NCERT Solutions Class 11 Maths Chapter 9
NCERT Solutions for Class 11 Mathematics Chapter 9 – Sequences and series
Mathematics is an important aspect of every field. It is majorly used in various disciplines of life. The applications of Mathematics make day-to-day dealings of life easier. Hence, it is included as a core subject in almost every academic curriculum.
You have already studied the various patterns of numbers and alphabets in the lower classes. It is nothing but an arrangement with certain repetitions considered as the part of sequence and series. The main topics covered in the chapter are the introduction to sequence and series, Arithmetic Progression (A.P.), Geometric Progression (G.P.), the relationship between A.M and G.M and sum to ‘n’ terms of special series.
NCERT Solutions for Class 11 Mathematics Chapter 9 has detailed coverage of every topic covered in the NCERT Class 9th Mathematics textbook. Students will be able to understand all the concepts related to sequence and series once you refer to NCERT Solutions. How the chapter is presented in the NCERT Solutions for Class 11 Mathematics Chapter 9 will help students gain interest in the subject and assist them in acquiring related knowledge. NCERT solutions will definitely prove fruitful to students in their class assignments, tests and preparation.
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Key Topics Covered In Chapter 9 Class 11 Mathematics
The arrangement of numbers, objects, and alphabets in a particular pattern are called sequence, whereas placing of numbers, objects, and alphabets one after the other without repetition of the pattern is called series. The chapter sequence and series are a combination of both. To get a particular series and understand its associated sequence, one needs to know certain rules based on it. Students will get a clear understanding of this concept in this chapter.
They would be able to derive a series with the help of a few formulas given in the chapter. This will make their calculations while finding a series easier and save their time. The various patterns covered for calculating series in this chapter are AP and GP. Every bit of this chapter is included in our NCERT Solutions for Class 11 Mathematics Chapter 9, available on the Extramarks’ website.
NCERT Solutions for Class 11 Mathematics Chapter 9 require students to use their critical thinking ability and apply a wide range of formulas they have learnt.
In this chapter, we will learn about the sequence and series. When we put a collection of objects in an orderly manner in a certain pattern using some sets of rules, it is called a sequence and series. The sequence and series have an important application in many human activities. When the sequence follows a pattern, it’s called a progression.
We have already studied the arithmetic progression in our previous classes. In this chapter, we will learn some more concepts about the arithmetic progression, series, geometric progression, the relationship between arithmetic and geometric progression and the sum of ‘n’ terms of the series.
In the section of this chapter, we will understand the sequence with the help of some examples listed below:
- 2, 4, 8, 16, 32, …………, 1024
- 3, .3, .33, .333, .3333……and so on.
The different elements in a sequence are called terms, for example (a1, a2, a3, a4,…… an)
There are two types of sequences:
- Infinite sequence
When the sequence contains an infinite number of elements, then it is called an infinite sequence.
- Finite sequence
When the sequence contains a finite number of elements, it’s called a finite sequence.
Fibonacci sequence :
A1 = A2 = 1,
A3 = A1 + A2,
An = An-2 + An-1,
n > 2
In this section, we will learn about the series.
There is an example given below that shows the series:
a1 + a2 + a3 + a4 + …… an + …..
The example of the series is associated with the sequence given above.
The important points you should note about the series are:
- The sequence of the series can be finite or infinite.
- The series is offered in the compact form called sigma notation, denoted by the ‘∑’.
Arithmetic progression (A.P)
In the section of this chapter, we will learn about the arithmetic progression or arithmetic sequence. There is a series given below which follows arithmetic progression:
a1 + a2 + a3 + a4 + …… an + …..
an+1 = an + d
Where a = first term, d = common difference of A.P
There are some properties that verify an A.P. They are as follows:
- If a constant is added to each term of A.P., then the resulting sequence will be in A.P
- If a constant is subtracted from each term of A.P., then the resulting sequence will be in A.P
- If a constant is multiplied by each term of A.P., then the resulting sequence will be in A.P
- If a constant/nonzero is divided fry each term of A.P., then the resulting sequence will be in A.P
The series given below follows arithmetic progression:
a, a + d, a + 2d, a + 3d…a + (n + 1)d
- The last number of A.P,
l = a + (n – 1)d
- The sum of n numbers in A.P,
Sn = n/2 [2a + (n – 1)d]
Sn = n/2 [a + l]
If we have a sequence, say a, b & c, then the arithmetic mean is given by b = (a + c) /2
Geometric progressions (G.P)
In this part of the chapter, we will learn about the Geometric progression or Geometric sequence.
There is a series given below which follows arithmetic progression:
a1 + a2 + a3 + a4 + …… an + …..
ak+1 / ak = r (constant), for k ≥ 1
Where, a = first term, r = common ratio of G.P
The series given below follows a Geometric progression:
a, ar, ar2, ar3,…..
- General ‘n’ term of a G.P,
an = arn-1
- The sum of n numbers in G.P,
Sn = a(1-rn) / (1-r)
Sn = a(rn-1) / (r-1)
If we have a sequence, say a, b & c, then the geometric mean is given by b = √a.c
Relationship between A.M and G.M
Let us check A.M & G.M.’s relationship of two given numbers a and b.
A.M = (a + b) /2
G.M = √a.b
Then, A.M – G.M = (a + b) /2 – √a.b = (√a – √b)2 / 2 ≥ 0.
This is the relationship between A.M ≥ G.M
Sum to ‘n’ terms of special series.
There is a special series, and we will find the sum of ‘n’ terms of this series.
- 1 + 2 + 3 + 4 +…….+n (sum of first ‘n’ numbers)
Sn = n(n + 1) / 2
- 12 + 22 + 32 + 42 +…….+n2 (sum of first square ‘n’ numbers)
Sn = n(n + 1)(2n + 1) / 6
- 13 + 23 + 33 + 43 +…….+n3(sum of first cube ‘n’ numbers)
Sn = [n(n + 1)]2 / 4
NCERT Solutions for Class 11 Mathematics Chapter 9 Exercise & Solutions
The NCERT Solutions for Class 11 Mathematics Chapter 9 is mostly practice-oriented. The more you practice, the better you will get. . Hence, multiple exercises are given in this chapter in the NCERT textbook. You must have all the accurate solutions to these exercises. As a result, we have provided a detailed solution to every question given in the chapter in the NCERT Solutions for Class 11 Mathematics Chapter 9 available on the Extramarks’ website. The solutions will give you a complete analysis of the steps you should follow when solving the problems related to sequence and series. Thus, helping you to develop clear mindsets while solving the problems. .
Click on the link below to view exercise-specific questions and solutions for NCERT Solutions for Class 11 Mathematics Chapter 9:
- Chapter 9 Class 11 Mathematics: Exercise 9.1
- Chapter 9 Class 11 Mathematics: Exercise 9.2
- Chapter 9 Class 11 Mathematics: Exercise 9.3
- Chapter 9 Class 11 Mathematics: Exercise 9.4
- Chapter 9 Class 11 Mathematics: Miscellaneous Exercise
Along with Class 11 Mathematics solutions, you can explore NCERT Solutions on our Extramarks’ website for all primary and secondary classes.
- NCERT Solutions Class 1,
- NCERT Solutions Class 2,
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- NCERT Solutions Class 9,
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NCERT Exemplar Class 11 Mathematics
NCERT Class 11th Mathematics lays a strong foundation for the aspirants preparing for competitive examinations like JEE, NEET, KVPY, WBJEE and many other engineering and medical-related examinations. To score well and be in the top percentile in these examinations, students must have a clear-cut understanding of every concept covered in Class 11 Mathematics.
To excel in it, students need to do rigorous practice. Hence, they must have access to NCERT-related questions. NCERT Exemplar Class 11th Mathematics thus has all NCERT-based questions with solutions. . Students can find each topic covered with various types of questions with varying difficulty levels that they will face in the examination.
Moreover, they will also find tricks to solve these questions in less time with greater accuracy and will become more proficient in their calculations. The subject matter experts have designed the book after analysing the past year’s papers and the competitive examinations. The challenging questions will aid in developing the strong mindsets among the students, and as a result, they will start thinking more rationally. Thus, the book builds students’ approach, making them capable of solving difficult questions with ease and becoming more confident in the process.
Key Features of NCERT Solutions for Class 11 Mathematics Chapter 9
The more you revise, the more you retain. Hence, Extramarks NCERT Solutions for Class 11 Mathematics Chapter 9 provides a complete revision guide to every student irrespective of their level. Along with the revision guides, students will also get access to study notes, solved questions from NCERT textbook and Exemplars, and so on.
The key features are as follows:
- The entire chapter has been summarised in a point-wise manner in our NCERT solutions.
- Our NCERT solutions are prepared by subject matter experts working conscientiously and diligently to prepare authentic, concise answers which students can trust and enjoy the process of learning.
- All the important formulas are listed in a structured way for the students to revise quickly.
- After completing the NCERT Solutions for Class 11 Mathematics Chapter 9, students will be able to apply all the concepts related to sequence and series in other chapters.
Q.1 Write the first five terms of the sequence whose nth terms is:
an = n(n+2)
Q.3 Write the first five terms of the sequence whose nth terms is:
an = 2n
Q.5 Write the first five terms of the sequence whose nth terms is an = (– 1)n–1 5n+1.
Q.10 Find the sum of odd integers from 1 to 2001.
Q.11 Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
Q.12 In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is –112.
Q.15 If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term.
Q.16 Find the sum to n terms of the A.P., whose kth term is 5k + 1.
Q.17 If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants, find the common difference.
Q.18 The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18th terms.
Q.19 If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.
Q.21 The ratio of the sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of mth and nth term is (2m – 1) : (2n – 1).
Q.22 If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.
Q.23 Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Q.25 Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A. P. and the ratio of 7th and (m – 1)th numbers is 5 : 9. Find the value of m.
Q.26 The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.
Q.28 Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.
Q.29 The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.
Q.30 The 4th term of a G.P. is square of its second term, and the first term is –3. Determine its 7th term.
Q.36 How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?
Q.37 The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128.
Determine the first term, the common ratio and the sum to n terms of the G.P.
Q.38 Given a G.P. with a = 729 and 7th term 64, determine S7.
Q.39 Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.
Q.40 If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.
Q.41 Find the sum to n terms of the sequence, 8, 88, 888, 8888… .
Q.43 Show that the products of the corresponding terms of the sequences a, ar, ar2, … arn – 1 and A, AR, AR2, … ARn–1 form a G.P, and find the common ratio.
Q.44 Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.
Q.46 If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab)n.
Q.48 If a, b, c and d are in G.P. show that (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2.
Q.49 Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Q.53 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 2nd hour, 4th hour and nth hour?
Q.54 What will â‚¹ 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
Q.55 If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
Q.56 Find the sum to n terms of each of the series in Exercises 1 to 7.
1. 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 +…
2. 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
3. 3 × 12 + 5 × 22 + 7 × 32 + …
5. 52 + 62 + 72 + … + 202
6. 3 × 8 + 6 × 11 + 9 × 14 + …
7. 12 + (12 + 22) + (12 + 22 + 32) + …
Q.57 Find the sum to n terms of the series in whose nth terms is given by:
1. n (n + 1) (n + 4).
2. n2 + 2n
3. (2n – 1)2
Q.59 Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.
Q.60 If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.
Q.61 Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that S3 = 3(S2 – S1).
Q.62 Find the sum of all numbers between 200 and 400 which are divisible by 7.
Q.63 Find the sum of integers from 1 to 100 that are divisible by 2 or 5.
Q.64 Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.
Q.66 The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.
Q.67 The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.
Q.68 The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.
Q.69 A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.
Q.70 The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.
Q.72 Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P2Rn = Sn.
Q.73 The pth, qth and rth terms of an A.P. are a, b, c, respectively. Show that (q – r )a + (r – p )b + (p – q )c = 0.
Q.75 If a, b, c, d are in G.P, prove that (an + bn), (bn + cn), (cn + dn) are in G.P.
Q.76 If a and b are the roots of x2 – 3x + p = 0 and c, d are roots of x2 – 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (q – p) = 17:15.
Q.79 Find the sum of the following series up to n terms:
(i) 5 + 55 + 555 + …
(ii) .6 +. 66 +. 666+…
Q.80 Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + … + n terms.
Q.81 Find the sum of the first n terms of the series: 3+ 7 + 13 + 21 + 31 +…
Q.85 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.
FAQs (Frequently Asked Questions)
1. How can I make my NCERT Class 11 Mathematics strong?
NCERT Class 11 Mathematics lays the foundation for NCERT Class 12 Mathematics. Students can strengthen their Class 11 Mathematics by building a solid conceptual understanding of all the topics covered in the NCERT textbook and by practising problems regularly. They can also refer to NCERT Solutions Class 11 Mathematics Chapter 9 available on the Extramarks’ website.
2. How will the NCERT Solutions for Class 11 Mathematics Chapter 9 benefit me?
Extramarks NCERT Solutions for Class 11 Mathematics Chapter 9 will benefit by providing you with detailed coverage of every concept covered in the chapter. You will find a lot of questions to practice that will help in building strong conceptual understanding and thereby will help you bring enhanced accuracy while doing calculations.
So along with the NCERT textbooks, students can confidently rely on Extramarks NCERT Solutions for their regular studies and exam preparation.