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NCERT Solutions Class 11 Maths Chapter 9 Sequences And Series – Here are all the NCERT solutions for Class 11 Maths Chapter 9. This solution contains questions, answers, images, explanations of the complete chapter 9 titled Of Sequences And Series taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Maths, then you must come across chapter 9 Sequences And Series After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Maths Chapter 9 Sequences And Series in one place.

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NCERT Solutions Class 11 Maths Chapter 9 Sequences And Series

Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 11 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 9 Sequences And Series , Maths, Class 11.

Class 11
Subject Maths
Book Mathematics
Chapter Number 9
Chapter Name

Sequences And Series

NCERT Solutions Class 11 Maths chapter 9 Sequences And Series

Class 11, Maths chapter 9, Sequences And Series solutions are given below in PDF format. You can view them online or download PDF file for future use.

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Question & Answer

Q.1: Write the first five terms of the sequences whose \(n^{th}\) term is \(a_{n}=n(n+2)\).

Ans : \(a_{n}=n(n+2)\) 
Substituting n = 1, 2, 3, 4, and 5, we obtain 
\(\begin{aligned} a_{1} &=1(1+2)=3 \\ a_{2} &=2(2+2)=8 \\ a_{3} &=3(3+2)=15 \\ a_{4} &=4(4+2)=24 \\ a_{5} &=5(5+2)=35 \end{aligned}\)
Therefore, the required terms are 3, 8, 15, 24, and 35. 
Q.2: Write the first five terms of the sequences whose \(n^{th}\) term is \(a_{n}=\frac{n}{n+1}\).

Ans : \(a_{n}=\frac{n}{n+1}\)
Substituting n = 1, 2, 3, 4, 5, we obtain
\(a_{1}=\frac{1}{1+1}=\frac{1}{2}, a_{2}=\frac{2}{2+1}=\frac{2}{3}, a_{3}=\frac{3}{3+1}=\frac{3}{4}, a_{4}=\frac{4}{4+1}=\frac{4}{5}, a_{5}=\frac{5}{5+1}=\frac{5}{6}\)
Therefore, the required terms are \(\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \text { and } \frac{5}{6}\) 
Q.3: Write the first five terms of the sequences whose \(n^{th}\) term is \(a_{n}=2^{n}\)

Ans : \(a_{n}=2^{n}\)
Substituting n = 1, 2, 3, 4, 5, we obtain
\(\begin{array}{l}{a_{1}=2^{1}=2} \\ {a_{2}=2^{2}=4} \\ {a_{3}=2^{3}=8} \\ {a_{4}=2^{4}=16} \\ {a_{5}=2^{5}=32}\end{array}\)
Therefore, the required terms are 2, 4, 8, 16, and 32. 
Q.4: Write the first five terms of the sequences whose \(n^{th}\) term is \(a_{n}=\frac{2 n-3}{6}\)

Ans : Substituting n = 1, 2, 3, 4, 5, we obtain 
\(\begin{aligned} a_{1} &=\frac{2 \times 1-3}{6}=\frac{-1}{6} \\ a_{2} &=\frac{2 \times 2-3}{6}=\frac{1}{6} \\ a_{3} &=\frac{2 \times 3-3}{6}=\frac{3}{6}=\frac{1}{2} \\ a_{4} &=\frac{2 \times 4-3}{6}=\frac{5}{6} \\ a_{5} &=\frac{2 \times 5-3}{6}=\frac{7}{6} \end{aligned}\)
Therefore, the required terms are \(\frac{-1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}, \text { and } \frac{7}{6}\) 
Q.5: Write the first five terms of the sequences whose \(n^{th}\) term is \(a_{n}=(-1)^{n-1} 5^{n+1}\)

Ans : Substituting n = 1, 2, 3, 4, 5, we obtain
\(\begin{array}{l}{a_{1}=(-1)^{1-1} 5^{1+1}=5^{2}=25} \\ {a_{2}=(-1)^{2-1} 5^{2+1}=-5^{3}=-125} \\ {a_{3}=(-1)^{3-1} 5^{3+1}=5^{4}=625} \\ {a_{4}=(-1)^{5-1} 5^{4+1}=-5^{5}=-3125} \\ {a^{5}=(-1)^{5-1} 5^{5+1}=5^{6}=15625}\end{array}\)
Therefore, the required terms are 25, –125, 625, –3125, and 15625.

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