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Class 11 Mathematics Case Study Questions

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The rationale behind teaching Mathematics

The general rationale to teach Mathematics at the senior secondary level is to assist students:

In knowledge acquisition and cognitive understanding of basic ideas, words, principles, symbols, and mastery of underlying processes and abilities, notably through motivation and visualization.
To experience the flow of arguments while demonstrating a point or addressing an issue.
To use the information and skills gained to address issues using several methods wherever possible.
To cultivate a good mentality in order to think, evaluate, and explain coherently.
To spark interest in the subject by taking part in relevant tournaments.
To familiarise pupils with many areas of mathematics utilized in daily life.
To pique students’ interest in studying mathematics as a discipline.

Case studies in Class 11 Mathematics

A case study in mathematics is a comprehensive examination of a specific mathematical topic or scenario. Case studies are frequently used to investigate the link between theory and practise, as well as the connections between different fields of mathematics. A case study will frequently focus on a specific topic or circumstance and will investigate it using a range of methodologies. These approaches may incorporate algebraic, geometric, and/or statistical analysis.

Sample Class 11 Mathematics case study questions

When it comes to preparing for Class 11 Mathematics, one of the best things Class 11 Mathematics students can do is to look at some Class 11 Mathematics sample case study questions. Class 11 Mathematics sample case study questions will give you a good idea of the types of Class 11 Mathematics sample case study questions that will be asked in the exam and help you to prepare more effectively.

Looking at sample questions is also a good way to identify any areas of weakness in your knowledge. If you find that you struggle with a particular topic, you can then focus your revision on that area.

myCBSEguide offers ample Class 11 Mathematics case study questions, so there is no excuse. With a little bit of preparation, Class 11 Mathematics students can boost their chances of getting the grade they deserve.

Some samples of Class 11 Mathematics case study questions are as follows:

Class 11 Mathematics case study question 1

9 km and 13 km
9.8 km and 13.8 km
9.5 km and 13.5 km
10 km and 14 km
x ≤ −1913
x < −1613
−1613 < x < −1913
There are no solution.
y ≤ 12 x+2
y > 12 x+2
y ≥ 12 x+2
y < 12 x+2

Answer Key:

(b) 9.8 km and 13.8 km
(a) −1913 ≤ x
(b) y > 12 x+2
(d) (-5, 5)

Class 11 Mathematics case study question 2

2 C 1 × 13 C 10
2 C 1 × 10 C 13
1 C 2 × 13 C 10
2 C 10 × 13 C 10
6 C 2 × 3 C 4 × 11 C 5
6 C 2 × 3 C 4 × 11 C 5
6 C 2 × 3 C 5 × 11 C 4
6 C 2 × 3 C 1 × 11 C 5
(b) (13) 4 ways
(c) 2860 ways.

Class 11 Mathematics case study question 3

Read the Case study given below and attempt any 4 sub parts: Father of Ashok is a builder, He planned a 12 story building in Gurgaon sector 5. For this, he bought a plot of 500 square yards at the rate of Rs 1000 /yard². The builder planned ground floor of 5 m height, first floor of 4.75 m and so on each floor is 0.25 m less than its previous floor.

Class 11 Mathematics case study question 4

Read the Case study given below and attempt any 4 sub parts: villages of Shanu and Arun’s are 50km apart and are situated on Delhi Agra highway as shown in the following picture. Another highway YY’ crosses Agra Delhi highway at O(0,0). A small local road PQ crosses both the highways at pints A and B such that OA=10 km and OB =12 km. Also, the villages of Barun and Jeetu are on the smaller high way YY’. Barun’s village B is 12km from O and that of Jeetu is 15 km from O.

Now answer the following questions:

5x + 6y = 60
6x + 5y = 60
(a) (10, 0)
(b) 6x + 5y = 60
(b) 60/√ 61 km
(d) 2√61 km

A peek at the Class 11 Mathematics curriculum

The Mathematics Syllabus has evolved over time in response to the subject’s expansion and developing societal requirements. The Senior Secondary stage serves as a springboard for students to pursue higher academic education in Mathematics or professional subjects such as Engineering, Physical and Biological Science, Commerce, or Computer Applications. The current updated curriculum has been prepared in compliance with the National Curriculum Framework 2005 and the instructions provided by the Focus Group on Teaching Mathematics 2005 in order to satisfy the rising demands of all student groups. Greater focus has been placed on the application of various principles by motivating the themes from real-life events and other subject areas.

Class 11 Mathematics (Code No. 041)


I.	Sets and Functions	60	23
II.	Algebra	50	25
III.	Coordinate Geometry	50	12
IV.	Calculus	40	08
V.	Statistics and Probability	40	12

	Internal Assessment

Design of Class 11 Mathematics exam paper

CBSE Class 11 mathematics question paper is designed to assess students’ understanding of the subject’s essential concepts. Class 11 mathematics question paper will assess their problem-solving and analytical abilities. Before beginning their test preparations, students in Class 11 maths should properly review the question paper format. This will assist Class 11 mathematics students in better understanding the paper and achieving optimum scores. Refer to the Class 11 Mathematics question paper design provided.

Class 11 Mathematics Question Paper Design


1.	Exhibit memory of previously learned material by recalling facts, terms, basic concepts, and answers.	44	55
1.	Demonstrate understanding of facts and ideas by organizing, comparing, translating, interpreting, giving descriptions, and stating main ideas	44	55
2	Solve problems to new situations by applying acquired knowledge, facts, techniques and rules in a different way.	20	25
3	Examine and break information into parts by identifying motives or causes. Make inferences and find evidence to support generalizations	16	20
3	Present and defend opinions by making judgments about information, validity of ideas, or quality of work based on a set of criteria.	16	20

No chapter-wise weightage. Care to be taken to cover all the chapters.
Suitable internal variations may be made for generating various templates keeping the overall weightage to different forms of questions and typology of questions the same.

Choice(s): There will be no overall choice in the question paper. However, 33% of internal choices will be given in all the sections.


Periodic Tests (Best 2 out of 3 tests conducted)	10 Marks
Mathematics Activities	10 Marks

Prescribed Books:

Mathematics Textbook for Class XI, NCERT Publications
Mathematics Exemplar Problem for Class XI, Published by NCERT
Mathematics Lab Manual class XI, published by NCERT

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Chapter 8 Class 11 Sequences and Series

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Updated for new NCERT - 2023-2024 Edition.

Solutions of Chapter 8 Sequences and Series of Class 11 NCERT book available free. All exercise questions, examples, miscellaneous are done step by step with detailed explanation for your understanding.

In this Chapter we learn about Sequences

Sequence is any group of numbers with some pattern.

Like 2, 4, 8, 16, 32, 64, 128, 256, ....

1, 2, 3, 4, 5, 6, 7, 8

In this chapter we learn

What a sequence is - and what is finite, infinite sequence, terms of a sequence
What a series is - it is the sum of a sequence
Denoting Sum by sigma Σ , and what it means
What is an AP (Arithmetic Progression) is, and finding its n th term and sum
Inserting AP between two numbers
What is Arithmetic Mean ( AM ), and how to find it
What is a GP (Geometric Progression) is, and finding its n th term and sum
Inserting GP between two numbers
What is Geometric Mean ( GM ), and how to find it
Relationship between AM and GM
Sum of special series
Finding sum of series when n th term is given
Finding sum of series when n th term is not given

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CBSE Class 11 Maths – Chapter 9 Sequences and Series- Study Materials

NCERT Solutions Class 11 All Subjects Sample Papers Past Years Papers

Sequence and Series : Notes and Study Materials -pdf

Concepts of Sequence and Series
Sequence and Series Master File
Sequence and Series Revision Notes
R D Sharma Solution of AP
R D Sharma Solution of GP
R D Sharma Solution of Special Series
NCERT Solution Sequence and Series
NCERT Exemplar Solution Sequence and Series
Sequence and Series : Solved Example 1

It means an arrangement of number in definite order according to some rule

Important point

The various numbers occurring in a sequence are called its terms. The number of terms is called the length of the series
Terms of a sequence are denoted generally by $a_1 , a_2, a_3, ….., a_n$ etc., The subscripts denote the position of the term. We can choose any other letter to denote it
The nth term is the number at the nth position of the sequence and is denoted by a n
The nth term is also called the general term of the sequence
A sequence can be regarded as a function whose domain is the set of natural numbers or some subset of it of the type {1, 2, 3…k}.
Many times is possible to express general term in terms of algebraic formula. But it may not be true in other cases. But we should be able to generate the terms of the sequence using some rules or theoretical scheme

Finite sequence A sequence containing a finite number of terms is called Finite sequence infinite sequence A sequence is called infinite if it is not a finite sequence.

Let a 1 , a 2 , a 3 ..be the sequence, then the sum expressed a 1 + a 2 +a 3 + ……. is called series.

A series is called finite series if it has got finite number of terms
A series is called infinite series if it has got infinite terms
Series are often represented in compact form, called sigma notation, using the Greek letter σ (sigma)
so , a 1 + a 2 +a 3 + …….a n can be expressed as $\sum_{k=1}^{n}a_k$

Example $1+2+3 + 4+ 5+….+100$ $= \sum_{k=1}^{n}k$

Types of Sequences

Arithmetic progression.

An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant Examples 1. $1,5,9,13,17….$ 2. $1,2,3,4,5,…..$ Important Notes about Arithmetic Progression 1.The difference between any successive members is a constant and it is called the common difference of AP 2. If a 1 , a 2 ,a 3 ,a 4 ,a 5 are the terms in AP then D=a 2 -a 1 =a 3 – a 2 =a 4 – a 3 =a 5 –a 4 3. We can represent the general form of AP in the form a,a+d,a+2d,a+3d,a+4d.. Where a is first term and d is the common difference 4. Nth term of Arithmetic Progression is given by n th term = a + (n – 1)d 5. Sum of nth item in Arithmetic Progression is given by $S_n =\frac {n}{2}[a + (n-1)d]$ Or $S_n =\frac {n}{2}[t_1+ t_n]$ Arithmetic Mean The arithmetic mean A of any two numbers a and b is given by (a+b)/2 i.e. a, A, b are in AP A -a = b- A or $A = \frac {(a+b)}{2}$ If we want to add n terms between A and B so that result sequence is in AP Let A 1 , A 2 , A 3 , A 4 , A 5 …. A n be the terms added between a and b Then b= a+ [(n+2) -1]d or $d = \frac {(b-a)}{(n+1)}$ So terms will be $a+ \frac {(b-a)}{(n+1)}, a + 2 \frac {(b-a)}{(n+1)} ,……. a+ n \frac {(b-a)}{(n+1)}$

Geometric Progressions

A sequence is said to be a geometric progression or G.P., if the ratio of any term to its preceding term is same throughout. Examples 1. $2,4,8,16,32….$ 2. $3,6,12,24,48,…..$ Important Notes about Geometric Progressions

The ratio of any term to its preceding term is same throughout. This constant factor is called the common ratio.
If a 1 , a 2 ,a 3 ,a 4 ,a 5 are the terms in AP then $r=\frac {a_2}{a_1} =\frac {a_3}{a_2} =\frac {a_4}{a_3}=\frac {a_5}{a_4}$
We can represent the general form of G.P in the form a,ar,ar 2 ,ar 3 …… Where a is first term and r is the common ratio
Nth term of Geometric Progression is given by n th term = ar n-1
Sum of nth item inGeometric Progression is given by $S_n =a \frac {r^n -1}{r-1}$ Or $S_n =a \frac {1-r^n }{ 1-r}$

Geometric Means The Geometric mean G of any two numbers a and b is given by (ab) 1/2 i.e. a, G, b are in G.P $\frac {G}{a} = \frac {b}{G}$ or $G=\sqrt {ab}$ If we want to add n terms between A and B so that result sequence is in GP Let G 1 , G 2 , G 3 , G 4 , G 5 …. G n be the terms added between a and b Then $b= ar^{n+1}$ or $r= (\frac {b}{a})^{\frac {1}{n+1}}$ So terms will be $a(\frac {b}{a})^{\frac {1}{n+1}}, a(\frac {b}{a})^{\frac {2}{n+1}} ,……. a(\frac {b}{a})^{\frac {n}{n+1}}$ Relationship Between A.P and G.P Let A and G be A.M. and G.M. of two given positive real numbers a and b, respectively. Then A ≥ G And $A – G = \frac {{\sqrt {a} – \sqrt {b}}^2}{2}$

Sum of Special Series

Sum of first n natural numbers $1 + 2 + 3 +….. + n = \frac {n(n+1}{2}$ $ \sum_{k=1}^{n} n =\frac {n(n+1}{2}$ sum of squares of the first n natural numbers $1^2+ 2^2+3^2 +…. + n^2 = \frac {n(n+1)(2n+1)}{6}$ $ \sum_{k=1}^{n} n^2= \frac {n(n+1)(2n+1)}{6}$ Sum of cubes of the first n natural numbers $1^3+ 2^3+3^3 +….. + n^3 = \frac {(n(n+1))^2}{4}$ $ \sum_{k=1}^{n} n^3= \frac {(n(n+1))^2}{4}$

Rules for finding the sum of Series

a. Write the nth term T n of the series b. Write the T n in the polynomial form of n $T_n= a n^3 + bn^2 + cn +d$ c. The sum of series can be written as $ \sum S = a \sum {n^3} + b \sum {n^2} + c \sum n + nd We already know the values if these standard from the formula given above and we can easily find the sum of the series Example Find the sum of the series $2^2 + 4^2 + 6^2 + …..(2n)^2} Solution Let nth term T n of the series Then $T_n = (2n)^2 = 4n^2$ Now $2^2 + 4^2 + 6^2 + …..(2n)^2 = \sum_{k=1}^{n} 4k^2 = 4 sum_{k=1}^{n} k^2 = 4 \frac {n(n+1)(2n+1)}{6}$ $=\frac {2n(n+1)(2n+1)}{3}$

Method of Difference

Many times , nth term of the series can be determined. For example 5 + 11 + 19 + 29 + 41…… If the series is such that difference between successive terms are either in A.P or G.P, then we can the nth term using method of difference $S_n = 5 + 11 + 19 + 29 + … + a_{n-1} + a_n$ or $S_n= 5 + 11 + 19 + … + + a_{n-1} + a_n$ On subtraction, we get $0 = 5 + [6 + 8 + 10 + 12 + …(n – 1) terms] – a_n$ Here 6,8,10 is in A.P,So $a_n = 5 + \frac {n-1}{2} [12 + (n-1)2]$ or $ a_n= n^2 + 3n + 1$ Now it is easy to find the Sum of the series $S_n = \sum_{k=1}^{n} {k^2 +3k +1}$ $=\frac {n(n+2)(n+4)}{3}$

Sequences and Series Class 11 MCQs Questions with Answers

Question 1. If a, b, c are in G.P., then the equations ax² + 2bx + c = 0 and dx² + 2ex + f = 0 have a common root if d/a, e/b, f/c are in (a) AP (b) GP (c) HP (d) none of these

Answer: (a) AP Hint: Given a, b, c are in GP ⇒ b² = ac ⇒ b² – ac = 0 So, ax² + 2bx + c = 0 have equal roots. Now D = 4b² – 4ac and the root is -2b/2a = -b/a So -b/a is the common root. Now, dx² + 2ex + f = 0 ⇒ d(-b/a)² + 2e×(-b/a) + f = 0 ⇒ db2 /a² – 2be/a + f = 0 ⇒ d×ac /a² – 2be/a + f = 0 ⇒ dc/a – 2be/a + f = 0 ⇒ d/a – 2be/ac + f/c = 0 ⇒ d/a + f/c = 2be/ac ⇒ d/a + f/c = 2be/b² ⇒ d/a + f/c = 2e/b ⇒ d/a, e/b, f/c are in AP

Question 2. If a, b, c are in AP then (a) b = a + c (b) 2b = a + c (c) b² = a + c (d) 2b² = a + c

Answer: (b) 2b = a + c Hint: Given, a, b, c are in AP ⇒ b – a = c – b ⇒ b + b = a + c ⇒ 2b = a + c

Question 3: Three numbers form an increasing GP. If the middle term is doubled, then the new numbers are in Ap. The common ratio of GP is (a) 2 + √3 (b) 2 – √3 (c) 2 ± √3 (d) None of these

Answer: (a) 2 + √3 Hint: Let the three numbers be a/r, a, ar Since the numbers form an increasing GP, So r > 1 Now, it is given that a/r, 2a, ar are in AP ⇒ 4a = a/r + ar ⇒ r² – 4r + 1 = 0 ⇒ r = 2 ± √3 ⇒ r = 2 + √3 {Since r > 1}

Question 4: The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is (a) n/(n+1) (b) 1/(n+1) (c) 1/n (d) None of these

Answer: (a) n/(n+1) Hint: Given series is: S = (1/1·2) + (1/2·3) + (1/3·4) – ………………. 1/n.(n+1) ⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -……… (1/n – 1/(n+1)) ⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1) ⇒ S = 1 – 1/(n+1) ⇒ S = (n + 1 – 1)/(n+1) ⇒ S = n/(n+1)

Question 5: If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then (a) a, b, c are in AP (b) a², b², c² are in AP (c) 1/1, 1/b, 1/c are in AP (d) None of these

Answer: (b) a², b², c² are in AP Hint: Given, 1/(b + c), 1/(c + a), 1/(a + b) ⇒ 2/(c + a) = 1/(b + c) + 1/(a + b) ⇒ 2b² = a² + c² ⇒ a², b², c² are in AP

Question 6: The sum of series 1/2! + 1/4! + 1/6! + ….. is (a) e² – 1 / 2 (b) (e – 1)² /2 e (c) e² – 1 / 2 e (d) e² – 2 / e

Answer: (b) (e – 1)² /2 e Hint: We know that, e x = 1 + x/1! + x² /2! + x³ /3! + x 4 /4! + ……….. Now, e 1 = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ……….. e -1 = 1 – 1/1! + 1/2! – 1/3! + 1/4! + ……….. e 1 + e -1 = 2(1 + 1/2! + 1/4! + ………..) ⇒ e + 1/e = 2(1 + 1/2! + 1/4! + ………..) ⇒ (e² + 1)/e = 2(1 + 1/2! + 1/4! + ………..) ⇒ (e² + 1)/2e = 1 + 1/2! + 1/4! + ……….. ⇒ (e² + 1)/2e – 1 = 1/2! + 1/4! + ……….. ⇒ (e² + 1 – 2e)/2e = 1/2! + 1/4! + ……….. ⇒ (e – 1)² /2e = 1/2! + 1/4! + ………..

Question 7: The third term of a geometric progression is 4. The product of the first five terms is (a) 4 3 (b) 4 5 (c) 4 4 (d) none of these

Answer: (b) 4 5 Hint: here it is given that T 3 = 4. ⇒ ar² = 4 Now product of first five terms = a.ar.ar².ar³.ar 4 = a 5 r 10 = (ar 2 ) 5 = 4 5

Question 8: Let Tr be the r th term of an A.P., for r = 1, 2, 3, … If for some positive integers m, n, we have Tm = 1/n and Tn = 1/m, then Tm n equals (a) 1/m n (b) 1/m + 1/n (c) 1 (d) 0

Answer: (c) 1 Hint: Let first term is a and the common difference is d of the AP Now, T m = 1/n ⇒ a + (m-1)d = 1/n ………… 1 and T n = 1/m ⇒ a + (n-1)d = 1/m ………. 2 From equation 2 – 1, we get (m-1)d – (n-1)d = 1/n – 1/m ⇒ (m-n)d = (m-n)/mn ⇒ d = 1/mn From equation 1, we get a + (m-1)/mn = 1/n ⇒ a = 1/n – (m-1)/mn ⇒ a = {m – (m-1)}/mn ⇒ a = {m – m + 1)}/mn ⇒ a = 1/mn Now, T mn = 1/mn + (mn-1)/mn ⇒ T mn = 1/mn + 1 – 1/mn ⇒ T mn = 1

Question 9. The sum of two numbers is 13/6 An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Then the number of means inserted is (a) 2 (b) 4 (c) 6 (d) 8

Answer: (c) 6 Hint: Let a and b are two numbers such that a + b = 13/6 Let A 1 , A 2 , A 3 , ………A 2n be 2n arithmetic means between a and b Then, A 1 + A 2 + A 3 + ………+ A 2n = 2n{(n + 1)/2} ⇒ n(a + b) = 13n/6 Given that A 1 + A 2 + A 3 + ………+ A 2n = 2n + 1 ⇒ 13n/6 = 2n + 1 ⇒ n = 6

Question 10. If the sum of the roots of the quadratic equation ax² + bx + c = 0 is equal to the sum of the squares of their reciprocals, then a/c, b/a, c/b are in (a) A.P. (b) G.P. (c) H.P. (d) A.G.P.

Answer: (c) H.P. Hint: Given, equation is ax² + bx + c = 0 Let p and q are the roots of this equation. Now p+q = -b/a and pq = c/a Given that p + q = 1/p² + 1/q² ⇒ p + q = (p² + q²)/(p² ×q²) ⇒ p + q = {(p + q)² – 2pq}/(pq)² ⇒ -b/a = {(-b/a)² – 2c/a}/(c/a)² ⇒ (-b/a)×(c/a)² = {b²/a² – 2c/a} ⇒ -bc²/a³ = {b² – 2ca}/a² ⇒ -bc²/a = b² – 2ca Divide by bc on both side, we get ⇒ -c /a = b/c – 2a/b ⇒ 2a/b = b/c + c/a ⇒ b/c, a/b, c/a are in AP ⇒ c/a, a/b, b/c are in AP ⇒ 1/(c/a), 1/(a/b), 1/(b/c) are in HP ⇒ a/c, b/a, c/b are in HP

Question 11. If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then (a) a, b, c are in AP (b) a², b², c² are in AP (c) 1/1, 1/b, 1/c are in AP (d) None of these

Question 12. The 35th partial sum of the arithmetic sequence with terms a n = n/2 + 1 (a) 240 (b) 280 (c) 330 (d) 350

Answer: (d) 350 Hint: The 35th partial sum of this sequence is the sum of the first thirty-five terms. The first few terms of the sequence are: a 1 = 1/2 + 1 = 3/2 a 2 = 2/2 + 1 = 2 a 3 = 3/2 + 1 = 5/2 Here common difference d = 2 – 3/2 = 1/2 Now, a 35 = a 1 + (35 – 1)d = 3/2 + 34 ×(1/2) = 17/2 Now, the sum = (35/2) × (3/2 + 37/2) = (35/2) × (40/2) = (35/2) × 20 = 35 × 10 = 350

Question 13. The sum of two numbers is 13/6 An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Then the number of means inserted is (a) 2 (b) 4 (c) 6 (d) 8

Question 14. The first term of a GP is 1. The sum of the third term and fifth term is 90. The common ratio of GP is (a) 1 (b) 2 (c) 3 (d) 4

Answer: (c) 3 Hint: Let first term of the GP is a and common ratio is r. 3rd term = ar² 5th term = ar 4 Now ⇒ ar² + ar 4 = 90 ⇒ a(r² + r 4 ) = 90 ⇒ r² + r 4 = 90 ⇒ r² ×(r² + 1) = 90 ⇒ r²(r² + 1) = 3² ×(3² + 1) ⇒ r = 3 So the common ratio is 3

Question 15. The sum of AP 2, 5, 8, …..up to 50 terms is (a) 3557 (b) 3775 (c) 3757 (d) 3575

Answer: (b) 3775 Hint: Given, AP is 2, 5, 8, …..up to 50 Now, first term a = 2 common difference d = 5 – 2 = 3 Number of terms = 50 Now, Sum = (n/2)×{2a + (n – 1)d} = (50/2)×{2×2 + (50 – 1)3} = 25×{4 + 49×3} = 25×(4 + 147) = 25 × 151 = 3775

Question 16. If 2/3, k, 5/8 are in AP then the value of k is (a) 31/24 (b) 31/48 (c) 24/31 (d) 48/31

Answer: (b) 31/48 Hint: Given, 2/3, k, 5/8 are in AP ⇒ 2k = 2/3 + 5/8 ⇒ 2k = 31/24 ⇒ k = 31/48 So, the value of k is 31/48

Question 17. The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is (a) n/(n+1) (b) 1/(n+1) (c) 1/n (d) None of these

Answer: (a) n/(n+1) Hint: Given series is: S = (1/1·2) + (1/2·3) + (1/3·4) – ……………….1/n.(n+1) ⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -………(1/n – 1/(n+1)) ⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1) ⇒ S = 1 – 1/(n+1) ⇒ S = (n + 1 – 1)/(n+1) ⇒ S = n/(n+1)

Question 18. If the third term of an A.P. is 7 and its 7 th term is 2 more than three times of its third term, then the sum of its first 20 terms is (a) 228 (b) 74 (c) 740 (d) 1090

Answer: (c) 740 Hint: Let a is the first term and d is the common difference of AP Given the third term of an A.P. is 7 and its 7th term is 2 more than three times of its third term ⇒ a + 2d = 7 ………….. 1 and 3(a + 2d) + 2 = a + 6d ⇒ 3×7 + 2 = a + 6d ⇒ 21 + 2 = a + 6d ⇒ a + 6d = 23 ………….. 2 From equation 1 – 2, we get 4d = 16 ⇒ d = 16/4 ⇒ d = 4 From equation 1, we get a + 2×4 = 7 ⇒ a + 8 = 7 ⇒ a = -1 Now, the sum of its first 20 terms = (20/2)×{2×(-1) + (20-1)×4} = 10×{-2 + 19×4)} = 10×{-2 + 76)} = 10 × 74 = 740

Question 19. If the sum of the first 2n terms of the A.P. 2, 5, 8, ….., is equal to the sum of the first n terms of the A.P. 57, 59, 61, ….., then n equals (a) 10 (b) 12 (c) 11 (d) 13

Answer: (c) 11 Hint: Given, the sum of the first 2n terms of the A.P. 2, 5, 8, …..= the sum of the first n terms of the A.P. 57, 59, 61, …. ⇒ (2n/2)×{2×2 + (2n-1)3} = (n/2)×{2×57 + (n-1)2} ⇒ n×{4 + 6n – 3} = (n/2)×{114 + 2n – 2} ⇒ 6n + 1 = {2n + 112}/2 ⇒ 6n + 1 = n + 56 ⇒ 6n – n = 56 – 1 ⇒ 5n = 55 ⇒ n = 55/5 ⇒ n = 11

Question 20. If a is the A.M. of b and c and G 1 and G 2 are two GM between them then the sum of their cubes is (a) abc (b) 2abc (c) 3abc (d) 4abc

Answer: (b) 2abc Hint: Given, a is the A.M. of b and c ⇒ a = (b + c) ⇒ 2a = b + c ………… 1 Again, given G 1 and G 1 are two GM between b and c, ⇒ b, G 1 , G 2 , c are in the GP having common ration r, then ⇒ r = (c/b) 1/(2+1) = (c/b) 1/3 Now, G 1 = br = b×(c/b) 1/3 and G 1 = br = b×(c/b) 2/3 Now, (G 1 )³ + (G 2 )3 = b³ ×(c/b) + b³ ×(c/b)² ⇒ (G 1 )³ + (G 2 )³ = b³ ×(c/b)×( 1 + c/b) ⇒ (G 1 )³ + (G 2 )³ = b³ ×(c/b)×( b + c)/b ⇒ (G 1 )³ + (G 2 )³ = b² ×c×( b + c)/b ⇒ (G 1 )³ + (G 2 )³ = b² ×c×( b + c)/b ………….. 2 From equation 1 2a = b + c ⇒ 2a/b = (b + c)/b Put value of(b + c)/b in eqaution 2, we get (G 1 )³ + (G 2 )³ = b² × c × (2a/b) ⇒ (G 1 )³ + (G 2 )³ = b × c × 2a ⇒ (G 1 )³ + (G 2 )³ = 2abc

NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series

June 20, 2022 by Sastry CBSE

NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series are part of NCERT Exemplar Class 11 Maths . Here we have given NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series.

Short Answer Type Questions:

Q1. The first term of an A.P. is a and the sum of the first p terms is zero, show that the sum of its next q terms $\frac { -a(p+q)q }{ p-1 } $

NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series Img 1

=> 6600 = 2a + 19 x 200 => 2a = 2800 ∴a = 1400

Q3. A man accepts a position with an initial salary of Rs. 5200 per month. It is understood that he will receive an automatic increase of Rs. 320 in the very next month and each month thereafter. (i) Find his salary for the tenth month. (ii) What is his total earnings during the first year? Sol: The man gets a fixed increment of Rs. 320 each month. Therefore, this forms an A.P. whose First term, a = 5200 and Common difference, d = 320

NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series Img 3

Q7. A side of an equilateral triangle is 20 cm long. A second equilateral triangle is inscribed in it by joining the mid-points of the sides of the first triangle. This process is continued for third, fourth, fifth, triangles. Find the perimeter of the sixth inscribed equilateral triangle. Sol: Let the given equilateral triangle be ∆ ABC with each side of 20 cm. By joining the mid-points of this triangle, we get another equilateral triangle of side equal to half of the length of side of ∆ABC. Continuing in this way, we get a set of equilateral triangles with side equal to half of the side of the previous triangle. Now, Perimeter of first triangle = 20 x 3 = 60 cm; Perimeter of second triangle = 10 x 3 = 30 cm; Perimeter of third triangle = 5×3 = 15 cm;

NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series Img 8

Long Answer Type Questions

NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series Img 15

Q15. 1f the sum of p terms of an AP. is q and the sum of q terms isp, then show that the sum ofp + q terms is —(p + q). Also, find the sum of first p — q terms (where, p > q). Sol: Let first term and common difference of the A.P. be a and d, respectively. Given, S p = q

NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series Img 17

Objective Type Questions:

NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series Img 20

Q21. If in an A.P., S n = qn 2 and S m = qm 2 , where S r denotes the sum of r terms of the AP, then S q equals

NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series Img 24

Q22. Let S n denote the sum of the first n terms of an A.P. If S 2n = 3S n , then S 3n : S n is equal to (a) 4 (b) 6 (c) 8 (d) 10 Sol: (b) Let first term be a and common difference be d. Then, S 2n = 3S n

NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series Img 25

Q25. If t n denotes the nth term of the series 2 + 3+ 6+11 + 18+…, then t 50 is (a) 49 2 – 1 (b) 49 2 (c) 50 2 +l (d) 49 2 +2

Fill in the Blanks

NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series Img 30

True/False Type Questions

Q30. Two sequences cannot be in both A.P. and G.P. together. Sol: False Consider the sequence 3,3,3; which is A.P. and G.P. both.

Q31. Every progression is a sequence but the converse, i.e., every sequence is also a progression need not necessarily be true. Sol: True – Consider the progression a, a + d, a + 2d, … and sequence of prime number 2, 3, 5, 7, 11,… Clearly, progression is a sequence but sequence is not progression because it does not follow a specific pattern.

NCERT Exemplar Class 11 Maths Chapter 9 Sequence and Series Img 33

NCERT Exemplar Class 11 Maths Solutions

Chapter 1 Sets
Chapter 2 Relations and Functions
Chapter 3 Trigonometric Functions
Chapter 4 Principle of Mathematical Induction
Chapter 5 Complex Numbers and Quadratic Equations
Chapter 6 Linear Inequalities
Chapter 7 Permutations and Combinations
Chapter 8 Binomial Theorem
Chapter 9 Sequence and Series
Chapter 10 Straight Lines
Chapter 11 Conic Sections
Chapter 12 Introduction to Three-Dimensional Geometry
Chapter 13 Limits and Derivatives
Chapter 14 Mathematical Reasoning
Chapter 15 Statistics
Chapter 16 Probability

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Sequences and Series Class 11 Notes CBSE Maths Chapter 9 (Free PDF Download)

Revision Notes
Chapter 9 Sequences And Series

Revision Notes for CBSE Class 11 Maths Chapter 9 (Sequences and Series) - Free PDF Download

A "sequence" is nothing but an ordered list of numbers. The numbers that are present in the ordered list are called as "elements" or "terms" of the sequence. When you add up all the terms in a sequence, you get a "series"; the addition, as well as the resulting value, is called the "sum" or "summation." For example, the sequence "1, 2, 3, 4" contains the terms "1", "2", "3", and "4"; the corresponding series is the sum "1 + 2 + 3 + 4", and the series' value is 10. The Sequence and Series Class 11 Notes is one of the important materials when it comes to understanding the basic topics and complex problems in the chapter. With the help of revision notes students can revise the syllabus in a concise manner, right from definitions of sequence, Series and Progressions to important problems from exam point of view. The first chapter includes sequences and series, as well as their key properties. Topics like increasing, decreasing, bounded, convergent, and divergent sequences are discussed at basic level, which is appropriate for a 11-grade student. A.P. and G.P. are explained in detail and important problems are addressed and solved. The famous Fibonacci type sequences are demonstrated, as well as different methods for finding formulae for the nth term of a recursive sequence and recursive formulas for other known series. You'll find new ways to find the nth term and partial sums for non-geometric and non-arithmetic sequences in Class 11 Maths Chapter 9 Notes.

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Also, check CBSE Class 11 Maths revision notes for all chapters:

CBSE Class 11 Maths Chapter-wise Notes




Chapter 9 Sequences and Series Notes

Sequences and Series Chapter-Related Important Study Materials It is a curated compilation of relevant online resources that complement and expand upon the content covered in a specific chapter. Explore these links to access additional readings, explanatory videos, practice exercises, and other valuable materials that enhance your understanding of the chapter's subject matter.

Sequences and Series Related Other Study Materials

Sequences and Series Class 11 Notes Maths - Basic Subjective Questions

Section–A (1 Mark Questions)

1. If the sum of n terms of an A.P. is given by S n =3n+2n 2 then find the common difference of the A.P.

Ans. Given, $S_n=3 n+2 n^2$

$$ \begin{aligned} & S_1=3(1)+2(1)^2=5=a_1 \\ & S_2=3(2)+2(2)^2=14=a_1+a_2 \\ & \therefore S_2-S_1=9=a_2 \\ & \therefore d=a_2-a_1=9-5=4 . \end{aligned} $$

2. If the third term of G.P. is 4, then what is the product of its first 5 terms.

Ans. Let $a$ and $r$ be the first term and common ratio of G.P., respectively.

Given that the third term is 4 .

$$\therefore a r^2=4$$

Product of first 5 terms

$$=a \cdot a r \cdot a r^2 \cdot a r^3 \cdot a r^4=a^5 r^{10}=\left(a r^2\right)^5=4^5\text {. }$$

3. The 17 th term from the end of A.P. -36,-31,-26,.....79 is ________.

Ans. Here, $a=-36$ and $d=-31-(-36)=5$

$$\begin{aligned}& l=79 \\& \therefore 17^{\text {th }} \text { term from the end }=l-(n-1) d \\& =79-(17-1)(5)=79-80=-1 .\end{aligned}$$

4. Sum of the series $3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+....$ n terms, is ________.

Ans. The formula for the summation of $n$ terms of an geometric series is $S_n=\frac{a\left(1-r^n\right)}{1-r}$, where $a$ is the first term in the series and $r$ is the rate of change between successive terms.

Here $a=3$ and $r=\frac{1}{3}$

$$S_n=\frac{3\left(1-\left(\frac{1}{3}\right)^n\right)}{1-\left(\frac{1}{3}\right)}=\frac{9}{}\left(1-\left(\frac{1}{3}\right)^n\right) \text {. }$$

5. The first two terms of the sequence defined by a 1 =3 and a n =3a n-1 + 2 for all n>1 ________.

Ans. Given: $a_1=3$ and $a_n=3 a_{n-1}+2$, for all $n>1$

When $n=2$ :

$$\begin{aligned}& a_2=3 a_{2-1}=3 a_1+2=3(3)+2 \\& =9+2=11 .\end{aligned}$$

Section-B (2 Marks Questions)

6. If 9 times the 9th term of an A.P. is equal to 13 times the 13th term, then prove that the 22nd term of the A.P. is zero.

Ans. Let the first term and common difference of given A.P. be $a$ and $d$, respectively.

It is given that $9 a_9=13 a_{13}$

$$ \begin{aligned} & \Rightarrow 9(a+8 d)=13(a+12 d) \\ & \Rightarrow 9 a+72 d=13 a+156 d \\ & \Rightarrow 4 a+84 d=0 \\ & \Rightarrow(a+21 d)=0 \\ & \Rightarrow a_{22}=0 . \end{aligned} $$

7. If a,b and c are in G.P., then find the value of $\frac{a-b}{b-c}$ .

Ans. Given that, $a, b$ and $c$ are in G.P.

$\Rightarrow b=a r$ and $c=a r^2$, where $r$ is the common ratio.

$$\Rightarrow \frac{a-b}{b-c}=\frac{a-a r}{a r-a r^2}=\frac{a(1-r)}{ar(1-r)}=\frac{1}{r}=\frac{a}{b} o r \frac{b}{c}$$

8. The sum of terms equidistant from the beginning and end in an A.P. is equal to ________.

Ans. Let $a$ be the first term and $d$ be the common difference of the A.P.

$a_r=r^{4 t}$ term from the beginning $=a+(r-1) d$

$a_r^{\prime}=r^{\text {di }}$ term from the end $=(a+(n-1) d)+(r-1)(-d)$

(as first term is $a_n=a+(n-1) d$ and common difference is ' $-d^{\prime}$ )

$$a_r+a^{\prime} r=a+(r-1) d+(a+(n-1) d)+(r-1)(-d)$$

$=2 a+(n-1) d$, which is independent of ' $r$ '

Thus, sum of the terms equidistant from the beginning and end in an A.P. is constant.

9. A man saved Rs. 66000 in 20 years. In each succeeding year after the first year, he saved Rs. 200 more than what he saved in the previous year. How much did he save in the first year?

Ans. Let us assume that the man saved Rs. $a$ in the first year.

In each succeeding year, an increment of Rs. 200 is made.

So, it forms an A.P. whose first term $=a$, common difference, $d=200$ and $n=20$

$$ \begin{aligned} & \therefore S_{20}=\frac{20}{2}[2 a+(20-1) d] \\ & \Rightarrow 66000=10[2 a+19 \times 200] \\ & \Rightarrow 6600=2 a+19 \times 200 \\ & \Rightarrow 2 a=2800 \\ & \therefore a=1400 . \end{aligned} $$

10. The sum of interior angles of a triangle is $180^{\circ}$. Show that the sum of the interior angles of polygons with 3, 4, 5, 6, ... sides form an arithmetic progression. Find the sum of the interior angles for a 21-sided polygon.

Ans. We know that the sum of interior angles of a polygon of side $n$ is $(n-2) \times 180^{\circ}$.

Let $a_n=(n-2) \times 180^{\circ}$

Since $a_n$ is linear in $n$, it is $n^{\text {,h }}$ term of some A.P.

$$a_3=(3-2) \times 180^{\circ}=180^{\circ}$$

Common difference, $d=180^{\circ}$

Sum of the interior angles for a 21-sided polygon is:

$$a_{21}=(21-2) \times 180^{\circ}=3420^{\circ} \text {. }$$

11. Find the $r^{th}$ term of an A.P., whose sum of first n terms is $2n+3n^{2}$ .

Ans. Sum of $n$ terms of A.P., $S_n=2 n+3 n^2$

$$a_n=S_n-S_{n-1}$$

$$\begin{aligned}& =\left(2 n+3 n^2\right)-\left[2(n-1)+3(n-1)^2\right] \\& =[2 n-2(n-1)]+\left[3 n^2-3(n-1)^2\right] \\& =2(n-n+1)+3(n-n+1)(n+n-1) \\& =2+3(2 n-1) \\& =6 n-1 \\& \therefore r^{\text {sn }} \text { term } a_r=6 r-1 .\end{aligned}$$

12. Show that the products of the corresponding terms of the sequences $a,ar,ar^{2},....ar^{n-1}$ and $A,AR,AR^{2},....AR^{n-1}$ form a G.P., and find the common ratio.

Ans. It has to be proved that the sequence, $a A, \operatorname{ar} A R, a r^2 A R^2, \ldots a r^{n-1} A R^{n-1}$, forms G.P

$\dfrac{\text { Second term }}{\text { First term }}=\dfrac{a r A R}{a A}=r R$

$\dfrac{\text { Third term }}{\text { Second term }}=\dfrac{a r^2 A R^2}{a r A R}=r R$

Thus, the above sequenee forms a G.P. and the common ratio is $r R$.

13. If the sum of n terms of an A.P. is $(pn+qn^{2})$, where p and q are constants, find the common difference.

Ans. It is known that,

$$S_n=\frac{n}{2}[2 a+(n-1) d]$$

According to the given condition,

$$\begin{aligned}& \frac{n}{2}[2 a+(n-1) d]=p n+q n^2 \\& \Rightarrow \frac{n}{2}[2 a+n d-d]=p n+q n^2 \\& \Rightarrow n a+n^2 \frac{d}{2}-n \cdot \frac{d}{2}=p n+qn^2\end{aligned}$$

Comparing the coefficients of $n^2$ on both sides, we obtain

$$\frac{d}{2}=q \Rightarrow d=2 q$$

Thus, the common difference of the A.P. is $2 q$.

PDF Summary - Class 11 Maths Sequences and Series Notes (Chapter 9)

1. Definition:

Any function with domain as a set of natural numbers is called sequence.

Real sequence: Sequence with range as subset of real numbers.

For example:-

If \[{a_1}\;,\;{a_2}\;,{a_3}\;,\;...............\;,\;{a_n}\]is a sequence, then \[{a_1}\; + \;{a_2}\; + \;{a_3}\; + \;...............\; + \;{a_n}\] is a series.

Progression: When terms of a sequence follow a certain pattern.

But it is not always necessary that terms of sequence follow a certain pattern.

1.1 Arithmetic Progression (AP):

An arithmetic progression is a sequence of numbers in which each successive term is a sum of its preceding term and a fixed number.

If this fixed number is positive, then it is an increasing AP and if this fixed number is negative, then it is a decreasing AP.

This fixed term is called common difference and is usually represented by ‘d’.

Let ‘a’ be the first term of an AP.

N th term of an AP: \[{t_n}\; = \;a\; + \left( {n\; - \;1} \right)d\;,\;{\text{where}}\;\;d\; = \;{a_n} - \;{a_{n - 1}}\]

Sum of first N terms of an AP: \[{S_n}\; = \;\dfrac{n}{2}\left[ {a\; + \;\left( {n\; - \;1} \right)d} \right]\; = \;\dfrac{n}{2}\left[ {a\; + \;l} \right]\;{\text{where}}\;{\text{,}}\;l\;{\text{is}}\;{\text{last}}\;{\text{term}}\;{\text{of}}\;{\text{an}}\;{\text{AP}}\].

Properties of an AP:

Increasing, Decreasing, Multiplying and dividing each term of an AP by a non-zero constant results into an AP.

3 numbers in an AP: \[a\; - \;d\;,\;a\;,\;a\; + \;d\]

4 numbers in an AP: \[a\; - \;3d\;,\;a\; - \;d\;,\;a\; + \;d\;,\;a\; + \;3d\]

5 numbers in an AP: \[a\; - \;2d\;,\;a\; - \;d\;,\;a\;,\;a\; + \;d\;,\;a\; + \;2d\]

6 numbers in an AP: \[a\; - \;5d\;,\;a\; - \;3d\;,\;a\; - \;d\;,\;a\; + \;d\;,\;a\; + \;3d\;,\;a\; + \;5d\]

An AP can have zero, positive or negative common difference.

The sum of the two terms of an AP equidistant from the beginning & end is constant and equal to the sum of first & last terms.

Any term of an AP (except the first) is equal to half the sum of terms which are equidistant from it.

\[ \Rightarrow \;\;{a_n}\; = \;\dfrac{1}{2}\left( {{a_{n\; - \;k}}\; + \;{a_{n\; + \;k}}} \right)\;\;,\;k\; < \;n\]

\[{t_r}\; = \;{S_r}\; - \;{S_{r\; - \;1}}\]

If three numbers are in AP : a, b, c are in AP \[ \Rightarrow \;2b\; = \;a\; + \;c\]

N th term of an AP is a linear expression in n: \[An\; + \;B\]where A is the common difference of an AP.

1.2 Geometric Progression (GP):

It is a sequence in which each term is obtained by multiplying the preceding term by a fixed number (which is constant) called common ratio. First term of GP is non zero.

Common ratio can be obtained by dividing a term by its consecutive preceding term.

If ‘a’ is the first term and ‘r’ is the common ratio then,

GP is \[a\;,ar\;,\;a{r^2}\;,\;a{r^3}\;,\;a{r^4}\;,\;.\;.\;.\;.\;.\;.\;.\;\]

N th term of a GP: \[{t_n}\; = \;a{r^{n - 1}}\]

Sum of first N terms of a GP: \[{s_n}\; = \;\dfrac{{a\left( {1\; - \;{r^n}} \right)}}{{\left( {1\; - \;r} \right)}}\;,\;r\; \ne \;1\]

Sum of infinite GP when \[|r|\; < \;1\;\;\& \;\;n\; \to \;\infty \]

\[|r|\; < \;1\;\; \Rightarrow \;{r^n}\; \to \;0\; \Rightarrow \;{S_\infty }\; = \;\dfrac{a}{{1\; - \;r}}\]

Properties of a GP:

Multiplying and dividing each term of a GP by a non- zero constant results into a GP.

Reciprocal of terms of GP is also GP.

3 consecutive terms in GP: \[\dfrac{a}{r}\;,\;a\;,\;ar\]

4 consecutive terms in GP: \[\dfrac{a}{{{r^2}}}\;,\;\dfrac{a}{r}\;,\;ar\;,\;a{r^2}\]

If three numbers are in GP : a, b, c are in GP \[ \Rightarrow \;{b^2}\; = \;ac\]

Each term of a GP raised to the same power also forms a G.P.

Choosing terms of GP at regular intervals also forms a GP.

The product of the terms equidistant from the beginning and the last is always same and is equal to the product of the first and the last term for a finite GP.

If \[{a_1}\;,\;{a_2}\;,{a_3}\;,\;...............\;,\;{a_n}\] forms GP with non-zero and non-negative terms then\[\log {a_1}\;,\;\log {a_2}\;\log ,{a_3}\;,\;...............\;,\;\log {a_n}\] are in GP or vice versa.

2.1 Arithmetic Mean

When three terms are in AP, the middle term is called AM between the other two.

If a, b, c are in AP, b is AM between a and c.

If n positive terms \[{a_1}\;,\;{a_2}\;,{a_3}\;,\;...............\;,\;{a_n}\] are in AP, then AM is:

\[A\; = \;\dfrac{{{a_1}\; + \;{a_2}\; + \;{a_3}\; + \;.......\; + \;{a_n}}}{n}\]

2.2 n-Arithmetic Means Between Two Numbers

If a, b are two numbers and \[a\;,\;{a_1}\;,\;{a_2}\;,{a_3}\;,\;...............\;,\;{a_n}\;,\;b\]are in an AP, then

\[{a_1}\;,\;{a_2}\;,{a_3}\;,\;...............\;,\;{a_n}\] are n AM’s between a and b.

\[{A_1}\; = \;a\; + \;d\;,\;{A_2}\; = \;a\; + \;2d\;,\;...........\;,\;{A_n}\; = \;a\; + \;nd\], where \[d\; = \;\dfrac{{b\; - \;a}}{{n\; + \;1}}\]

NOTE: Sum of n AM’s inserted between a and b is equal to n times a single AM between a and b.\[\Rightarrow \; \sum\limits_{r\; = \;1}^n {{A_r}}\; = \;nA\]

2.3 Geometric Mean

If \[a,{\text{ }}b,{\text{ }}c\]are in GP, then b is called GM between a and c.

So, \[{b^2}\; = \;ac\;\;or\;\;b\; = \;\sqrt {ac} \;;\;a > 0\;,\;b > 0\]

2.4 n-Geometric Means between two numbers

If a, b are two numbers and \[a\;,\;{G_1}\;,\;{G_2}\;,{G_3}\;,\;...............\;,\;{G_n}\;,\;b\]are in a GP, then

\[{G_1}\;,\;{G_2}\;,{G_3}\;,\;...............\;,\;{G_n}\] are n GM’s between a and b.

\[{G_1}\; = \;ar\;,\;{G_2}\; = \;a{r^2}\;,..........,\;{G_n}\; = \;a{r^{n\; - \;1}}\], where \[r\; = \;{\left( {\dfrac{b}{a}} \right)^{\dfrac{1}{{n\; + \;1}}}}\]

NOTE: Product of n GM’s inserted between a and b is equal to n th power of single GM between a and b. \[ \Rightarrow \;{\prod\limits_{r\; = \;1}^n {{G_r}\; = \;\left( G \right)} ^n}\]

2.5 Arithmetic, Geometric and Harmonic means between two given numbers

Let A, G and H be the arithmetic, geometric and harmonic mean between two integers numbers a and b.

\[ \Rightarrow \;A\; = \;\dfrac{{a\; + \;b}}{2}\;,\;G\; = \;\sqrt {ab} \;,\;H\; = \;\dfrac{{2ab}}{{a\; + \;b}}\]

The three means have following three properties:

\[A\; \geqslant \;G\; \geqslant \;H\]

\[{G^2}\; = \;AH\] which means that A, G, H forms a GP.

Equation \[{x^2}\; - \;2Ax\; + \;{G^2}\; = \;0\] have a and b as its roots.

If A, G, H are corresponding means between three given numbers a, b and c, then the equation having a, b, c as its roots is \[{x^3}\; - \;3A{x^2}\; + \;\dfrac{{3{G^2}}}{H}x\; - \;{G^3}\; = \;0\]

NOTE: Some important properties of Arithmetic & Geometric Means between two quantities:

If A and G are arithmetic and geometric mean between a and b then Quadratic equation \[{x^2}\; - \;2Ax\; + \;{G^2}\; = \;0\]has a and b as its roots.

If A and G are AM and GM between two numbers a and b, then

\[a\; = \;A\; + \;\sqrt {{A^2}\; - \;{G^2}} \], \[b\; = \;A\; - \;\sqrt {{A^2}\; - \;{G^2}} \]

3. Sigma Notations:

3.1 Theorems

(i) \[\sum\limits_{r\; = \;1}^n {({a_r}\; + \;{b_r})\; = \;\sum\limits_{r\; = \;1}^n {{a_r}\; + \;\sum\limits_{r\; = \;1}^n {{b_r}} } } \]

(ii) \[\sum\limits_{r\, = \;1}^n {ka\; = \;k\;\sum\limits_{r\, = \;1}^n {{a_r}} } \]

(iii) \[\sum\limits_{r\, = \;1}^n k \; = \;nk\]

4. Sum of n Terms of Some Special Sequences

4.1 Sum of first n natural numbers

\[\sum\limits_{k\, = \;1}^n k \; = \;1\; + \;2\; + \;3\; + \;.......\; + \;n\; = \;\dfrac{{n\left( {n\; + \;1} \right)}}{2}\]

4.2 Sum of squares of first n natural numbers

\[\sum\limits_{k\, = \;1}^n {{k^2}} \; = \;{1^2}\; + \;{2^2}\; + \;{3^2}\; + \;.......\; + \;{n^2}\; = \;\dfrac{{n\left( {n\; + \;1} \right)\left( {2n\; + \;1} \right)}}{6}\]

4.3 Sum of cubes of first n natural numbers

\[\sum\limits_{k\, = \;1}^n {{k^3}} \; = \;{1^3}\; + \;{2^3}\; + \;{3^3}\; + \;.......\; + \;{n^3}\; = \;{\left[ {\dfrac{{n\left( {n\; + \;1} \right)}}{2}} \right]^2}\; = \;{\left[ {\sum\limits_{k\, = \;1}^n k } \right]^2}\]

5. Arithmetico-Geometric series

An arithmetic-geometric progression (A.G.P.) is a progression in which each term can be represented as the product of the terms of an arithmetic progression (AP) and a geometric progression (GP).

\[AP\;:\;1\;,\;3\;,\;5\;,\;..........\] and \[GP\;;\;1\;,\;x\;,\;{x^2}\;,........\]

\[ \Rightarrow \;\;AGP\;:\;1\;,\;3x\;,\;5{x^2},........\]

5.1 Sum of n terms of an Arithmetico-Geometric Series

${{\text{S}}_n} = {\text{a}} + ({\text{a}} + {\text{d}}){\text{r}} + ({\text{a}} + 2\;{\text{d}}){{\text{r}}^2} + \ldots \ldots + $ $[a + (n - 1)d]{r^{n - 1}}$

then ${S_n} = \dfrac{a}{{1 - r}} + \dfrac{{dr\left( {1 - {r^{n - 1}}} \right)}}{{{{(1 - r)}^2}}} - \dfrac{{[a + (n - 1)d]{r^n}}}{{1 - r}},r \ne 1$

5.2 Sum to Infinity

If $|r| < 1{\text{ \& }}\;n \to \infty $, then $\mathop {\lim }\limits_{n \to \infty } = 0.\;{S_\infty } = \dfrac{a}{{1 - r}} + \dfrac{{dr}}{{{{(1 - r)}^2}}}$.

6. Harmonic Progression (HP)

A sequence, reciprocal of whose terms forms an AP is called HP.

If the sequence ${a_1},{a_2},{a_3}, \ldots \ldots \ldots \ldots \ldots ,{a_n}$ is an HP, then

$\dfrac{1}{{{a_1}}}\;,\;\dfrac{1}{{{a_2}}}\;,\;\dfrac{1}{{{a_3}}}\;,\;.......\;,\;\dfrac{1}{{{a_n}}}$ is an AP or vice versa. There is no formula for the sum of the $n$ terms of an HP. For HP with first terms is a and second term is $b$, then ${n^{{\text{th }}}}$ term is ${t_n} = \dfrac{{ab}}{{b + (n - 1)(a - b)}}$

If \[a,\;b,\;c\] are in ${\text{HP}} \Rightarrow {\text{b}} = \dfrac{{2{\text{ac}}}}{{{\text{a}} + {\text{c}}}}$ or $\dfrac{{\text{a}}}{{\text{c}}} = \dfrac{{{\text{a}} - {\text{b}}}}{{{\text{b}} - {\text{c}}}}$.

7. Harmonic Mean

If \[a,\;b,\;c\] are in HP then, \[b\] is the HM between \[a\;\& \;c\] \[ \Rightarrow \;b\; = \;\dfrac{{2ac}}{{a\; + \;c}}\].

Sequence and Series Class 11 Notes

Preparing from CBSE Sequence and Series Notes helps students to understand the important topics such as A.P, G.P, harmonic progressions, the arithmetic-geometric mean, and harmonic mean. These notes help students to get a good score in examinations. Topics are explained in very easy language which helps the students to understand and revise syllabus with almost no time in Sequences and Series Revision Notes. Students can solve any MCQs and Subjective question paper, once they are thorough with the notes. So students are advised to study Class 11 Maths Chapter 9 Notes without any confusion. Let’s look at the topics covered in these notes.

Meaning of Sequence

What is a sequence in Math?

Finite Sequence

Infinite Sequence

Types of Sequence

Arithmetic sequence, geometric sequence.

Fibonacci Sequence

Meaning of Series

Notation of Series

Finite and Infinite Series

Types of Series

Arithmetic Series

Geometric Series

Meaning of Geometric Progression (G.P.)

Meaning of Arithmetic Progression (A.P.)

Arithmetic Mean

Geometric Mean

Relation between A.M. and G.M.

Special Series

Sum to n terms of Special Series

A sequence is nothing but a group of objects that follow some particular pattern. If we have some objects which are listed in some kind order so that it has 1st term, 2nd term and so on, then it is a sequence.

What is a Sequence in Math?

In Mathematics, it is defined as a group of numbers which are in an ordered form which follows a particular pattern is called Sequence. There are Finite Sequences and Infinite Sequences. The sequence which has a finite number of terms (Limited terms) is called Finite Sequence. The sequence that has an unlimited number of terms (Infinite terms) is called Infinite Sequence.

There are 3 types of sequences:

In any sequence, if the difference between every successive term is a constant then it is defined as Arithmetic Sequence. It can be in ascending or descending order, but it has to be according to a constant number.

In any sequence, if the ratio between each successive term is constant then it is known as Geometric Sequence. It can be in ascending or descending order according to the constant ratio.

Like we discussed a few topics above, Class 12 Maths, Chapter 9 Sequences and Series is a difficult subject with many problems and concepts. Many of the definitions are thoroughly clarified. As a result, learning all of these will require some extra effort, and students will need to keep revising and practising in order to completely master the subject. Though students may not have enough time to prepare notes on their own, we at Vedantu provide well-organized CBSE Class 11 Maths Notes Chapter 9 Sequences and Series that will assist them in their examination preparation as well as increase their interest in the concepts. Refer to the free PDF of CBSE Sequence and Series Notes for the complete notes.

Tips to Prepare for Exams Using CBSE Sequence and Series Notes

You must complete the previous year's questions after you have completed the concepts and numerical. With the previous year's problems, you'll be able to see just where you're missing and how to progress accordingly.

To improve your pace and accuracy, take online mock tests on a regular basis. This activity will be particularly beneficial in JEE Mains.

Understand your strengths and weaknesses, and work to improve both.

If you notice any questions that seem to be critical when practicing, make a note of them. You must solve the question again later when revising this chapter; this will help you brush up on your concepts.

For this chapter, you should make a small formula notebook/flashcards and revise them weekly to keep them fresh in your mind.

The Sequence and Series Class 11 Notes prepared by Vedantu is helpful for students to score good marks in their board exams. These solutions are prepared based on important questions from the NCERT curriculum by the top faculty of Vedantu. The practice problems provided in CBSE Sequence and Series Notes will help students to revise the concepts and ace their exams. The solutions and concepts are prepared by experts to provide top-notch learning content to students. Experts have done a lot of research on the preparation of solutions to provide a unique and fun learning experience to students.

FAQs on Sequences and Series Class 11 Notes CBSE Maths Chapter 9 (Free PDF Download)

1. What are the 4 Types of Sequences?

The 4 types of sequences are:

Arithmetic Sequences

Geometric Sequences

Harmonic Sequences

Fibonacci Numbers

2. What is Sequence and Series?

3. How do I get the Free PDF of CBSE Class 11 Maths Notes Chapter 9 Sequences and Series?

Students can download the free PDF of CBSE Class 11 Maths Notes Chapter 9 Sequences and Series on Vedantu’s website which provides free PDF on different topics of Mathematics. The solutions and concepts are prepared by experts to provide top-notch learning content to students and is helpful for students to score good marks in their board exams.

4. Is Chapter 9 of Class 11 Maths tough?

Mathematics isn't as difficult as it seems. One can be a star in the CBSE Maths exam with excellent grades. You must solve all the questions from Chapter 9 of the Class 11 Maths NCERT textbook. You should also refer to Vedantu’s Revision Notes for Chapter 9 of Class 11 Maths. Mathematics can’t be mugged up as it depends upon your accuracy and precision. You should also solve previous years or sample papers available on Vedantu .

5. Which concepts are discussed in this chapter?

The relevance of sequences,' which play a key part in a range of human activities, is explained in chapter 9 Sequences and Series. In our daily lives, we encounter many examples of sequences, such as the human population, money placed in banks, the value of any product over some time, and so on. When a collection is arranged in such a way that its members are labeled as first, second, third, and so on, it is said to be listed in the form of a sequence.'

6. Write the first five terms of the sequences and obtain the corresponding series:

a1 = -1, an = an-1/n, n ≥ 2

Given, an = an-1/n and a1 = -1

a2 = a1/2 = -½

a3 = a2/3 = -⅙

a4 = a3/4 = -1/24

a5 = a4/5 = -1/120

Thus, the first 5 terms we obtained are -1, -½, -⅙, -1/24, -1/120.

7. Are the Revision Notes for Chapter 9 of Class 11 Maths important for the students?

The Revision Notes for Chapter 9 of Class 11 Maths are important for students as Class 11 is the nurturing and base for future competitive examinations. Good grades will result in direct admission to prestigious institutions. Most examinations are based on NCERT and going through Vedantu’s Chapter 9 of Class 11 Maths Revision Notes is important for a better score in the examination. You can easily find the Revision Notes online on Vedantu. These notes are accurate and reliable. These are also great when you want to do quick revisions before the exam.

8. From where I can download Revision Notes of Chapter 9 “Sequence and Series” of Class 11 Maths?

You can download Revision Notes of Chapter 9 “Sequence and Series” of Class 11 Maths from Vedantu’s official website (vedantu.com). On this site, there are notes for all the chapters of Class 11 Maths. Visit their website to download the notes that are available free of cost, as these will help you in revising the concepts in Chapter 9 which you’ve studied. These notes will save your time before the exams as you will have all the important points handy.

CBSE Study Materials for Class 11

NCERT Books

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series

Practicing NCERT Solutions of Class 11 Maths Chapter 9 helps students to have a thorough understanding of concepts in Sequences and Series. The 11th Class Maths NCERT Solutions Ch 9 Sequences and Series prepared by subject experts help students to attempt the difficult questions too with utmost confidence. You can get new methods of solving problems in less time that can improve your performance in the actual exam.

You can rely on the NCERT Solutions for Class 11 Maths Exercises 9.1, 9.2, 9.3, 9.4 provided as all of them are given after extensive research. Furthermore, you can download the 11th Class Maths NCERT Solutions PDF through the quick links available for free of cost. Begin your preparation right away with the handy NCERT Solutions of Class 11 Maths and crack the exam with high scores. Make the most out of the 11th Class Maths NCERT Solutions and ace up your preparation level.

Class 11 Maths NCERT Solutions Chapter 9 Sequences and Series

Class 11 Maths Chapter 9 includes various topics like Sequences, Series, Arithmetic Progression, Geometric Progression, Relationship between A.P and G.P, Sum to n terms of Special Series. NCERT Solutions for Class 11 Maths Chapter 9 will help students to improve their performance in the Chapter Sequences and Series. You can revise the complete syllabus in a smart way by covering the 11th Class Maths NCERT Solutions PDF.

Make the most out of Class 11 Maths NCERT Solutions of Ch 9 Sequences and Series Ex 9.1, Ex 9.2, Ex 9.3, Ex 9.4 , and Miscellaneous Exercise to score well in the exam.

Class	11
Book	Mathematics
Subject	Maths
Chapter Number	9
Name of the Chapter	Sequences and Series

NCERT Class 11 Maths Solutions of Chapter 9 Sequences and Series – Solved Exercises

Students of UP, MP, Uttarakhand, Bihar, Gujarat boards can refer to the Class 11 Maths NCERT Solutions of Ch 9 . Refer to the NCERT Solutions of Class 11 Maths Chapter 9 provided in Hindi & English Mediums. Brainstorm and Mindmap the Formulas, Notes, Important Questions prevailing here in order to score maximum marks. You can use them during your last-minute preparation too and score high in the exam.

Sequences and Series Class 11 Ex 9.1
Sequences and Series Class 11 Ex 9.2
Sequences and Series Class 11 Ex 9.3
Sequences and Series Class 11 Ex 9.4

11th Class Maths NCERT Solutions provided Exercise wise helps you get a good grip on all the concepts as all of them are provided in detail. Access the Maths NCERT Solutions of Class 11 for Sequences and Series PDF through the quick links and prepare anywhere and everywhere. With regular practice, you can attempt the questions regarding Sequences and Series in the exam with confidence. In turn, you can score good marks in the exam.

Final Words

Hope, the information prevailing on our page has been useful in clarifying your queries to the fullest possible. For more queries, you can always drop us a comment and we will guide you at the earliest possibility. Stay in touch with our site to avail the latest info concerning NCERT Solutions of Class 11 Maths, Books, Study Material, Previous Papers, etc.

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series PDF Download

NCERT Solutions are essential for all students who want to score well in their exams. Our subject matter experts have developed these solutions to help the students score good marks in their examinations. These solutions are well-explained and include each and every topic. Another essential benefit of NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series is that it is created as per the latest pattern of CBSE (Central Board of Secondary Education). These solutions also ensure that a student gets a clear understanding of each and every concept.

In the Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Solutions, each and every type of question is included. From easy to hard, the students will get an idea of all the types of questions which can appear in the exam. Also, apart from this, the students will also get to know about the most common questions which have a high chance of appearing in the exam.

It can be assumed that a student will score good marks in their exams if they will study the Maths Class 11 Chapter 9 Sequences and Series PDF NCERT Solution on a regular basis. Our subject matter experts at Selfstudys have invested a good amount of time in creating them so that the students do not find any difficulty studying from these solutions. As the experts have been in the education line for quite some time, they are aware of the learning potential of every student, which is why they have created these solutions in a simple way so that no student faces any difficulty studying from them. Students can also use these solutions at the time of revision.

The NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series can be easily downloaded from the website of Selfstudys i.e. Selfstudys.com.

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series PDF

The NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series PDF can be easily downloaded from the website of Selfstudys. As they are in PDF format, it becomes easier for the students to access them. They are also mobile-friendly.

How to Download the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series?

Downloading the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series is if you know the right steps. The steps to download them are as follows:

Visit the official website of Selfstudys i.e. Selfstudys.com.

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Once the website is completely loaded, you need to click on the three lines which you will find on the upper left side. Following that, you have to click on the ‘NCERT Books and Solutions’ column.

After clicking on the option of ‘NCERT Books and Solutions’, a new list will open where you need to select the option of ‘NCERT Solutions’.

Following that, you will be redirected to a page where you need to select the class and the subject for which you want to download the NCERT Solutions.
In this case, you need to select class 10 and the subject Maths to download the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series.

What are the Features of the Maths Class 11 Chapter 9 Sequences and Series PDF NCERT Solution?

There are many features of the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series which can be very beneficial for all the students. The most important features include:

Well-Elaborated: The NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series is well-elaborated to make it easier for the students to understand all the concepts with in-depth knowledge.
Includes every topic: Every important topic is covered in the Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Solutions which increases the confidence of the students as they feel that they know every topic and will eventually score good marks in their exams.
24*7 Online Availability: One more feature of the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series is that it is available 24*7. This is very beneficial for all the students as they can access these notes anytime either day or night.
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Well-organized Format: One thing which makes our Maths Class 11 Chapter 9 Sequences and Series PDF NCERT Solution different from other solutions is that they are developed in a well-structured format. These solutions are organized systematically, following the order of the chapters and topics in the textbooks.
Accuracy and reliability: Our subject matter experts have invested a good amount of time in developing these NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series. They are accurate and reliable for the preparation of exams. These notes can also be helpful for the students who do self-study.

What are the Benefits of the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series?

The NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series has several benefits to offer to the students. Some of the most important of them are

Clear Understanding of the Topic: The Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Solutions provides in-depth knowledge of each and every topic which is present in the syllabus. Each chapter is explained in detail in a step-by-step process. To make learning easier and more interesting for all the students, diagrams and examples are also given.
Authentic Solutions: The Maths Class 11 Chapter 9 Sequences and Series PDF NCERT Solution are completely authentic and can be trusted 100%. The highly qualified subject matter experts at Selfstudys have created these solutions in an authentic way which makes it simple for all the students to learn all the concepts.
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Boosts Confidence: If a student regularly goes through the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series, there is no doubt that the students will gain confidence in their maths power. This confidence can motivate them to explore more difficult topics and make a strong base for all of them.

What are the tips to study from the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series?

There are various tips that can help the students to study effectively from the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series:

Go through the chapter thoroughly: It is very important for all the students to first go through the chapter as it will help them to understand the basics. If their basics will get strong, they will automatically be able to grasp the other concepts in a better way.
Practice Problem solving step-by-step: It is advisable for all the students to solve the problems step-by-step which are present in the Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Solutions without looking at the answers. This will allow them to develop problem-solving skills and test their understanding of the concepts.
Create a study schedule: It is important for students to create a study schedule and follow it consistently to stay focused. Devote at least 45 minutes to grasp the concepts thoroughly. Give enough time to each topic covered in NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series and review them daily. Sticking to the schedule is recommended to ensure timely completion of all the topics.
Do not get distracted: All the students should stay focused to concentrate better and should not get distracted. They should stay away from their mobile phones and other electronic gadgets, for example, TV, etc. to avoid all the distractions while studying Maths Class 11 Chapter 9 Sequences and Series PDF NCERT Solution.
Practice Time Management: While doing exam preparation from the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series, it is advisable for all the students to practice time management by setting a time duration for solving problems. This will help you develop efficiency and improve your speed in solving problems accurately.

How to Master the Maths Class 11 Chapter 9 Sequences and Series PDF NCERT Solution ?

There are various ways to master NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series. Some of the most important of them are:

Study the NCERT Solutions: The students can start by studying the Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Solutions. It is advisable for them to read the solutions carefully and understand them stepwise manner used to solve each problem. They should use them regularly as it will help them to understand the concepts in a clear manner.
Do regular practice of these solutions: It is advisable for all the students to regularly practice these NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series as it helps the students to do effective preparation for their exams. The conceptual knowledge of the students is also boosted by regularly practicing these solutions.
Do group discussions with your peers: It is advisable for all the students to do group discussions with their peers as discussing the concepts, problem-solving techniques, and solutions with others can help you understand them in a better way. Teaching and explaining concepts to others can deepen your understanding of all the concepts.
Revise regularly: Regularly revise the concepts and solutions from NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series. Review your notes, summaries, and solved problems. Revisit the solutions and attempt to solve them again without referring to the provided solutions. This revision will help reinforce your understanding and improve retention.
Take note of your mistakes and work on them: It is advisable for all the students to take note of their mistakes while solving the Maths Class 11 Chapter 9 Sequences and Series PDF NCERT Solution and work on them. Find your weaknesses and identify the specific concepts or steps that caused confusion. This will help you improve your problem-solving skills.

How to Maximise Your Learning Potential While Solving the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series?

To maximize your learning potential while solving the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series, here are the tips which can help you:

Do a thorough reading of the chapter: All the students should read Maths Class 11 Chapter 9 Sequences and Series PDF NCERT Solution thoroughly to maximize their learning potential as it will help them to understand the basic concepts which will make it easier for them to grasp all the concepts.
Practice regularly: One of the most important tips which every student should follow in order to maximize their learning potential is to practice these NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series regularly. It includes practicing the questions regularly in the Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Solutions. This will help them in understanding the concepts with a deep understanding which can help them to score high marks in their exams.
Revise the concepts on a regular basis: It is very important for all the students to Revise all the concepts from the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series as it helps to recall all the information which they have studied during the preparation for the exam. You can also use different methods to revise the concepts on a regular basis, for example, making flashcards, notes, etc.

What Are the Methods To Study From NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series?

Studying for the exam requires focus and concentration. Below we will discuss various methods of how students can study for an Exam with the help of NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series:

Making a strong grip of basics: It is very important for all students to create a strong base for basics and learn them for NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series as it helps a student to get a clear understanding of the concepts. This makes it easier for the students to learn them fast.
Study the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series on a regular basis: It is advisable for all the students to study Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Solutions on a regular basis as it will help them to do effective preparation for their exams and also increases the chances of the students to score good marks in their exams.
Always read the summary of your textbook: It is advisable for all students to always read the summary which is included in the Maths Class 11 Chapter 9 Sequences and Series PDF NCERT Solution. The students should read them on a regular basis as it will help them to get a thorough knowledge of all the topics and will also help them to prepare well for the exams. This will increase their chances of scoring good marks in the exams.
Identify your strong and weak areas: It is very important for all the students to identify their strong and weak areas as it will help them to understand where they need to put more effort in order to do well in their exams.
Do timely revision: All the students should do timely revision from the NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series if they want to do extremely well in their exams as it will ensure that you have created a strong base of all the topics in their minds.

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CBSE NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Exercise 9.3 Free PDF

The NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Exercise 9.3 are provided below, in detailed and free to download PDF format. The solutions are latest , comprehensive , confidence inspiring , with easy to understand explanation . To download NCERT Class 11 Solutions PDF for Free, just click ‘ Download pdf ’.

Class 11 Maths NCERT Solutions Chapter 9 Sequences and Series

CBSE NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Exercise 9.3 PDF

Check out solutions to other chapters.

NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations
NCERT Solutions for Class 11th Maths Chapter 8 Binomial Theorem
NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections

How should I study for my upcoming exams?

First, learn to sit for at least 2 hours at a stretch

Solve every question of NCERT by hand, without looking at the solution.

Solve NCERT Exemplar (if available)

Sit through chapter wise FULLY INVIGILATED TESTS

Practice MCQ Questions (Very Important)

Practice Assertion Reason & Case Study Based Questions

Sit through FULLY INVIGILATED TESTS involving MCQs. Assertion reason & Case Study Based Questions

After Completing everything mentioned above, Sit for atleast 6 full syllabus TESTS.

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NCERT Solutions for Class 11 Maths Chapter 9 – Sequences and Series

Ncert solutions for class 11 maths chapter 9 – sequences and series pdf.

Free PDF of NCERT Solutions for Class 11 Maths Chapter 9 – Sequences and Series includes all the questions provided in NCERT Books prepared by Mathematics expert teachers as per CBSE NCERT guidelines from Mathongo.com. To download our free pdf of Chapter 9 – Sequences and Series Maths NCERT Solutions for Class 11 to help you to score more marks in your board exams and as well as competitive exams.

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Assertion Reason Questions for Class 11 Maths Chapter 9 Sequence and Series

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[PDF] Download Assertion Reason Questions for Class 11 Maths Chapter 9 Sequence and Series

Here we are providing assertion reason questions for class 11 maths. In this article, we are covering Class 11 Maths Chapter 9 Sequence and Series Assertion Reason Questions. Detailed Solutions are also provided at the end of questions so that students can check their answers after completing all the questions.

What is assertion reason questions.

Assertion Reason Questions are a type of question that can be asked in Class 11 Maths exams. These questions usually consist of two parts:

Assertion: A statement that presents a claim or proposition that needs to be verified or evaluated.
Reason: A statement that provides a logical explanation or justification for the assertion.

The student is required to evaluate both the assertion and the reason and determine whether they are true or false. Here are some examples of Assertion Reason Questions in Class 11 Maths:

Example 1: Assertion: The sum of the angles of a triangle is 180 degrees. Reason: The angles of a triangle are in a ratio of 1:2:3.

Solution: The assertion is true as it is a well-known fact in geometry that the sum of the angles of a triangle is 180 degrees. However, the reason is false as the angles of a triangle are not necessarily in the ratio of 1:2:3. Therefore, the correct option is (a) Assertion is true, but the reason is false.

Example 2: Assertion: The equation x^2 + 1 = 0 has no real solutions. Reason: The square of any real number is always non-negative.

Solution: The assertion is true as x 2 + 1 = 0 has no real solutions. The reason is also true as the square of any real number is always non-negative. Therefore, the correct option is (c) Assertion is true, and the reason is also true, and the reason is a correct explanation for the assertion.

Importance of Practicing Assertion Reason Questions for Class 11 Maths

Practicing Assertion Reason Questions in Class 11 Maths has several benefits:

Develops critical thinking skills: Assertion Reason Questions require students to analyze and evaluate statements logically. By practicing such questions, students can develop their critical thinking skills and learn to reason effectively.
Improves problem-solving ability: Assertion Reason Questions often involve complex concepts and require students to apply their knowledge to solve problems. Practicing such questions can help students improve their problem-solving ability and build a strong foundation in Maths.
Helps in exam preparation: Assertion Reason Questions are frequently asked in Class 11 Maths exams. By practicing such questions regularly, students can get familiar with the exam pattern and improve their chances of scoring well in the exam.
Enhances conceptual understanding: Assertion Reason Questions involve analyzing the relationship between concepts, which can help students understand the underlying principles better. By practicing such questions, students can improve their conceptual understanding of Maths.
Builds confidence: Regular practice of Assertion Reason Questions can help students build confidence in their ability to solve complex problems and reason effectively. This can help them perform better in exams and in their future academic and professional pursuits.

Why CBSE Students Fear Assertion Reason Questions?

CBSE students may fear assertion-reasoning questions because they require a deeper level of understanding and analytical skills compared to regular multiple-choice questions. Assertion-reasoning questions consist of two statements: an assertion and a reason. The student must determine if both statements are true and if the reason explains the assertion.

These questions require the student to not only know the facts but also understand the relationships between them. Students may fear these types of questions if they are not confident in their ability to analyze and evaluate the information provided. Additionally, these questions often carry a good weightage, which can add to the pressure of performing well on the exam.

However, with practice and a solid understanding of the concepts, students can overcome their fear of assertion-reasoning questions and perform well on their exams. It is important for students to read the questions carefully, understand the meaning of each statement, and analyze the relationship between the two statements before selecting their answer.

What are the best ways to prepare for assertion reason questions?

Preparing for assertion-reasoning questions can be challenging, but with the right approach and strategies, you can improve your skills and perform well on exams. Here are some ways to prepare for assertion-reasoning questions:

Understand the concepts: To answer assertion-reasoning questions correctly, you need to have a clear understanding of the concepts involved. Make sure you study the topic thoroughly and have a strong grasp of the fundamental principles.
Practice regularly: Practice is key to improving your skills. Look for sample assertion-reasoning questions in textbooks or online resources and practice answering them. Try to identify the relationship between the two statements and determine if both are true and if the reason explains the assertion.
Develop analytical skills: Assertion-reasoning questions require you to analyze and evaluate the information provided. Develop your analytical skills by reading and analyzing different types of texts, such as news articles or scientific research papers.
Take notes: Taking notes while studying can help you remember important information and make connections between different concepts. Write down key points, definitions, and examples to refer to later.
Seek help: If you are struggling with assertion-reasoning questions, don’t hesitate to seek help from your teacher or a tutor. They can provide you with additional resources and guidance on how to approach these types of questions.

By following these strategies and putting in the effort to practice, you can improve your skills and perform well on assertion-reasoning questions.

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* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 8.

The NCERT Solutions of the first exercise of Class 11 Chapter 9 are available here. These solutions can be downloaded in PDF format as well. Chapter 9 Sequences and Series of Class 11 Maths is categorised under the CBSE Syllabus for 2023-24. Exercise 9.1 of NCERT Solutions for Class 11 Maths Chapter 9- Sequences and Series is based on the following topics:

Introduction to Sequences and Series

The NCERT textbook contains numerous questions intended for the students to solve and practise. To obtain high marks in the Class 11 examination, solving and practising the NCERT Solutions for Class 11 Maths is a must.

NCERT Solutions for Class 11 Maths Chapter 9 – Sequences and Series Exercise 9.1

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Exercise 9.2 Solutions 18 Questions

Exercise 9.3 Solutions 32 Questions

Exercise 9.4 Solutions 10 Questions

Miscellaneous Exercise on Chapter 9 Solutions 32 Questions

Also explore – NCERT Class 11 Solutions

Access Solutions for Class 11 Maths Chapter 9 Exercise 9.1

Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

1. a n = n (n + 2)

n th term of a sequence a n = n (n + 2)

On substituting n = 1, 2, 3, 4, and 5, we get the first five terms

a 1 = 1(1 + 2) = 3

a 2 = 2(2 + 2) = 8

a 3 = 3(3 + 2) = 15

a 4 = 4(4 + 2) = 24

a 5 = 5(5 + 2) = 35

Hence, the required terms are 3, 8, 15, 24, and 35.

2. a n = n/n+1

Given n th term, a n = n/n+1

On substituting n = 1, 2, 3, 4, 5, we get

Hence, the required terms are 1/2, 2/3, 3/4, 4/5 and 5/6.

3. a n = 2 n

Given n th term, a n = 2 n

On substituting n = 1, 2, 3, 4, 5, we get

a 1 = 2 1 = 2

a 2 = 2 2 = 4

a 3 = 2 3 = 8

a 4 = 2 4 = 16

a 5 = 2 5 = 32

Hence, the required terms are 2, 4, 8, 16, and 32.

4. a n = (2n – 3)/6

Given n th term, a n = (2n – 3)/6

On substituting n = 1, 2, 3, 4, 5, we get

Hence, the required terms are -1/6, 1/6, 1/2, 5/6 and 7/6..

5. a n = (-1) n-1 5 n+1

Given n th term, a n = (-1) n-1 5 n+1

Hence, the required terms are 25, –125, 625, –3125, and 15625.

On substituting n = 1, 2, 3, 4, 5, we get first 5 terms

Hence, the required terms are 3/2, 9/2, 21/2, 21 and 75/2.

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose n th terms are:

7. a n = 4n – 3; a 17 , a 24

n th term of the sequence is a n = 4n – 3

On substituting n = 17, we get

a 17 = 4(17) – 3 = 68 – 3 = 65

Next, on substituting n = 24, we get

a 24 = 4(24) – 3 = 96 – 3 = 93

8. a n = n 2 /2 n ; a 7

n th term of the sequence is a n = n 2 /2 n

Now, on substituting n = 7, we get

a 7 = 7 2 /2 7 = 49/ 128

9. a n = (-1) n-1 n 3 ; a 9

n th term of the sequence is a n = (-1) n-1 n 3

On substituting n = 9, we get

a 9 = (-1) 9-1 (9) 3 = 1 x 729 = 729

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.1 - 6

On substituting n = 20, we get

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.1 - 7

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:

11. a 1 = 3, a n = 3a n-1 + 2 for all n > 1

Given, a n = 3a n-1 + 2 and a 1 = 3

a 2 = 3a 1 + 2 = 3(3) + 2 = 11

a 3 = 3a 2 + 2 = 3(11) + 2 = 35

a 4 = 3a 3 + 2 = 3(35) + 2 = 107

a 5 = 3a 4 + 2 = 3(107) + 2 = 323

Thus, the first 5 terms of the sequence are 3, 11, 35, 107 and 323.

Hence, the corresponding series is

3 + 11 + 35 + 107 + 323 …….

12. a 1 = -1, a n = a n-1 /n, n ≥ 2

a n = a n-1 /n and a 1 = -1

a 2 = a 1 /2 = -1/2

a 3 = a 2 /3 = -1/6

a 4 = a 3 /4 = -1/24

a 5 = a 4 /5 = -1/120

Thus, the first 5 terms of the sequence are -1, -1/2, -1/6, -1/24 and -1/120.

-1 + (-1/2) + (-1/6) + (-1/24) + (-1/120) + …….

13. a 1 = a 2 = 2, a n = a n-1 – 1, n > 2

a 1 = a 2 , a n = a n-1 – 1

a 3 = a 2 – 1 = 2 – 1 = 1

a 4 = a 3 – 1 = 1 – 1 = 0

a 5 = a 4 – 1 = 0 – 1 = -1

Thus, the first 5 terms of the sequence are 2, 2, 1, 0 and -1.

The corresponding series is

2 + 2 + 1 + 0 + (-1) + ……

14. The Fibonacci sequence is defined by

1 = a 1 = a 2 and a n = a n – 1 + a n – 2 , n > 2

Find a n+1 /a n , for n = 1, 2, 3, 4, 5

1 = a 1 = a 2

a n = a n – 1 + a n – 2 , n > 2

a 3 = a 2 + a 1 = 1 + 1 = 2

a 4 = a 3 + a 2 = 2 + 1 = 3

a 5 = a 4 + a 3 = 3 + 2 = 5

a 6 = a 5 + a 4 = 5 + 3 = 8

NCERT Solutions Class 11 Mathematics Chapter 9 ex.9.1 - 8

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Class 11 Mathematics Case Study Questions
Class 11 Mathematics case study question 2. A state cricket authority has to choose a team of 11 members, to do it so the authority asks 2 coaches of a government academy to select the team members that have experience as well as the best performers in last 15 matches. They can make up a team of 11 cricketers amongst 15 possible candidates.
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Updated for new NCERT - 2023-2024 Edition. Solutions of Chapter 8 Sequences and Series of Class 11 NCERT book available free. All exercise questions, examples, miscellaneous are done step by step with detailed explanation for your understanding. In this Chapter we learn about Sequences. Sequence is any group of numbers with some pattern.
Important Questions for Class 11 Maths Chapter 9 with Solutions
Class 11 Chapter 9 - Sequences and Series Important Questions with Solutions. Practice class 11 chapter 9 sequences and series problems provided here, which are taken from the previous year question papers. Question 1: The sums of n terms of two arithmetic progressions are in the ratio 5n+4: 9n+6. Find the ratio of their 18 th terms.
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The chapter Sequences and Series belongs to the unit Algebra under the first term Class 11 Maths CBSE Syllabus 2023-24, which adds up to 30 marks of the total 80 marks. There are 4 exercises, along with a miscellaneous exercise, in this chapter to help students understand the concepts related to Sequences and Series clearly.
NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series
NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1. Write the first five terms of the sequence whose nth term is a n = n (n + 2). Therefore, the required terms are 3, 8, 15, 24 and 35. Write the first five terms of the sequence whose nth term is a n = n n+1. a n = 22+1 = 23.
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Comparing the coefficients of n^2 on both sides, we obtain. \frac {d} {2}=q \Rightarrow d=2 q. Thus, the common difference of the A.P. is 2 q. PDF Summary - Class 11 Maths Sequences and Series Notes (Chapter 9) 1. Definition: Any function with domain as a set of natural numbers is called sequence. Real sequence: Sequence with range as subset of ...
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In Sequences and Series class 11 chapter 9 deals with the study of sequences which follow a specific pattern called progression. In this chapter, the concepts such as arithmetic progression (A.P), geometric mean, arithmetic mean, the relationship between A.M. and G.M., special series in forms of the sum to n terms of consecutive natural numbers, sum to n terms of squares and cubes of natural ...
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Access Solutions for Class 11 Maths Chapter 9 Miscellaneous Exercise. 1. Show that the sum of (m + n)th and (m - n)th terms of an A.P. is equal to twice the mth term. Solution: Let's take a and d to be the first term and the common difference of the A.P., respectively. We know that the kth term of an A. P. is given by.
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Free PDF of NCERT Solutions for Class 11 Maths Chapter 9 - Sequences and Series includes all the questions provided in NCERT Books prepared by Mathematics expert teachers as per CBSE NCERT guidelines from Mathongo.com. To download our free pdf of Chapter 9 - Sequences and Series Maths NCERT Solutions for Class 11 to help you to score more ...
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Download PDF - Chapter 9 Sequence and Series MCQs. Students should choose the appropriate option and check their answers against the solutions on our website. Take a look at the important questions for class 11 Maths as well. 1) If a, 4, b are in Arithmetic Progression; a, 2, b are in Geometric Progression; then a, 1, b are in. A.P.
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Here are some examples of Assertion Reason Questions in Class 11 Maths: Example 1: Assertion: The sum of the angles of a triangle is 180 degrees. Reason: The angles of a triangle are in a ratio of 1:2:3. Solution: The assertion is true as it is a well-known fact in geometry that the sum of the angles of a triangle is 180 degrees.
NCERT Solutions for Class 11 Maths Chapter 9- Sequences and Series
The NCERT textbook contains numerous questions intended for the students to solve and practise. To obtain high marks in the Class 11 examination, solving and practising the NCERT Solutions for Class 11 Maths is a must. NCERT Solutions for Class 11 Maths Chapter 9 - Sequences and Series Exercise 9.1. Download PDF. carouselExampleControls112

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