CBSE Class 11 Maths Chapter 9: Sequences and Series NCERT Solutions

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NCERT Solutions PDF Class 11 PDF

This resource provides detailed NCERT Solutions for Chapter 9, Sequences and Series, for CBSE Class 11 Mathematics. It focuses on understanding and calculating the first five terms of various sequences when the nth term is given. The solutions cover different types of nth term formulas, including polynomial, rational, and exponential expressions. Students will learn to substitute values of n (from 1 to 5) into the given formula to generate the initial terms of each sequence. This chapter is fundamental for grasping arithmetic progressions, geometric progressions, and other series concepts introduced later. These step-by-step solutions are designed to aid students in comprehending the basic principles of sequences and preparing effectively for their examinations.

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Quick info

BoardCBSE
ClassClass 11
Subjectगणित
Session2026
LanguageEnglish
TypeNCERT Solutions
Chapter9. अनुक्रम तथा श्रेणी

Chapter summary

Chapter 9, Sequences and Series, introduces the fundamental concept of sequences and how to determine their terms. This section of NCERT Solutions specifically addresses finding the first five terms of sequences based on their general nth term formula. It covers various forms of the nth term, requiring students to perform substitutions and calculations. Mastering these basics is crucial for understanding the properties and applications of different types of series.

Learning outcomes

  • Understand the definition of a sequence and its nth term.
  • Calculate the first five terms of a sequence given its nth term formula.
  • Apply substitution methods to find specific terms of a sequence.
  • Recognize and work with different types of nth term expressions (polynomial, rational, exponential).
  • Develop foundational skills for studying arithmetic and geometric progressions.

Topics covered

Paper topics

  • Sequences
  • nth term of a sequence
  • Calculating sequence terms
  • Polynomial nth terms
  • Rational nth terms
  • Exponential nth terms
  • Introduction to series

Important topics

  • Understanding the nth term formula
  • Calculating the first five terms
  • Substitution in algebraic expressions
  • Working with fractions in sequences
  • Working with exponents in sequences

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Questions and Solutions

Question 1

Write the first five terms of the sequence whose nth term is given by $a_n = n(n+2)$.
Solution:

We are given the nth term of the sequence as $a_n = n(n+2)$. To find the first five terms, we substitute $n = 1, 2, 3, 4, 5$ into the formula:

  1. For $n=1$: $a_1 = 1(1+2) = 1(3) = 3$
  2. For $n=2$: $a_2 = 2(2+2) = 2(4) = 8$
  3. For $n=3$: $a_3 = 3(3+2) = 3(5) = 15$
  4. For $n=4$: $a_4 = 4(4+2) = 4(6) = 24$
  5. For $n=5$: $a_5 = 5(5+2) = 5(7) = 35$

Thus, the first five terms of the sequence are 3, 8, 15, 24, and 35.

Question 2

Write the first five terms of the sequence whose nth term is given by $a_n = \frac{n}{n+1}$.
Solution:

The nth term of the sequence is given by $a_n = \frac{n}{n+1}$. We will find the first five terms by substituting $n = 1, 2, 3, 4, 5$:

  1. For $n=1$: $a_1 = \frac{1}{1+1} = \frac{1}{2}$
  2. For $n=2$: $a_2 = \frac{2}{2+1} = \frac{2}{3}$
  3. For $n=3$: $a_3 = \frac{3}{3+1} = \frac{3}{4}$
  4. For $n=4$: $a_4 = \frac{4}{4+1} = \frac{4}{5}$
  5. For $n=5$: $a_5 = \frac{5}{5+1} = \frac{5}{6}$

The first five terms of the sequence are $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \text{ and } \frac{5}{6}$.

Question 3

Write the first five terms of the sequence whose nth term is given by $a_n = 2^n$.
Solution:

We are given the nth term of the sequence as $a_n = 2^n$. To find the first five terms, we substitute $n = 1, 2, 3, 4, 5$ into the formula:

  1. For $n=1$: $a_1 = 2^1 = 2$
  2. For $n=2$: $a_2 = 2^2 = 4$
  3. For $n=3$: $a_3 = 2^3 = 8$
  4. For $n=4$: $a_4 = 2^4 = 16$
  5. For $n=5$: $a_5 = 2^5 = 32$

Therefore, the first five terms of the sequence are 2, 4, 8, 16, and 32.

Question 4

Write the first five terms of the sequence whose nth term is given by $a_n = \frac{2n-3}{6}$.
Solution:

The nth term of the sequence is given by $a_n = \frac{2n-3}{6}$. We will find the first five terms by substituting $n = 1, 2, 3, 4, 5$:

  1. For $n=1$: $a_1 = \frac{2(1)-3}{6} = \frac{2-3}{6} = \frac{-1}{6}$
  2. For $n=2$: $a_2 = \frac{2(2)-3}{6} = \frac{4-3}{6} = \frac{1}{6}$
  3. For $n=3$: $a_3 = \frac{2(3)-3}{6} = \frac{6-3}{6} = \frac{3}{6} = \frac{1}{2}$
  4. For $n=4$: $a_4 = \frac{2(4)-3}{6} = \frac{8-3}{6} = \frac{5}{6}$
  5. For $n=5$: $a_5 = \frac{2(5)-3}{6} = \frac{10-3}{6} = \frac{7}{6}$

The first five terms of the sequence are $\frac{-1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}, \text{ and } \frac{7}{6}$.

Common mistakes

  • Errors in basic arithmetic calculations during substitution.
  • Incorrectly simplifying fractions or exponents.
  • Misinterpreting the nth term formula.
  • Forgetting to calculate all five required terms.

Revision tips

  • Practice substituting n=1, 2, 3, 4, 5 carefully into each formula.
  • Double-check all arithmetic operations, especially with fractions and exponents.
  • Write down each step clearly to avoid calculation errors.
  • Review the relationship between the nth term formula and the resulting sequence terms.

Practice MCQs

Q1. If the nth term of a sequence is given by $a_n = n(n+2)$, what is the 3rd term?

Q2. For the sequence with nth term $a_n = \frac{n}{n+1}$, what is the 4th term?

Q3. What is the 5th term of the sequence where $a_n = 2^n$?

Q4. If $a_n = \frac{2n-3}{6}$, which term is equal to $\frac{1}{2}$?

Frequently asked questions

What is the main objective of these NCERT Solutions for Class 11 Maths Chapter 9?

The main objective is to help students understand how to find the first five terms of a sequence when the general formula for the nth term is provided. It covers various types of formulas and calculation methods.

How do these solutions help in preparing for exams?

These solutions provide clear, step-by-step methods for solving problems related to sequences. Practicing these will build a strong foundation for understanding more complex series concepts and improve calculation accuracy for exam questions.

What types of nth term formulas are covered in these exercises?

The exercises cover nth term formulas that are polynomial (like $n(n+2)$), rational (like $\frac{n}{n+1}$), and exponential (like $2^n$), among others.

What is the role of substituting 'n' values in these problems?

Substituting values of 'n' (specifically 1, 2, 3, 4, and 5 as requested) into the nth term formula allows us to calculate the corresponding terms of the sequence.

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