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

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NCERT Solutions PDF Class 11 PDF

CBSE Class 11 Mathematics Chapter 9, Sequences and Series, introduces students to the fundamental concept of ordered lists of numbers. This section of the NCERT Solutions focuses on Exercise 9.1, providing a clear pathway to understanding how to generate the initial terms of various sequences. By working through problems that utilize nth term formulas, students learn to substitute values of 'n' (typically from 1 to 5) into given expressions. These expressions can be algebraic, exponential, or fractional, showcasing the diverse ways sequences can be defined. The systematic approach detailed in these solutions helps solidify the understanding that a general formula dictates the terms of a sequence. Mastering these basics is vital for progressing to more complex topics like arithmetic and geometric progressions, making these solutions an invaluable tool for exam preparation and building a robust mathematical foundation.

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Quick info

BoardCBSE
ClassClass 11
SubjectMathematics
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterChapter 9

Chapter summary

Chapter 9, Sequences and Series, introduces students to the fundamental concept of sequences and how to generate their terms. This section of NCERT Solutions specifically covers Exercise 9.1, focusing on writing the first five terms of sequences given their nth term formulas. The problems involve various types of formulas, including polynomial, rational, and exponential expressions. Students will learn to substitute n=1, 2, 3, 4, 5 into these formulas to find the initial sequence elements. This exercise is key to understanding the definition of a sequence and its relationship with its general term.

Learning outcomes

  • Understand the concept 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 different types of sequence formulas (algebraic, fractional, exponential).
  • Develop problem-solving skills for sequence generation.

Topics covered

Paper topics

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

Important topics

  • Understanding the nth term formula
  • Calculating the first five terms
  • Substitution in sequence formulas
  • Basic arithmetic operations for sequences

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Questions and Solutions

Question 1

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

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

The \(n^{th}\) term of the sequence is defined by \(a_n = \frac{n}{n+1}\). We find the first five terms by substituting \(n = 1, 2, 3, 4,\) and \(5\):

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

Thus, 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 \(a_n = 2^n\).
Solution:

The formula for the \(n^{th}\) term is \(a_n = 2^n\). To determine the first five terms, we substitute \(n = 1, 2, 3, 4,\) and \(5\):

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 \(n=5\): \(a_5 = 2^5 = 32\)

The first five terms of this 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 \(a_n = \frac{2n-3}{6}\).
Solution:

We are given the \(n^{th}\) term formula \(a_n = \frac{2n-3}{6}\). We calculate the first five terms by substituting \(n = 1, 2, 3, 4,\) and \(5\):

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 basic arithmetic calculations when substituting values of n.
  • Incorrectly simplifying fractions or exponents.
  • Misinterpreting the nth term formula, leading to wrong substitutions.
  • Forgetting to list all the required first five terms.

Revision tips

  • Practice substituting different values of 'n' into various nth term formulas.
  • Ensure all calculations, especially with fractions and exponents, are double-checked.
  • Write down each step clearly, from substitution to the final term calculation.
  • Review the definition of a sequence and its nth term before starting the problems.

Practice MCQs

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

Q2. The first five terms of the sequence \(a_n = \frac{n}{n+1}\) are:

Q3. Which sequence has the nth term \(a_n = 2^n\)?

Q4. What is the 4th term of the sequence \(a_n = \frac{2n-3}{6}\)?

Frequently asked questions

What is the main goal of Exercise 9.1 in Chapter 9?

The main goal of Exercise 9.1 is to help students understand how to generate the first five terms of a sequence when the formula for the nth term is provided.

How do I find the terms of a sequence if I'm given the nth term formula?

To find the terms, you substitute the position number (n=1 for the first term, n=2 for the second, and so on) into the given nth term formula and calculate the result.

Are there any specific types of sequences covered in these solutions?

Yes, these solutions cover sequences defined by algebraic expressions like \(n(n+2)\), fractional expressions like \(\frac{n}{n+1}\), and exponential expressions like \(2^n\).

Why is it important to rewrite the solutions?

Rewriting solutions ensures clarity, provides more detailed explanations, and helps students understand the underlying concepts better, rather than just copying an answer.

What mathematical skills are practiced in these problems?

These problems primarily practice substitution skills and basic arithmetic operations, including working with fractions and exponents.

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