 3.
Answer
$\frac{2 ( 3 ^{5} - 1 )}{2} = 242$ .
Q5 · 2 marks · Very short answerHarder exam level
Write the first five terms of $a_{n} = (- 1)^{n - 1} 2^{n}$ and the corresponding series.
Hint
Alternating signs.
Answer
2, −4, 8, −16, 32; $2 - 4 + 8 - 16 + 32 - \dots$
Q6 · 2 marks · Very short answerHarder exam level
Find the sum to n terms of $1 + 2 b + 4 b^{2} + 8 b^{3} + \dots$ ( $b \neq = \frac{1}{2}$ ) and check it for b = 1, n = 4.
Hint
r = 2b.
Answer
$S_{n} = \frac{( 2 b ) ^{n} - 1}{2 b - 1}$ . For b = 1, n = 4: $\frac{16 - 1}{1} = 15 = 1 + 2 + 4 + 8$ ✓
Q7 · 2 marks · Very short answerHarder exam level
Find the sum of the products of the corresponding terms of 1, 3, 9, 27 and 81, 27, 9, 3.
Hint
Each product is the same.
Answer
Each product is 81, so the sum is $4 \times 81 = 324$ .
Q8 · 3 marks · Short answerHarder exam level
If a, b, c, d are in G.P., prove that $(b - c)^{2} + (c - a)^{2} + (d - b)^{2} = (a - d)^{2}$ .
Hint
Write b = ar, c = ar², d = ar³.
Answer
LHS $= a^{2} [(r - r^{2})^{2} + (r^{2} - 1)^{2} + (r^{3} - r)^{2}] = a^{2} [r^{2} - 2 r^{3} + r^{4} + r^{4} - 2 r^{2} + 1 + r^{6} - 2 r^{4} + r^{2}] = a^{2} (r^{6} - 2 r^{3} + 1) = a^{2} (1 - r^{3})^{2} = (a - d)^{2}$ .
Q9 · 3 marks · Short answerHarder exam level
A salary starts at ₹20000 a month and rises by 10% each year. Write the monthly salary in the 5th year as a term of a G.P. and find it.
Hint
a = 20000, r = 1.1.
Answer
$a_{5} = 20000 \times (1.1)^{4} = 20000 \times 1.4641 = ₹29282$ .
Q10 · 3 marks · Short answerHOTS / competency
Prove that for any two distinct positive numbers, the A.M. exceeds the G.M., and show that the difference equals $\frac{( a - b ) ^{2}}{2}$ .
Hint
Write a and b as squares.
Answer
$A - G = \frac{a + b}{2} - ab = \frac{( a ) ^{2} - 2 a b + ( b ) ^{2}}{2} = \frac{( a - b ) ^{2}}{2} > 0$ since $a \neq = b$ .
Q11 · 3 marks · Short answerHOTS / competency
The sum of an infinite G.P. is 20 and the sum of the squares of its terms is 100. Find the first term and the common ratio.
Hint
Squares form a G.P. with ratio $r^{2}$ .
Answer
$\frac{a}{1 - r} = 20$ and $\frac{a ^{2}}{1 - r ^{2}} = 100$ . Dividing the square of the first by the second: $\frac{1 + r}{1 - r} = 4 \Rightarrow r = \frac{3}{5}$ ; then $a = 20 \times \frac{2}{5} = 8$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the n-th term of a G.P.?Show answer
$a r^{n - 1}$ .
2.Sum of n terms when r = 1?Show answer
na.
3.When does an infinite G.P. have a sum?Show answer
When $∣ r ∣ < 1$ .
4.G.M. of 4 and 25?Show answer
10.
5.A.M. of 7 and 19?Show answer
13.
6.Write $0. \overline{3}$ as a fraction.Show answer
$\frac{1}{3}$ .
7.When is A.M. equal to G.M.?Show answer
When the two numbers are equal.
8.Condition for a, b, c to be in G.P.?Show answer
$b^{2} = a c$ .

Best order to study it

Learn sequences, general terms and recursive rules.
Learn series and sigma notation.
Revise the arithmetic mean.
Learn the G.P. definition and n-th term; practise finding a and r.
Derive and use the sum of n terms.
Learn the infinite G.P. sum, recurring decimals and bouncing balls.
Learn the G.M., inserting G.M.s and the A.M. ≥ G.M. proof.
Finish with mixed word problems and proofs.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you write terms from a formula or a recursive rule?
Can you evaluate a sigma sum by splitting it?
Can you find a and r from any two terms of a G.P.?
Can you find which term of a G.P. equals a given number?
Can you derive $S_{n}$ for a G.P.?
Can you find how many terms give a certain sum?
Can you sum 5 + 55 + 555 + … to n terms?
Can you find the sum to infinity and convert recurring decimals?
Can you insert G.M.s between two numbers?
Can you prove A.M. ≥ G.M. and find numbers from their means?

Sequences and Series: frequently asked questions

Is arithmetic progression (A.P.) in the 2026-27 Class 11 syllabus?

The full A.P. section was removed from the Class 11 book because A.P.s are covered in Class 10. The arithmetic mean is still in the syllabus, mainly for the A.M.-G.M. relation.

Are the sums $\sum k$, $\sum k^2$ and $\sum k^3$ in the exam?

No. CBSE lists these special sums as assessed only formatively for 2026-27, and the section on special series was removed from the NCERT book.

Is the infinite G.P. in the syllabus if the NCERT chapter has no section on it?

Yes. The CBSE syllabus names 'infinite G.P. and its sum', so learn S_\infty = \frac{a}{1 - r} for |r| < 1 and its uses with recurring decimals and repeated processes.

Why must the numbers be positive for the G.M.?

\sqrt{ab} is real only if ab is non-negative, and the A.M.-G.M. proof uses \sqrt{a} and \sqrt{b}, which need a, b \ge 0.

How is the Class 11 Maths exam set?

It is a school annual exam of 80 marks (3 hours) plus 20 marks of internal assessment; Unit II, Algebra, which includes this chapter, carries 25 marks. CBSE gives no Class 11 sample paper, so schools usually mix MCQs, assertion-reason, short and long answers and case-based questions.

Based on: CBSE Mathematics (041) syllabus 2026-27 (Classes XI-XII); NCERT Mathematics, Textbook for Class XI, Chapter 8 (Reprint 2026-27)