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

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NCERT Solutions PDF Class 11 PDF

CBSE Class 11 Mathematics Chapter 9: Sequences and Series introduces students to the fundamental concepts of sequences. This section of the NCERT Solutions focuses on Exercise 9.1, guiding learners through the process of finding the initial terms of a sequence when its general (nth) term is provided. The solutions meticulously demonstrate how to substitute values for 'n' into various formulas, which may include algebraic expressions and exponents, to calculate each term sequentially. This step-by-step approach ensures a clear understanding of sequence notation and the mechanics of term generation. Mastering these basics is essential for progressing to more advanced topics in series. These explanations are crafted to build a solid foundation, aiding students in their preparation for examinations and ensuring they can confidently tackle sequence-related problems.

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BoardCBSE
ClassClass 11
SubjectMathematics
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterChapter 9: Sequences and Series

Chapter summary

Chapter 9, Sequences and Series, begins with an introduction to sequences in Exercise 9.1. The NCERT Solutions for this exercise focus on understanding the definition of a sequence and how to find its initial terms. Students will practice calculating the first five terms of sequences defined by explicit formulas for the nth term. This foundational exercise helps in grasping the concept of a sequence as an ordered list of numbers and the relationship between the term number and its value.

Learning outcomes

  • Understand the definition of a sequence.
  • Determine the nth term of a sequence given its formula.
  • Calculate the first five terms of a sequence.
  • Apply algebraic and exponential formulas to find sequence terms.
  • Represent sequences as ordered lists of numbers.

Topics covered

Paper topics

  • Sequences
  • nth term of a sequence
  • Generating sequence terms
  • Algebraic sequence formulas
  • Exponential sequence formulas
  • Exercise 9.1

Important topics

  • Understanding the nth term formula
  • Calculating the first five terms
  • Substitution of values for n
  • Basic sequence generation

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Questions and Solutions

Question 1

Write the first five terms of the sequence whose \(n^{th}\) term is given by the formula \(a_n = n(n + 2)\).
Solution:

The \(n^{th}\) term of the sequence is given by \(a_n = n(n + 2)\). To find the first five terms, we substitute \(n = 1, 2, 3, 4,\) and \(5\) into the formula:

For \(n=1\): \(a_1 = 1(1 + 2) = 1(3) = 3\)

For \(n=2\): \(a_2 = 2(2 + 2) = 2(4) = 8\)

For \(n=3\): \(a_3 = 3(3 + 2) = 3(5) = 15\)

For \(n=4\): \(a_4 = 4(4 + 2) = 4(6) = 24\)

For \(n=5\): \(a_5 = 5(5 + 2) = 5(7) = 35\)

Therefore, 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 \(n^{th}\) term is given by the formula \(a_n = \frac{n}{n+1}\).
Solution:

The \(n^{th}\) term of the sequence is given by \(a_n = \frac{n}{n+1}\). To find the first five terms, we substitute \(n = 1, 2, 3, 4,\) and \(5\) into the formula:

For \(n=1\): \(a_1 = \frac{1}{1+1} = \frac{1}{2}\)

For \(n=2\): \(a_2 = \frac{2}{2+1} = \frac{2}{3}\)

For \(n=3\): \(a_3 = \frac{3}{3+1} = \frac{3}{4}\)

For \(n=4\): \(a_4 = \frac{4}{4+1} = \frac{4}{5}\)

For \(n=5\): \(a_5 = \frac{5}{5+1} = \frac{5}{6}\)

Therefore, the first five terms of the sequence are \(\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5},\) and \(\frac{5}{6}\).

Question 3

Write the first five terms of the sequence whose \(n^{th}\) term is given by the formula \(a_n = 2^n\).
Solution:

The \(n^{th}\) term of the sequence is given by \(a_n = 2^n\). To find the first five terms, we substitute \(n = 1, 2, 3, 4,\) and \(5\) into the formula:

For \(n=1\): \(a_1 = 2^1 = 2\)

For \(n=2\): \(a_2 = 2^2 = 4\)

For \(n=3\): \(a_3 = 2^3 = 8\)

For \(n=4\): \(a_4 = 2^4 = 16\)

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 \(n^{th}\) term is given by the formula \(a_n = \frac{2n-3}{6}\).
Solution:

The \(n^{th}\) term of the sequence is given by \(a_n = \frac{2n-3}{6}\). To find the first five terms, we substitute \(n = 1, 2, 3, 4,\) and \(5\) into the formula:

For \(n=1\): \(a_1 = \frac{2(1) - 3}{6} = \frac{2 - 3}{6} = \frac{-1}{6}\)

For \(n=2\): \(a_2 = \frac{2(2) - 3}{6} = \frac{4 - 3}{6} = \frac{1}{6}\)

For \(n=3\): \(a_3 = \frac{2(3) - 3}{6} = \frac{6 - 3}{6} = \frac{3}{6} = \frac{1}{2}\)

For \(n=4\): \(a_4 = \frac{2(4) - 3}{6} = \frac{8 - 3}{6} = \frac{5}{6}\)

For \(n=5\): \(a_5 = \frac{2(5) - 3}{6} = \frac{10 - 3}{6} = \frac{7}{6}\)

Therefore, the first five terms of the sequence are \(\frac{-1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6},\) and \(\frac{7}{6}\).

Common mistakes

  • Errors in substituting values of 'n' into the formula.
  • Calculation mistakes, especially with fractions or exponents.
  • Incorrectly simplifying fractions.
  • Misinterpreting the 'nth' term formula.

Revision tips

  • Practice substituting different values of 'n' into the given formulas.
  • Double-check all arithmetic calculations, especially for fractions and powers.
  • Write down each step clearly to avoid errors.
  • Review the definition of a sequence and its notation.

Practice MCQs

Q1. What is the 3rd term of the sequence with nth term \(a_n = n(n+2)\)?

Q2. If the nth term of a sequence is \(a_n = \frac{n}{n+1}\), what is the 5th term?

Q3. Which of the following is the 4th term of the sequence \(a_n = 2^n\)?

Q4. What is the first term of the sequence \(a_n = \frac{2n-3}{6}\)?

Q5. For the sequence \(a_n = n(n+2)\), the difference between the 2nd and 1st term is:

Frequently asked questions

What is the main focus of these NCERT Solutions for Chapter 9?

These solutions focus on Exercise 9.1 of Chapter 9: Sequences and Series for CBSE Class 11 Mathematics. They provide detailed steps to find the first five terms of sequences when the general nth term is given.

How do these solutions help in understanding sequences?

The solutions demonstrate how to substitute values of 'n' (from 1 to 5) into the given nth term formula and calculate the corresponding term of the sequence, making the concept of sequence generation clear.

Are the questions in these solutions the same as in the NCERT textbook?

Yes, the questions are kept exactly the same as in the NCERT textbook, including their numbering and problem statement. The wording has been expanded for clarity where needed.

What types of formulas are used for the nth term in this exercise?

This exercise includes formulas that are algebraic (like \(n(n+2)\) or \(\frac{2n-3}{6}\)) and exponential (like \(2^n\)).

How can these solutions be used for exam preparation?

These solutions offer a clear, step-by-step approach to solving basic sequence problems, which are fundamental for understanding more advanced topics in series. Practicing these will build confidence for exam questions on sequences.

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