CBSE Class 11 Maths 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, delves into the fascinating world of ordered numbers and their sums. This chapter is all about understanding how sequences are formed and how to find specific terms within them. You'll learn to use the nth term formula to calculate the initial terms of different sequences, whether they're defined by simple algebraic rules or more complex exponential patterns. The NCERT Solutions offer clear, step-by-step explanations, making it easy to grasp the notation and the process of generating terms. Mastering these concepts is key to building a solid foundation for future mathematical explorations, especially in areas like progressions and series, which are vital for your exams.

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

Chapter summary

Chapter 9, "Sequences and Series," in NCERT Class 11 Maths, covers the basics of sequences. The provided solutions focus on Exercise 9.1, guiding students to calculate the first five terms of a sequence when the general nth term (a_n) is given. This involves substituting n = 1, 2, 3, 4, and 5 into the formula and simplifying the results. The exercises cover various types of nth term formulas, including polynomial, rational, and exponential forms.

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 to find specific terms of a sequence.
  • Simplify results obtained from sequence formulas.
  • Recognize different types of sequence formulas (algebraic, exponential).

Topics covered

Paper topics

  • Sequences
  • nth term of a sequence
  • Generating sequence terms
  • Algebraic sequences
  • Exponential sequences
  • Rational sequences

Important topics

  • Understanding the nth term formula
  • Calculating the first five terms
  • Substitution in formulas
  • Simplification of terms

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Questions and Solutions

Question 1

For the sequence where the nth term is given by $a_n = n(n+2)$, write the first five terms.
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

For the sequence where the nth term is given by $a_n = \frac{n}{n+1}$, write the first five terms.
Solution:

The nth term of the sequence is given by the formula $a_n = \frac{n}{n+1}$. 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 this sequence are $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}$.

Question 3

For the sequence where the nth term is given by $a_n = 2^n$, write the first five terms.
Solution:

We are provided with the nth term of the sequence as $a_n = 2^n$. To determine the first five terms, we substitute $n = 1, 2, 3, 4, 5$ into this 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

For the sequence where the nth term is given by $a_n = \frac{2n-3}{6}$, write the first five terms.
Solution:

The nth term of the sequence is given by the formula $a_n = \frac{2n-3}{6}$. We will calculate 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 this sequence are $\frac{-1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}, \frac{7}{6}$.

Common mistakes

  • Errors in substituting values of 'n' into the formula.
  • Mistakes in arithmetic calculations, especially with exponents or fractions.
  • Incorrect simplification of the resulting terms.
  • Misinterpreting the 'nth term' concept for generating sequence elements.

Revision tips

  • Practice substituting 'n' values systematically for each question.
  • Double-check all arithmetic calculations, especially for fractions and powers.
  • Write down each step clearly to avoid confusion.
  • Review the formula for the nth term before starting calculations for each sequence.

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 $a_n = \frac{n}{n+1}$, what is the 4th term?

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

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

Frequently asked questions

What is the main goal of Exercise 9.1 in NCERT Class 11 Maths?

The main goal of Exercise 9.1 is to help students understand how to find the first five terms of a sequence when the general formula for the nth term ($a_n$) is provided.

How do I find the terms of a sequence if the nth term is given?

To find the terms of a sequence, you substitute the desired term number (like 1, 2, 3, etc.) for 'n' in the given nth term formula and then calculate the result.

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

This exercise uses various types of formulas for the nth term, including those that are polynomial (like $n(n+2)$), rational (like $\frac{n}{n+1}$), and exponential (like $2^n$).

Why is it important to keep the mathematical expressions exact in the solutions?

Keeping mathematical expressions exact ensures that the method and the final answer are accurate and verifiable, helping students learn the correct procedure without any alterations.

How can these solutions help with exam preparation?

These solutions provide clear, step-by-step explanations for each problem, reinforcing the understanding of how to generate sequence terms. Practicing these will build confidence and accuracy for exam questions on sequences.

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