case study on arithmetic progression class 11

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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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CBSE Class 10 Maths Case Study Questions for Maths Chapter 5 - Arithmetic Progression (Published by CBSE)

Case study questions on cbse class 10 maths chapter 5 - arithmetic progression are provided here. these questions are published by cbse to help students prepare for their maths exam..

CBSE Class 10 Case Study Questions for Maths Chapter 5 - Arithmetic Progression are available here with answers. All the questions have been published by the CBSE board. Students must practice all these questions to prepare themselves for attempting the case study based questions with absolute correctness and obtain a high score in their Maths Exam 2021-22.

Case Study Questions for Class 10 Maths Chapter 5 - Arithmetic Progression

CASE STUDY 1:

India is competitive manufacturing location due to the low cost of manpower and strong technical and engineering capabilities contributing to higher quality production runs. The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in 6th year and 22600 in 9th year.

Based on the above information, answer the following questions:

1. Find the production during first year.

2. Find the production during 8th year.

3. Find the production during first 3 years.

4. In which year, the production is Rs 29,200.

5. Find the difference of the production during 7th year and 4th year.

2. Production during 8th year is (a+7d) = 5000 + 2(2200) = 20400

3. Production during first 3 year = 5000 + 7200 + 9400 = 21600

4. N = 12 5.

Difference = 18200 - 11600 = 6600

CASE STUDY 2:

Your friend Veer wants to participate in a 200m race. He can currently run that distance in 51 seconds and with each day of practice it takes him 2 seconds less. He wants to do in 31 seconds.

1. Which of the following terms are in AP for the given situation

a) 51,53,55….

b) 51, 49, 47….

c) -51, -53, -55….

d) 51, 55, 59…

Answer: b) 51, 49, 47….

2. What is the minimum number of days he needs to practice till his goal is achieved

Answer: c) 11

3. Which of the following term is not in the AP of the above given situation

Answer: b) 30

4. If nth term of an AP is given by an = 2n + 3 then common difference of an AP is

Answer: a) 2

5. The value of x, for which 2x, x+ 10, 3x + 2 are three consecutive terms of an AP

Answer: a) 6

CASE STUDY 3:

Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of Rs 1,18,000 by paying every month starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month , answer the following:

1. The amount paid by him in 30th installment is

Answer: a) 3900

2. The amount paid by him in the 30 installments is

Answer: b) 73500

3. What amount does he still have to pay offer 30th installment?

Answer: c) 44500

4. If total installments are 40 then amount paid in the last installment?

Answer: a) 4900

5. The ratio of the 1st installment to the last installment is

Answer: b) 10:49

Also Check:

CBSE Case Study Questions for Class 10 Maths - All Chapters

Tips to Solve Case Study Based Questions Accurately

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Download Case Study Questions for Class 11 Maths

• Last modified on: 9 months ago
• Reading Time: 9 Minutes

[PDF] Download Case Study Questions for Class 11 Maths

Here we are providing case study questions for Class 11 Maths. In this article, we are sharing links for Class 11 Maths All Chapters. All case study questions of Class 11 Maths are solved so that students can check their solutions after attempting questions.

Click on the chapter to view.

Class 11 Maths Chapters

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 Permutation and Combinations Chapter 8 Binomial Theorem Chapter 9 Sequences 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

What is meant by Case Study Question?

In the context of CBSE (Central Board of Secondary Education), a case study question is a type of question that requires students to analyze a given scenario or situation and apply their knowledge and skills to solve a problem or answer a question related to the case study.

Case study questions typically involve a real-world situation that requires students to identify the problem or issue, analyze the relevant information, and apply their understanding of the relevant concepts to propose a solution or answer a question. These questions may involve multiple steps and require students to think critically, apply their problem-solving skills, and communicate their reasoning effectively.

Importance of Solving Case Study Questions for Class 11 Maths

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. Here are some reasons why case study questions are important in Class 11 maths education:

• Real-world application: Case study questions allow students to see how the concepts they are learning in mathematics can be applied in real-life situations. This helps students understand the relevance and importance of mathematics in their daily lives.
• Higher-order thinking: Case study questions require students to think critically, analyze data, and make connections between different concepts. This helps develop higher-order thinking skills, which are essential for success in both academics and real-life situations.
• Collaborative learning: Case study questions often require students to work in groups, which promotes collaborative learning and helps students develop communication and teamwork skills.
• Problem-solving skills: Case study questions require students to apply their knowledge and skills to solve complex problems. This helps develop problem-solving skills, which are essential in many careers and in everyday life.
• Exam preparation: Case study questions are included in exams and tests, so practicing them can help students prepare for these assessments.

Overall, case study questions are an important component of Class 11 mathematics education, as they help students develop critical thinking, problem-solving, and analytical skills, which are essential for success in both academics and real-life situations.

Feature of Case Study Questions on This Website

Here are some features of a Class 11 Maths Case Study Questions Booklet:

Many Case Study Questions: This website contains many case study questions, each with a unique scenario and problem statement.

Different types of problems: The booklet includes different types of problems, such as optimization problems, application problems, and interpretation problems, to test students’ understanding of various mathematical concepts and their ability to apply them to real-world situations.

Multiple-choice questions: Questions contains multiple-choice questions to assess students’ knowledge, understanding, and critical thinking skills.

Focus on problem-solving skills: The questions are designed to test students’ problem-solving skills, requiring them to identify the problem, select appropriate mathematical tools, and analyze and interpret the results.

Emphasis on practical applications: The case studies in the booklet focus on practical applications of mathematical concepts, allowing students to develop an understanding of how mathematics is used in real-life situations.

Comprehensive answer key: The booklet includes a comprehensive answer key that provides detailed explanations and step-by-step solutions for all the questions, helping students to understand the concepts and methods used to solve each problem.

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CBSE 10th Standard Maths Subject Arithmetic Progressions Case Study Questions With Solution 2021

By QB365 on 22 May, 2021

QB365 Provides the updated CASE Study Questions for Class 10 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams

QB365 - Question Bank Software

10th Standard CBSE

Final Semester - June 2015

Case Study Questions

(ii) Find the sum of common difference of the two progressions.

(iii) Find the 19 th term of the progression written by Geeta.

(iv) Find the sum of first 10 terms of the progression written by Geeta.

(v) Which term of the two progressions will have the same value?

(ii) What is the total amount saved by Anuj in 8 days?

(iii) What is the amount saved by Anuj on 30 th day?

(iv) What is the total amount saved by him in the month of June, if he starts savings from 1 st June?

(v) On which day, he save tens times as much as he saved on day-I?

(ii) The number on first card is

(iii) What is the number on the 19 th card?

(iv) What is the number on the 23 rd card?

(v) The sum of numbers on the first 15 cards is 

A sequence is an ordered list of numbers. A sequence of numbers such that the difference between the consecutive terms is constant is said to be an arithmetic progression (A.P.). On the basis of above information, answer the following questions. (i) Which of the following sequence is an A.P.?

(ii) If x, y and z are in A.P., then

(iii) If a 1  a 2 , a 3  ..... , a n are in A.P., then which of the following is true?

(iv) If the n th term (n > 1) of an A.P. is smaller than the first term, then nature of its common difference (d) is

(v) Which of the following is incorrect about A.P.?

(ii) What is the first term?

(iii) Which term of the A.P. is -160?

(iv) Which of the following is not a term of the given A.P.?

(v) What is the 75 th term of the A.P.?

*****************************************

Cbse 10th standard maths subject arithmetic progressions case study questions with solution 2021 answer keys.

Geeta's A.P. is -5, -2, 1,4, ... Here, first term (a 1 ) = -5 and common difference (d 1 ) = -2 + 5 = 3 Similarly, Madhuri's A.P. is 187, 184, 181, ... Here first term (a 2 ) = 187 and common difference (d 2 ) = 184 - 187 = -3 (i) (b): t 34 = a 2 + 33d 2 = 187 + 33(-3) = 88 (ii) (d): Required sum = 3 + (-3) = 0 (iii) (a): t 19  = a 1  + 18d 1  = (-5) + 18(3) = 49 (iv) (a) :  \(S_{10}=\frac{n}{2}\left[2 a_{1}+(n-1) d_{1}\right]=\frac{10}{2}[2(-5)+9(3)]=85\) (v) (b): Let n th terms of the two A.P:s be equal. \(\therefore\) -5 + (n - 1)3 = 187 + (n - 1)(-3) \(\Rightarrow\) 6(n - 1) = 192 \(\Rightarrow\) n = 33

Here the savings form an A.P. i.e., Rs 2.75, Rs 3, Rs 3.25, ... So, a = 2.75, d = 3 - 2.75 = 0.25 (i) (b): Amount saved by Anuj on 14 th day = t 14 = a + 13d = 2.75 + 13(0.25) = ₹ 6 (ii) (d): Total amount saved by Anuj in 8 days \(=S_{8}=\frac{8}{2}[2(2.75)+7(0.25)]=₹ 29\) (iii) (a): Amount saved by Anuj on 30 th day = t 30 = a + 29d = 2.75 + 29(0.25) = ₹ 10 (iv) (b): Number of days in June = 30 \(\therefore S_{30}=\frac{30}{2}[2(2.75)+29(0.25)]=₹ 191.25\) (v) (d): Let on nth day, he saves 10 times as he saves on 1 st day. t n = 10(2.75) \(\Rightarrow\) a + (n - 1)d = 27.5 \(\Rightarrow\) n = 100.

Let the numbers on the cards be a, a + d, a + Zd, ... According to question, We have (a + 5d) + (a + 13d) = -76 \(\Rightarrow\) 2a+18d = -76 \(\Rightarrow\) a + 9d= -38 ... (1) And (a + 7d) + (a + 15d) = -96 \(\Rightarrow\) 2a + 22d = -96 \(\Rightarrow\) a + 11d = -48 ...(2) From (1) and (2), we get 2d= -10 \(\Rightarrow\) d= -5 From (1), a + 9(-5) = -38 \(\Rightarrow\) a = 7 (i) (b): The difference between the numbers on any two consecutive cards = common difference of the A.P.=-5 (ii) (d): Number on first card = a = 7 (iii) (b): Number on 19th card = a + 18d = 7 + 18(-5) = -83 (iv) (a): Number on 23rd card = a + 22d = 7 + 22( -5) = -103 (v) (d):  \(S_{15}=\frac{15}{2}[2(7)+14(-5)]=-420\)

(i) (c) (ii) (c) (iii) (d) (iv) (b) (v) (c)

We have, 3 rd term = 4 and 9 th term = -8 i.e., a + 2d = 4 ........(i) and a + 8d = -8 .........(ii) Solving (1) and (2), we get d = -2, a = 8 (i) (c) (ii) (d) (iii) (b): Let t n = -160 \(\Rightarrow\) a + (n - 1) d = -160 \(\Rightarrow\) 8 + (n - 1)(-2) = -160 \(\Rightarrow\) (n - 1)(-2) = -168 \(\Rightarrow\) n - 1 = 84 \(\Rightarrow\) n = 85 So, t 85  = -160 (iv) (a) (v) (a): t 75 = a + 74d = 8 + 74( -2) = -140

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• RS Aggarwal Class 11 Solutions Chapter-11 Arithmetic Progression
• RS Aggarwal Solutions

Class 11 RS Aggarwal Chapter-11 Arithmetic Progression Solutions - Free PDF Download

CBSE students who are preparing for their Class 11th examination can take reference from arithmetic progression Class 11 RS Aggarwal. This book is phenomenal for its study materials that help students to understand every topic with clarity. A chapter-wise solution is outlined in the textbooks which makes the concept easy and intriguing. It is the right tool for students to overcome the difficulty section in the chapter.

RS Aggarwal Class 11 Maths Chapter 11 Solutions Arithmetic Progression

Progression is a term used to describe a succession whose terms follow particular patterns. Arithmetic Progression in RS Aggarwal Solutions Class 11 Maths Chapter 11 is an important chapter. It is a sequence of numbers in which the preceding term adds a constant quantity to form the consequent term. It states that the difference between the preceding term and the consequent term is a constant quantity. For example, in all the natural numbers: 1,2,3,4,5,6...the difference between two successive numbers is always 1. Also, in odd or even numbers we observe a similar thing is happening, where the difference between the two numbers always remains as 2. 

bn+1- bn = constant (=d) for all n∈ N Arithmeticc progression is about a sequence of numbers in which the difference between two terms remains constant. 

The sequence {5, 10, 15, 20, 25, 30, …………………} is an Arithmetic Progression where the common difference is 5, since

Why is it Beneficial to Learn Arithmetic Progression?

An arithmetic progression is a sequence of terms with a common difference between them that is a constant value. It's a term that's used to generalize a group of patterns that we see in our daily lives.

The capacity to notice and generate patterns aids us in making predictions based on our observations is a crucial mathematical talent. Pattern recognition helps children’s acquisition and understanding of complex numerical concepts and mathematical processes. Patterns let us recognize connections and make generalizations.

What are the Properties of Arithmetic Progression Class 11 RS Aggarwal?

The properties of arithmetic progression discussed in RS Aggarwal class 11 maths chapter 11 solutions are as follows:

Property 1: When you add or subtract any constant term to the given AP in each number, the resulting sequence will be an arithmetic progression. 

Property 2: When you multiply or divide each term of a given AP with a non-zero constant, the result will also create an arithmetic progression. 

Property 3: In a finite term of numbers of an arithmetic progression, the total summation of two numbers equidistant from the end and starting will be the numbers is the same as the total summation of the last term and first term. 

Property 4: If 2b = a+c then the three numbers are said to be in an arithmetic progression.

Property 5: A sequence will be established as an arithmetic progression if the nth term is a linear expression. 

Property 6: A sequence will be an arithmetic progression if the summation of the 1st n terms is of An2 + Bn, where A and B are said to be two constant quantities independent of n. 

Note: The effect of adding or subtracting a constant from each term of an AR is an AP with the same common difference. The resulting sequence is also an AP if each term of an AP is multiplied or divided by a non-zero constant.

What are the Benefits of  Preparing from RS Aggarwal Class 11 Maths Arithmetic Progression Exercise?

There are several benefits of preparing from RS Aggarwal Arithmetic Progression exercises. 

Exercise 1- The first exercise contains questions, where one needs to find the number of terms in the sequence. Suppose the exact term like the 23rd term of the sequence, identifying the terms and finding the differences, etc. This exercise is a starter exercise that doesn’t include a lot of difficult questions. This exercise exists to make students used to the process of identifying the terms and finding the differences.

Exercise 2- The second exercise has questions related to finding some differences as asked in the questions, r term of AP, the value of x, the sum of n terms of AP, last term of Ap, etc. The second exercise is one step up. The exercises in the RS Aggarwal reference book build up the difficulty level with every question you solve, ensuring that one is prepared for any kind of question that might come in the exam.

Exercise 3- In this exercise, some questions need to be solved using the various formulas of an arithmetic progression. This is a comparatively basic exercise. Helps in making the basics strong. 

Exercise 4- Here, students will find different questions, where one has to find the arithmetic mean. The questions in this exercise will help you brush up on your concepts.

Exercise 5- Here students have to prove different situations that are asked in the questions. This exercise requires a good presence of mind and decent application skills. It will require you to apply the concepts that you’ve already learned and practiced differently.

Exercise 6- The final exercise comprises various questions that are present in the overall chapter. Solving this exercise will help you assess where you stand on your overall preparation for this chapter. 

Solved Exercise Question from Arithmetic Progression

Question Given that: the nth term of series = (5a+2) 

Sol: Putting a= 1,3,5,7 in the nth term, we obtain, 

First-term a1 = (5 x 1 + 2) = 7

Second term a2 = (5 x 3 + 2) = 17

Third term a3 = (5 x 5 + 2) = 27

Fourth term a4 = (5 x 7 + 2) = 37

Therefore, the first four terms of the series are (7, 17, 27, 37)

Did You Know?  

The behaviour of an arithmetic progression depends upon the common difference. If the difference is positive, it progresses towards positive infinity, if negative it goes towards negative infinity.  The distinction between arithmetic and a geometric sequence is that the difference between two consecutive terms in an arithmetic sequence remains constant, whereas the ratio between two consecutive terms in a geometric series remains constant.

The chapter covers all the important questions along with the answers that serve as a ready reference to comprehend and solve various questions for examination. Students can now include it in their daily practice schedule and home task to grasp the chapter very well. RS Aggarwal Class 11 Maths Arithmetic Progression PDF by Vedantu is made according to the latest question pattern that follows in the examination. So, it will be a good resource for scoring higher marks.

FAQs on RS Aggarwal Class 11 Solutions Chapter-11 Arithmetic Progression

1. Explain the Term Sequence Mentioned in Arithmetic Progression Class 11 RS Aggarwal?

Sequence refers to a group of numbers in a series that can be differentiated through a constant number. For example, the even numbers (2,4,6,8,...) represents an arithmetic progression with a difference from a constant of 2. Also, in this chapter of RS Aggarwal, you can learn about types of sequences like real sequences and what function it plays in arithmetic progression. The difference of this constant is denoted by d. To find out the constant, the given example below will make us understand the concept in a better way. 

AP= 2,5,8,11,15,...,etc. 

A given sequence in an arithmetic progression, if the difference between the two terms is constant. 

Therefore, d = n2-n1 where n3-n2 is also the same. 

Here, 5-2=3, and 8-5=3

So, you can easily conclude the sequence represents an arithmetic progression. 

2. How to Score Goods Marks in Arithmetic Progression?

To prepare for exams and score good marks in arithmetic progression students need to devote a lot of time in practice. It will improve their problem-solving skills and help to get rid of the factual errors. Always, keep yourself handy with the various concepts and formula of arithmetic progression so that you can quickly solve the question. This will save time in the examination. For this, students have to go through different exercises and solve all the problems. This will expose them to various sums present in the exercises. Knowing the concept is the first step, but practising trains you for the examination. 

3. Are class - 11 RS Aggarwal Chapter - 11 Arithmetic Progression Solutions helpful for me?

Yes, RS Aggarwal Solutions for Class 11 Arithmetic Progression are helpful for you as students studying maths as it gives you an advantage over the other students. Preparing for NCERT is very important but if you are also referring to these solutions then you’ll have solid preparation for your exam. These solutions not only make you practice the concepts you’ve already learned. But also give you an idea about some mind-bending questions to enhance your problem-solving skills. So, if you’re someone who wants to score 100/100 in his/her maths exam, these solutions available on Vedantu.com are the way to go!

4. What does the term Series in Arithmetic Progression mean?

The sum of an arithmetic sequence is an arithmetic series. We calculate the sum by combining the first and last terms, a1 and an, dividing by 2 to obtain the mean of the two values, then multiplying by the number of values, n: Sn=n2(a1+an). The concept of arithmetic progression or sequence can be applied to any element of our existence. All we have to do now is examine how it is used in our daily lives. Knowing about this type of sequence can provide us with a different perspective on how things happen in our life.

5. What are some other kinds of progressions in maths?

Other than Arithmetic Progression, there also exists Harmonic Progression and Geometric Progression. Harmonic Progression. A harmonic progression (or harmonic sequence) is a mathematical progression that is created by multiplying the reciprocals of an arithmetic progression. When each term is the harmonic mean of the nearby terms, a sequence is equivalently a harmonic progression. Whereas, Geometric Progression is a sequence in which each subsequent element is derived by multiplying the preceding element by a constant known as the common ratio, which is indicated by the letter r.

• Math Article

Arithmetic Progression

Arithmetic Progression (AP) is a sequence of numbers in order, in which the difference between any two consecutive numbers is a constant value. It is also called Arithmetic Sequence. For example, the series of natural numbers : 1, 2, 3, 4, 5, 6,… is an Arithmetic Progression, which has a common difference between two successive terms (say 1 and 2) equal to 1 (2 -1). Even in the case of odd numbers and even numbers, we can see the common difference between two successive terms will be equal to 2.

Check: Mathematics for Grade 11

If we observe in our regular lives, we come across Arithmetic progression quite often. For example, Roll numbers of students in a class, days in a week or months in a year. This pattern of series and sequences has been generalized in Maths as progressions.

What is Arithmetic Progression?

In mathematics, there are three different types of progressions. They are:

• Arithmetic Progression (AP)
• Geometric Progression (GP)
• Harmonic Progression (HP)

A progression is a special type of sequence for which it is possible to obtain a formula for the nth term.  The Arithmetic Progression is the most commonly used sequence in maths with easy to understand formulas. 

Definition 1: A mathematical sequence in which the difference between two consecutive terms is always a constant and it is abbreviated as AP.

Definition 2: An arithmetic sequence or progression is defined as a sequence of numbers in which for every pair of consecutive terms, the second number is obtained by adding a fixed number to the first one.

The fixed number that must be added to any term of an AP to get the next term is known as the common difference of the AP.  Now, let us consider the sequence, 1, 4, 7, 10, 13, 16,…

It is considered as an arithmetic sequence (progression) with a common difference 3. 

Notation in Arithmetic Progression

In AP, we will come across some main terms, which are denoted as:

• First term (a)
• Common difference (d)
• nth Term (a n )
• Sum of the first n terms (S n )

All three terms represent the property of Arithmetic Progression. We will learn more about these three properties in the next section.

First Term of AP

The AP can also be written in terms of common differences, as follows;

where  “a” is the first term of the progression. 

Common Difference in Arithmetic Progression

In this progression, for a given series, the terms used are the first term, the common difference and nth term. Suppose,  a 1 , a 2 , a 3 , ……………., a n is an AP, then;   the common difference “ d ” can be obtained as;

Where “d” is a common difference. It can be positive, negative or zero.

Also, check:

General Form of an AP

Consider an AP to be:  a 1 , a 2 , a 3 , ……………., a n

Arithmetic Progression Formulas

• The nth term of AP
• Sum of the first n terms

nth Term of an AP

The formula for finding the n-th term of an AP is:

a = First term

d = Common difference

n = number of terms

a n  = nth term

Example: Find the nth term of AP: 1, 2, 3, 4, 5….,  a n , if the number of terms are 15.

Solution: Given, AP: 1, 2, 3, 4, 5….,  a n

By the formula we know, a n = a+(n-1)d

First-term, a =1

Common difference, d=2-1 =1

Therefore, a n  = a 15 = 1+(15-1)1 = 1+14 = 15

Note:  The behaviour of the sequence depends on the value of a common difference.

• If the value of “d” is positive, then the member terms will grow towards positive infinity
• If the value of “d” is negative, then the member terms grow towards negative infinity

Types of AP

Finite AP: An AP containing a finite number of terms is called finite AP . A finite AP has a last term.

For example: 3,5,7,9,11,13,15,17,19,21

Infinite AP: An AP which does not have a finite number of terms is called infinite AP. Such APs do not have a last term.

For example: 5,10,15,20,25,30, 35,40,45………………  

Sum of N Terms of AP

For an AP, the sum of the first n terms can be calculated if the first term, common difference and the total terms are known. The formula for the arithmetic progression sum is explained below:

Consider an AP consisting “n” terms.

This is the AP sum formula to find the sum of n terms in series.

Proof:  Consider an AP consisting “n” terms having the sequence a, a + d, a + 2d, …………., a + (n – 1) × d

Sum of first n terms = a + (a + d) + (a + 2d) + ………. + [a + (n – 1) × d] ——————-(i)

Writing the terms in reverse order, we have:

S n = [a + (n – 1) × d] + [a + (n – 2) × d] + [a + (n – 3) × d] + ……. (a) ———–(ii)

Adding both the equations term wise, we have:

2S n = [2a + (n – 1) × d] + [2a + (n – 1) × d] + [2a + (n – 1) × d] + …………. + [2a + (n – 1) ×d] (n-terms)

2S n = n × [2a + (n – 1) × d]

S n = n/2[2a + (n − 1) × d]

Example: Let us take the example of adding natural numbers up to 15 numbers.

AP = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15

Given, a = 1, d = 2-1 = 1 and a n = 15

Now, by the formula we know;

Hence, the sum of the first 15 natural numbers is 120.

Sum of AP when the Last Term is Given

Formula to find the sum of AP when first and last terms are given as follows:

List of Arithmetic Progression Formulas

The list of formulas is given in a tabular form used in AP. These formulas are useful to solve problems based on the series and sequence concept.

Arithmetic Progressions Solved Examples

Below are the problems to find the nth term and the sum of the sequence, which are solved using AP sum formulas in detail. Go through them once and solve the practice problems to excel in your skills.

Example 1: Find the value of n, if a = 10, d = 5, a n  = 95.

Solution:  Given, a = 10, d = 5, a n = 95

From the formula of general term, we have:

a n = a + (n − 1) × d

95 = 10 + (n − 1) × 5

(n − 1) × 5 =  95 – 10 = 85

(n − 1) =  85/ 5

(n − 1) = 17

Example 2: Find the 20 th term for the given AP: 3, 5, 7, 9, ……

Solution:  Given, 

3, 5, 7, 9, ……

a = 3, d = 5 – 3 = 2, n = 20

a 20 = 3 + (20 − 1) × 2

a 20 = 3 + 38

Example 3: Find the sum of the first 30 multiples of 4.

The first 30 multiples of 4 are: 4, 8, 12, ….., 120

Here,  a = 4, n = 30, d = 4

S 30 = n/2 [2a + (n − 1) × d]

S 30 = 30/2[2 (4) + (30 − 1) × 4]

S 30 = 15[8 + 116]

S 30 = 1860

Practice Problems on AP

Find the below questions based on Arithmetic sequence formulas and solve them for good practice.

Question 1: Find the a n and 10th term of the progression: 3, 1, 17, 24, ……

Question 2: If a = 2, d = 3 and n = 90. Find a n    and S n .

Question 3: The 7th term and 10th terms of an AP are 12 and 25. Find the 12th term.

Frequently Asked Questions on Arithmetic Progression

What is the general form of arithmetic progression, what is arithmetic progression give an example., how to find the sum of arithmetic progression, what are the types of progressions in maths, what is the use of arithmetic progression.

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

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The n the term of an an arithmetic progression is an=4n+5 then the 3rd term is

The n the term of an arithmetic progression is an=4n+5. For the 3rd term, n = 3. Therefore, a3 = 4(3) + 5= 12+5 = 17

Very good explanation with various series option.

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• Arithmetic Progression
• NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions
• Class 10 NCERT Solutions- Chapter 5 Arithmetic Progressions - Exercise 5.2
• Class 10 NCERT Solutions- Chapter 5 Arithmetic Progressions - Exercise 5.1
• Program to print Arithmetic Progression series
• Progression - Aptitude Questions and Answers
• Longest Arithmetic Progression
• Aptitude | Progressions | Question 2
• Aptitude | Progressions | Question 4
• Aptitude | Progressions | Question 3
• Aptitude | Progressions | Question 6
• Aptitude | Progressions | Question 1
• Aptitude | Progressions | Question 5
• Aptitude | Progressions | Question 8
• Aptitude | Progressions | Question 7
• Program for N-th term of Arithmetic Progression series

Practice Questions on Arithmetic Progression

Arithmetic Progression is an important topic for students of class 9 and class 10. It is used in our everyday lives with some real-life applications. So, to learn them and understand them is essential to ease out our several tasks.

Important Formulas on Arithmetic Progression

Various formulas on arithmetic progression are:

• a n is nth term
• a 1 is first term
• d is common difference
• S n is sum of the first n terms
• S n-1 is sum of the first n-1 terms
• n is number of terms

Q 1. In an AP, if d = 8, n = 10 and a n = 80, then Find the value of a?

Q 2. Find the 19th term of the AP: 5, 11, 17, 23, …?

Q 3. What is the sum of first 30 even natural numbers?

Q 4. Which term of the arithmetic progression 13, 18, 23, ….. is 80 more than its 20th term?

Q 5. If the sum of first p terms of an A.P., is ap 2 + bp, find its common difference.

Q 6. The first term of an A.P. is p and its common difference is q. Find its 10th term

Q 7. Find the 17th term from the end of the AP: 1, 6, 11, 16 … … 211, 216.

Q 8. If seven times the 7th term of an A.P. is equal to eleven times the 11th term, then what will be its 18th term?

Q 9. How many terms of the A.P. 18, 16, 14, … … be taken so that their sum is zero?

Q 10. 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.

Practice Questions on Arithmetic Progression with Solution

Q 1. what is the sum of first 20 odd natural numbers .

First odd natural number is 1 Common Difference is 2 So, twentieth term is a 20 = a 1 + (n-1)d a 20 = 1 + (20-1)2 a 20 = 1 + 38 a 20 = 39 So, Sum = n/2(a 1 + a 20 ) Sum = 20/2(1 + 39) Sum = 10 × 40 Sum = 400 So, sum of first 20 odd numbers = 400

Q 2. Find the common difference from the given AP: 1, 6, 11, 16, 21.

To find common difference we need to subtract consecutive terms d = a n – a n-1 d = 6 – 1 = 5 So, common difference of the given AP is 5.

Q 3. Find the general term of the series 4,7, 10,13……

In the given AP, we have a 1 = 4, d = 3 So, general term of the given AP is a n = a 1 + (n-1)d a n = 4 + (n-1)3 a n = 4 + 3n – 3 a n = 3n + 1 General term = 3n + 1.

Q 4. Which term of the arithmetic progression 30, 27, 24,…. is 0?

In the given AP, a 1 = 30 d = -3 a n = 0 Now, to find n a n = a 1 + (n-1)d 0 = 30 + (n-1)(-3) 0 = 30 – 3n + 3 3n = 33 n = 11 So, the 11th term is 0

Q 5. Find the first negative term of the arithmetic progression 36, 30, 24,…..?

In the given AP, we have a 1 = 36 d = -6 To find the first negative term, 36 + (n−1)(−6) < 0 36 + 6 – 6n < 0 42 – 6n < 0 -6n < – 42 n > 7 So, the 8th term is the first negative term of the given AP.

Q 6. If the common difference of an A.P. is 4, then 𝑎20 − 𝑎15 is

So, the given common difference = 4. Now, we want to calculate a20 – a15, So, a 20 = a + 19d a 15 = a + 14d Then, a 20 – a 15 = a+ 19d – a – 14d a 20 – a 15 = 5d a 20 – a 15 = 5 × 4 = 20 So the value of a 20 – a 15 is 20.

Q 7. Find the middle term of the A.P. 6, 13, 20, … , 216.

In the given AP, First term a 1 = 6 Last term a n = 216 Common difference = 7 Now, to find the number of terms, n = (a n – a 1 )/d + 1 n = (216 – 6)/7 + 1 n = 210/7 + 1 n = 30 + 1 n = 31 So, middle term is (n + 1) / 2 = (31 + 1)/2 = 16 Now to calculate the middle term a 16 = a1 + 15×d a 16 = 6 + 15 × 7 a 16 = 6 + 105 = 111 So the middle term of the given AP is 111

Q 8. In an AP, if 𝑑 = −2, 𝑛 = 5 and 𝑎𝑛 = 0, then find the value of 𝑎 ?

Given values are: d = -2 n = 5 a n = 0 So, to calculate follow these steps: a n = a + (n-1)d 0 = a + (5-1)(-2) 0 = a – 8 a = 8

Q 9. Which term of the arithmetic progression 3, 8, 13, ….. is 55 more than its 20th term?

In the given AP, we have following values a = 3 d = 5 Let the unknown term is n term, Now, an = a20 + 55 a1 + (n-1)d = a1 + 19d + 55 3 + (n-1)d = 3 + 19×5 + 55 5n – 5 = 95 + 55 5n = 155 n = 31 So, the 31st term is 55 more than the 20th term.

Q 10. Determine the sum of the first 18 terms, of the arithmetic progression 12b, 8b, 4b,……

In the given AP we have, a 1 = 12b d = -4b So, the sum of first 18 terms is Sum = n/2[2a 1 + (n-1)d] Sum = 9[2×12b + (18-1)(-4b)] Sum = 9[24b – 68b] Sum = 9 × (-44b) Sum = -396b So sum of first 18 terms is -396b.

FAQs on Arithmetic Progression

Who is known as father of arithmetic.

Brahmagupta is known as father of Arithmetic

What is sum of first 5 natural numbers?

a 1 = 1 d = 1 sum = n/2[2a 1 + (n-1)d] sum = 5/2 × [2×1 + (5-1)d] sum = 5/2 × [2 + 4] = 5×3 = 15 So, the sum of first 5 natural numbers is 15.

Find the common difference in the following AP: 1, 3, 5, …

Common difference in the following AP: 1, 3, 5, … is 2

Write Formula to find the sum of first n terms

Sum of n terms = n/2[2a1 + (n-1)d]

Can sum of an AP be zero.

Yes, sum of an AP may be zero sometimes.

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CBSE Assertion Reason Questions for Class 11 Maths Sequences and Series Free PDF

Mere Bacchon, you must practice the CBSE Assertion Reason Questions for Class 11 Maths Sequences and Series  in order to fully complete your preparation . They are very very important from exam point of view. These tricky Assertion Reason Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Assertion Reason Questions , just click ‘ Download PDF ’.

CBSE Assertion Reason Questions for Class 11 Maths Sequences and Series PDF

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