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Class 11 Mathematics Case Study Questions

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If you’re seeking a comprehensive and dependable study resource with Class 11 mathematics case study questions for CBSE, myCBSEguide is the place to be. It has a wide range of study notes, case study questions, previous year question papers, and practice questions to help you ace your examinations. Furthermore, it is routinely updated to bring you up to speed with the newest CBSE syllabus. So, why delay? Begin your path to success with myCBSEguide now!

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)

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

  • 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.

  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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Category: Case Study Questions for Class 11 Maths

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  • Sequences and Series

Sequence and Series Word Problems | Class 11 Maths

Sequences and series have several important applications in several spheres of human activities. When sequences follow some specific patterns, they are usually called progressions. Arithmetic and Geometric progressions are some examples of commonly occurring progressions. Let’s see some problems on these progressions to understand them better. 

Growth Patterns

Such types of number sequence problems first describe how a sequence of numbers is generated. Some terms of the sequence are given, and we need to figure out the patterns in them and then the next terms of the sequence. 

Solving such sequences: 

  • Look for a pattern between the given numbers.
  • Decide whether to use +, -, × or ÷
  • Use the pattern to solve the sequence.

Increasing Pattern

Increasing Patterns as the name suggests will always be increasing in nature. The next term and the term just before that will all be similarly related with the help of operations (×, -, +).

Question: 6,13,27,55, ….. In the given sequence, what is the value of the next term? 

On looking carefully into the pattern, one can see that 13 = 6 × 2 + 1  27 = 13 × 2 + 1  55 = 27 × 2 + 1  This shows that every term is twice the preceding term plus one. So, let the next term be “a”.  a = 55 × 2 + 1     = 110 + 1     = 111  Hence, the next term is 111. 

Decreasing Pattern

In decreasing Patterns, the next will be lesser than the previous term and two consecutive terms will follow a certain pattern.

Question: What is the next term in the series: 220,100,40, ….

Answer:  

The series follow the pattern: 100 = (220×0.5)-10                                                 40 = (100×0.5)-10 Therefore, the next term will be (40×0.5)-10 = 10

Arithmetic Progressions 

In Arithmetic Progressions, the consecutive terms will have the same difference and are denoted as ‘d’, the first term is called as ‘a’, and the number of terms is denoted as ‘n’. 

The formula for n th term is: T n = a+ (n-1)d

Question 1: 2, 5, 8, 11, …. Find the next term of the sequence. 

Answer: 

It can be noticed by carefully studying the terms of the sequence that the difference between each consecutive term remains the same. For example:  5 – 2 = 3  8 – 5 = 3 11 – 8 = 3  So, the next will be at a difference of three from the last term. Since the last term of the sequence is 11. The next terms will be 14. OR The formula for n th term in Arithmetic progression can also be used here, 5 th term is required here: a = 2 d = (3) T 5 = 2+(5-1)(3)  = 2+12 = 14

Question 2: 15, 12, 9, … __. Find the next term.

In this problem also, all the terms have a difference of three between them. The difference is that the sequence in decreasing in nature. Since the last term is 9, the next term will be 3 less than the last term. So, the last term will be 6. 

Fibonacci Sequence

Sometimes there are sequences for which pattern is not visible, the Fibonacci sequence is an example of such a sequence. It is a very commonly occurring sequence in the field of mathematics and computer science.

The number is arranged as, 1, 1, 2, 3, 5, 8 …. Here the pattern is not visible, this sequence proceeds in a manner that depends on its history. 

Let a n be the nth term for the sequence. In this sequence a n = a n-1 + a n-2 . 

For example, 

a 2 = a 1 + a 0 i.e 2 = 1 + 1,  a 3 = a 1 + a 2 i.e 3 = 2 + 1,  a 4 = a 3 + a 2 i.e 5 = 2 + 3 and,  a 5 = a 4 + a 3 i.e 8 = 5 + 3. 

Question: What is the next term in the Fibonacci series: 1,1,2,3,5,8,13,……

In Fibonacci Series, the next term is the sum of the last two terms. Therefore, the next term will be (8+13)= 21

Before starting out with the problems related to a geometric progression. Let’s take a quick recap of the formulas for the sum and nth term of a GP. 

The general form of a geometric progression is a, ar, ar 2 , ar 3 …… where a = first term, r = common ratio and a n be the nth term. 

  • The nth term of the progression: a n = ar n-1 .

S_{n} = a[\frac{r^{n}-1}{r-1}] \text{ where } r \ne 1

Finite GP problems

These kinds of problems include the geometric series where there are a finite number of terms. 

Question 1: The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of the 2nd hour, 4th hour, and nth hour? 

The growth of the bacteria makes a GP, 30, 60, 120,….. and so on.  In this GP, a = 30, r = 2. Let the number of bacteria at nth hour be a n .  At n = 2, a 2 = ar 2-1                       = (30)(2) 2-1 = 30 × 2 = 60  At n = 4, a 4 = ar 4-1                       = (30)(2) 4-1 = 30 × 2 3 = 240 At nth step an = ar (n-1) = 30 × 2 n-1

Question 2: A person has 2 parents, 4 grandparents, 8 great-grandparents, and so on. Find the number of his ancestors during the ten generations preceding his own. 

This is a problem of finite GP. The sequence can be thought like this,  2, 4, 8, 16, …..  So, the total number of ancestors in 10 generation of his family.  2, 4, 8, 16, 32, …..10 terms.  Here, a = 2, r = 2 and n = 10  S10 = a 

Infinite GP Problems

An infinite geometric series is the sum of an infinite geometric sequence. This series would have no last term

Question 1: A monkey is swinging from a tree. On the first swing, she passes through an arc of 24m. With each swing, she passes through an arc of 24m. With each swing, she passes through an arc half the length of the previous swing. What is the total distance traveled by the monkey when she completes her 100000th swing? 

Now this movement represents a GP with a = 24 and r = 1/2. Now, since the GP is decreasing and the question asks for the sum till 100000th term. To save the calculation , we can consider it an infinite GP and round of the answer we get.  Sum of an infinite GP =  Here, a = 24 and r = 1/2. Let the sum be S  So the monkey travels almost 24m in these many swings. 

Question 2: A ball was dropped from a 24-inch high table. The ball rebounds and always reaches three fourth of the distance fallen. What is the approximate distance the ball travels before finally coming at rest on the ground. 

It should be noticed that this problem actually involves two infinite geometric series. The first series involves ball falling and the other series involves ball rising after rebounding from the ground.  Falling: a 1 = 24 , r = 3/4  Rising: a 2 = 24(3/4) = 18 ,r = 3/4  Using the formula for infinite geometric series,  S =  Let S be the total distance travelled: S = S rising + S falling S rising = S falling =  Now, S = S rising + S falling = 72 + 96 = 168

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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

  • Class 11 Maths
  • Chapter 9: Sequences Series

Important Questions for Class 11 Maths Chapter 9 - Sequences and Series

Important questions for class 11 Maths Chapter 9 – sequences and series are given here. Sequences and series have several applications in our daily life. The important questions cover all the topics and subtopics in the NCERT textbook. The solutions for the questions are given in a step by step procedure so that students can understand them in a better way. Go through the important questions provided at BYJU’S and achieve excellent marks in the annual examination. Get all the important Maths questions from class 11 chapters here.

Class 11 Maths Chapter 9 – Sequences and series cover the following important concepts such as:

  • Introduction to Sequences and Series
  • Arithmetic Progression
  • Geometric Progression
  • Sum of n terms of the special series
  • The relation between the arithmetic and geometric mean

Also, Check:

  • Important 1 Mark Questions for CBSE Class 11 Maths
  • Important 4 Marks Questions for CBSE Class 11 Maths
  • Important 6 Marks Questions for CBSE Class 11 Maths

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.

Let a 1 , a 2 and d 1 , d 2 be the first term and the common difference of the first and second

arithmetic progression respectively.

(Sum of n terms of the first A.P)/(Sum of n terms of the second A.P) = (5n+4)/(9n+6)

⇒ [ (n/2)[2a 1 + (n-1)d 1 ]]/ [(n/2)[2a 2 + (n-1)d 2 ]]= (5n+4)/(9n+6)

Cancel out (n/2) both numerator and denominator on L.H.S

⇒[2a 1 + (n-1)d 1 ]/[2a 2 + (n-1)d 2 ]= (5n+4)/(9n+6) …(1)

Now susbtitute n= 35 in equation (1), {Since (n-1)/2 = 17}

Then equation (1) becomes

⇒[2a 1 + 34d 1 ]/[2a 2 + 34d 2 ]= (5(35)+4)/(9(35+6)

⇒[a 1 + 17d 1 ]/[a 2 + 17d 2 ]= 179/321 …(2)

Now, we can say that.

18th term of first AP/ 18th term of second AP = [a 1 + 17d 1 ]/[a 2 + 17d 2 ]….(3)

Now, from (2) and (3), we can say that,

18th term of first AP/ 18th term of second AP = 179/321

Hence, the ratio of the 18th terms of both the AP’s is 179:321.

Question 2:

Insert five numbers between 8 and 26 such that resulting sequence is an A.P.

Assume that A 1 , A 2 , A 3 , A 4 , and A 5 are the five numbers between 8 and 26, such that the sequence of an A.P becomes 8, A 1 , A 2 , A 3 , A 4 , A 5 , 26

Here, a= 8, l =26, n= 5

Therefore, 26= 8+(7-1)d

Hence it becomes,

6d = 26-8 = 18

A 1 = a+d = 8+ 3 =11

A 2 = a+2d = 8+ 2(3) =8+6 = 14

A 3 = a+3d = 8+ 3(3) =8+9 = 17

A 4 = a+4d = 8+ 4(3) =8+12 = 20

A 5 = a+5d = 8+ 5(3) =8+15 = 23

Hence, the required five numbers between the number 8 and 26 are 11, 14, 17, 20, 23.

Question 3:

The 5th, 8th, and 11th terms of a GP are p, q and s respectively. Prove that q 2 = ps

Given that:

5th term = P

8th term = q

11th term = s

To prove that: q 2 = ps

By using the above information, we can write the equation as:

a 5 = ar 5-1 = ar 4 = p ….(1)

a 8 = ar 8-1 = ar 7 = q ….(2)

a 11 = ar 11-1 = ar 10 =s …(3)

Divide the equation (2) by (1), we get

r 3 = q/p …(4)

Divide the equation (3) by (2), we get

r 3 = s/q …(5)

Now, equate the equation (4) and (5), we get

It becomes, q 2 = ps

Hence proved.

Question 4:

Show that the sum of (m + n) th and (m – n) th terms of an A.P. is equal to twice the m th term.

Let a and d be the first term and the common difference of the A.P. respectively. It is known

that the k th term of an A.P. is given by

a k = a +(k -1)d

Therefore, a m+n = a +(m+n -1)d

a m-n = a +(m-n -1)d

a m = a +(m-1)d

Hence, the sum of (m + n) th and (m – n) th terms of an A.P is written as:

a m+n + a m-n = a +(m+n -1)d + a +(m-n -1)d

= 2a +(m + n -1+ m – n -1)d

=2a+(2m-2)d

=2a + 2(m-1)d

Therefore, the sum of (m + n) th and (m – n) th terms of an A.P. is equal to twice the m th term.

Question 5:

Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

The integers from 1 to 100, which are divisible by 2, are 2, 4, 6 ….. 100.

This forms an A.P. with both the first term and common difference equal to 2.

⇒ 100=2+(n-1)2

Therefore, the sum of integers from 1 to 100 that are divisible by 2 is given as:

= (50/2)(4+98)

The integers from 1 to 100, which are divisible by 5, 10…. 100

This forms an A.P. with both the first term and common difference equal to 5.

Therefore, 100= 5+(n-1)5

= (20/2)(10+95)

Hence, the integers from 1 to 100, which are divisible by both 2 and 5 are 10, 20, ….. 100.

This also forms an A.P. with both the first term and common difference equal to 10.

Therefore, 100= 10+(n-1)10

⇒ n= 100/10

= (10/2)(20+90)

Therefore, the required sum is:

= 2550+ 1050 – 550

Hence, the sum of the integers from 1 to 100, which are divisible by 2 or 5 is 3050.

Practice Problems for Class 11 Maths Chapter 9 Sequences and Series

Go through class 11 chapter 9 sequences and series concepts and solve the practice problems provided below:

  • Prove that the sequence t n = 3n + 5 is an Arithmetic Progression. Find its common difference.
  • 3+7+ 11+…..
  • 1 + 6 + 11 +16
  • Is 309 a term of the AP 11, 17, 23….?
  • The sum of 3 numbers of an arithmetic progression is 24 and their product is 44. Find the three numbers.
  • If the sum of first n terms of a progression is a quadratic expression in n, show that it is an arithmetic progression.
  • If the AM between the a th and b th terms of an AP be equal to the AM between the c th and d th term of it then prove that a + b = c + d
  • Find three numbers in GP whose product is 1728 and sum is 38.
  • What will Rs 5000 amount to in 10 years, compounded annually at 10 % per annum?
  • Find two positive numbers m and n whose AM and GM are 34 and 16 respectively.
  • Find the sum of the series – 1 + 5 + 12 +….to n terms.
  • 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?
  • A carpenter was hired to build 192 window frames. The first day he made five frames and each day, thereafter he made two more frames than he made the day before. How many days did it take him to finish the job?
  • Find the sum of the series (3 3 – 2 3 ) + (5 3 – 4 3 ) + (7 3 – 6 3 ) + … to (i) n terms (ii) 10 terms

Download BYJU’S – The Learning App today to practice many problems in class 11 Maths to score good marks in the final examination.

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CBSE Case Study Questions for Class 11 Maths Sets Free PDF

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Mere Bacchon, you must practice the CBSE Case Study Questions Class 11 Maths Sets in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

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CBSE Case Study Questions for Class 11 Maths Sets PDF

Checkout our case study questions for other chapters.

  • Chapter 2 Relations and Functions Case Study Questions
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COMMENTS

  1. CBSE Case Study Questions For Class 11 Maths Sequences And Series Free

    Level 1 First, learn to sit for at least 2 hours at a stretch Level 2 Solve every question of NCERT by hand, without looking at the solution. Level 3 Solve NCERT Exemplar (if available) Level 4 Sit through chapter wise FULLY INVIGILATED TESTS Level 5 Practice MCQ Questions (Very Important) Level 6

  2. Case Study Questions for Class 11 Maths Chapter 9 Sequence and Series

    Case study questions are an important aspect of mathematics education at the Class 11 level. These questions require students to apply their knowledge and skills to real-world scenarios, helping them develop critical thinking, problem-solving, and analytical skills.

  3. Class 11 Mathematics Case Study Questions

    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.

  4. Case Studies on the topic Sequences & Series

    Case Studies on the topic Sequences & Series - Class 11 Mathematics Conquer Mathematics 2.93K subscribers Subscribe Subscribed Like Share 7K views 2 years ago CLASS 11 - Mathematics...

  5. CBSE 11th : Case study Based Questions (4) : Sequences and series

    CBSE 11th : Case study Based Questions (4) : Sequences and series cool guru 1.06K subscribers Subscribe Subscribed 7.5K views 2 years ago 11th Case Study based Questions Case study...

  6. Sequence and series

    Unit 1 Sets Unit 2 Relations and functions Unit 3 Trigonometric functions Unit 4 Complex numbers Unit 5 Linear inequalities Unit 6 Permutations and combinations Unit 7 Binomial theorem Unit 8 Sequence and series Unit 9 Straight lines Unit 10 Conic sections Unit 11 Introduction to three dimensional geometry Unit 12 Limits Unit 13 Derivatives

  7. CBSE 11th : Case study Based Questions (12) : Sequence and Series

    Case study based question from chapter " Sequence and Series" based on Geometric Progression.Next in playlist:https://youtu.be/H4ZsFcqxIR4

  8. Chapter 8 Class 11 Sequences and Series

    What's in it? 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

  9. Sequences and Series Class 11 Notes CBSE Maths Chapter 9 [PDF]

    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 ...

  10. Sequences and Series Class 11 Notes Maths Chapter 9

    CBSE Class 11 Maths Notes Chapter 9 Sequences and Series Sequence A succession of numbers arranged in a definite order according to a given certain rule is called sequence. A sequence is either finite or infinite depending upon the number of terms in a sequence. Series

  11. Sequences and Series Class 11 Chapter 9 Notes and Examples

    The topics and subtopics covered in chapter 9 - Sequences and Series class 11 concepts are: Introduction Sequences Series Arithmetic Progression (A.P.) Arithmetic mean Geometric Progression (G.P.) The general term of a G.P. Sum to n terms of a G.P. Geometric Mean (G.M.) Relationship Between A.M. and G.M. Sum to n Terms of Special Series

  12. Category: Case Study Questions for Class 11 Maths

    Case Study Questions for Class 11 Maths Chapter 9 Sequence and Series. February 25, 2023 February 25, 2023 Physics Gurukul Leave a Comment on Case Study Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations.

  13. CBSE 11th: Case study based question (14th) : Sequence and series

    Case study based question from "Sequence and Series" based on Arithmetic Progression.Next in playlist:https://youtu.be/rNQLfALbEvc

  14. Sequence and Series Word Problems

    Answer: The series follow the pattern: 100 = (220×0.5)-10 40 = (100×0.5)-10 Therefore, the next term will be (40×0.5)-10 = 10 Arithmetic Progressions In Arithmetic Progressions, the consecutive terms will have the same difference and are denoted as 'd', the first term is called as 'a', and the number of terms is denoted as 'n'.

  15. CBSE Class 11 Maths

    Sequence and Series : Notes and Study Materials -pdf. ... 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

  16. NCERT Solutions Class 11 Maths Chapter 9 Sequences and Series

    Practising the NCERT Solutions Class 11 Chapter 9 Sequences and Series can help the students develop a thorough understanding of the topics explained in the Chapter.

  17. Important Questions for Class 11 Maths Chapter 9 with Solutions

    Maths Class 11 Maths Chapter 9: Sequences Series Important Questions for Class 11 Maths Chapter 9 - Sequences and Series Important questions for class 11 Maths Chapter 9 - sequences and series are given here. Sequences and series have several applications in our daily life.

  18. CBSE Notes Class 11 Maths Sequences and Series

    So, go ahead and check the Important Notes for CBSE Class 11 Maths Sequences and Series from this article. 1. Sequence: Sequence is a function whose domain is a subset of natural numbers. It represents the images of 1, 2, 3,… ,n, as f 1, f 2, f 3, …., f n , where f n = f (n). 2.

  19. CBSE Case Study Questions for Class 11 Maths Sets Free PDF

    First, learn to sit for at least 2 hours at a stretch. Level 2. Solve every question of NCERT by hand, without looking at the solution. Level 3. Solve NCERT Exemplar (if available) Level 4. Sit through chapter wise FULLY INVIGILATED TESTS. Level 5. Practice MCQ Questions (Very Important)

  20. CBSE 11th Standard CBSE all Case study Questions Updated

    CBSE 11th Standard CBSE all question papers, important notes , study materials , Previuous Year questions, Syllabus and exam patterns. Free 11th Standard CBSE all books and syllabus online. Practice Online test for free in QB365 Study Material. Important keywords, Case Study Questions and Solutions. Updates about latest education news and ...

  21. CBSE 11th: Case study based question (20th) : "Sequence and Series"

    Case study based question from the chapter "Sequence and Series"Next in playlist:https://youtu.be/BO2H7g-sPaY