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Looking for Ex 9.1 Class 11 Maths NCERT Solutions? You’ve come to the right place! This section provides detailed, step-by-step solutions for all questions from Exercise 9.1 of Chapter 9 – Sequences and Series. Aligned with the latest NCERT syllabus, these solutions help you understand the basics of sequences, general terms, and patterns that form the foundation of arithmetic and geometric progressions. Whether you’re solving problems from the Class 11 Ch 9 Exercise 9.1, using the NCERT Exemplar Class 11 Maths, or cross-checking your answers with RD Sharma Exercise 9.1 Solutions, this resource is perfect for concept clarity and exam preparation. Download or view the complete NCERT Solutions for Class 11 Maths Chapter 9 and strengthen your grasp on Sequences and Series Class 11 today!
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Exercise 9.1
1. Write the first five terms of the sequences whose \( {n}^{\text {th}} \) term is \( {a}_{{n}}={n}({n}+2) \).
Answer
\(a_{n}=n(n+2)\)
substituting \( {n}=1,2,3,4 \), and \(5 ,\) we obtain
\(a_{1}=1(1+2)=3\)
\(a_{2}=2(2+2)=8\)
\(a_{3}=3(3+2)=15\)
\(a_{4}=4(4+2)=24\)
\(a_{5}=5(5+2)=35\)
Therefore, the required terms are \( 3,8,15,24,35 \).
2. Write the first five terms of the sequences whose \( {n}^{\text {th}} \) term is \( {a}_{{n}}=\frac{n}{n+1} \)
Answer
\({a}_{{n}}=\frac{n}{n+1}\)
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 \( \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5} \), and \( \frac{5}{6} \)
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3. Write the first five terms of the sequences whose \( {n}^{\text {th}} \) term is \( {a}_{{n}}=2^{{n}} \)
Answer
\(a_{n}=2^{n}\)
substituting \( {n}=1,2,3,4,5 \), we obtain
\(\begin{array}{l}
a_{1}=2^{1}=2 \\
a_{2}=2^{2}=4 \\
a_{3}=2^{3}=8 \\
a_{4}=2^{4}=16 \\
a_{5}=2^{5}=25
\end{array}\)
therefore, the required terms are \( 2,4,8,16 \), and \(32 \).
4. Write the first five terms of the sequences whose \( {n}^{\text {th}} \) term is \( \frac{2 n-3}{6} \)
Answer
Substituting \( {n}=1,2,3,4,5 \), we obtain
\( {a}_{1}=\frac{2 \times 1-3}{6}=\frac{-1}{6} \)
\( {a}_{2}=\frac{2 \times 2-3}{6}=\frac{1}{6} \)
\( {a}_{3}=\frac{2 \times 3-3}{6}=\frac{1}{2} \)
\( {a}_{4}=\frac{2 \times 4-3}{6}=\frac{5}{6} \)
\( {a}_{5}=\frac{2 \times 5-3}{6}=\frac{7}{6} \)
Therefore, the required terms are \( , \frac{-1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6} \) and \( \frac{7}{6} \)
5. Write the first five terms of the sequences whose \( {n}^{\text {th}} \) term is \(a_{n}=(-1)^{n-1} 5^{n+1}\)
Answer
Substituting \( {n}=1,2,3,4,5 \), we obtain
\(a_{1}=(-1)^{1-1} 5^{1+1}=5^{2}=25\)
\(a_{2}=(-1)^{2-1} 5^{2+1}=-5^{3}=-125\)
\(a_{3}=(-1)^{3-1} 5^{3+1}=5^{4}=625\)
\(a_{4}=(-1)^{4-1} 5^{4+1}=-5^{5}=-3125\)
\(a_{5}=(-1)^{5-1} 5^{5+1}=5^{6}=15625\)
therefore, the required terms are \( 25,-125,625,-3125 \), and \(15625 \).
6. Write the first five terms of the sequences whose \( {n}^{\text {th}} \) term is
\({a}_{{n}}={n} \frac{n^{2}+5}{4}\)
Answer
Substituting \( {n}=1,2,3,4,5 \), we obtain
\(a_{1}=1 \cdot \frac{1^{2}+5}{4}=\frac{6}{4}=\frac{3}{2}\)
\(a_{2}=2 \cdot \frac{2^{4}+5}{4}=2 \cdot \frac{9}{4}=\frac{9}{2}\)
\(a_{3}=3 \cdot \frac{3^{2}+5}{4}=2 \cdot \frac{14}{4}=\frac{21}{2}\)
\(a_{4}=4 \cdot \frac{4^{2}+5}{4}=21\)
\(a_{5}=5 \cdot \frac{5^{2}+5}{4}=5 \cdot \frac{30}{4}=\frac{75}{2}\)
therefore, the required terms are \( \frac{3}{2}, \frac{9}{2}, \frac{21}{2}, 21 \), and \( \frac{75}{2} \).
7. Find the \( 17^{\text {th}} \) term in the following sequence whose \( {n}^{\text {th}} \) term is \( a_{n}=4 n-3 ; a_{17}, a_{24} \)
Answer
Substituting \( {n}=17 \), we obtain
\( a_{17}=4(17)-3=68-3=65 \)
Substituting \( {n}=24 \), we obtain \( {a}_{24}=4(24)-3=96-3=93 \)
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8. Find the \( 7^{\text {th}} \) term in the following sequence whose \( {n}^{\text {th}} \) term is \( {a}_{{n}}=\frac{n^{2}}{2 n} \); \( {a}_{7} \)
Answer
Substituting \( {n}=7 \), we obtain
\(a_{7}=\frac{7^{2}}{2^{7}}=\frac{49}{128}\)
9. Find the \( 9^{\text {th}} \) term in the following sequence whose \( {n}^{\text {th}} \) term is \( a_{n}=(-1)^{n-1} n^{3} ; a_{9} \)
Answer
Substituting \( {n}=9 \), we obtain
\(a_{9}=(-1)^{9-1}(9)^{3}=(9)^{3}=729\)
10. Find the \( 20^{\text {th}} \) term in the following sequence whose \( {n}^{\text {th}} \) term is
\({a}_{{n}}=\frac{n(n-2)}{n+3} ; {a}_{20}\)
Answer
Substituting \( {n}=20 \), we obtain
\({a}_{20}=\frac{20(20-2)}{20+3}=\frac{20(18)}{23}=\frac{360}{23}\)
11. Write the first five terms of the following sequence and obtain the corresponding series:
\(a_{n}=3 a_{n-1}+2 \text { for all } n > 1\)
\(a_{n}=3 a_{n-1}+2 \text { for all } n > 1\)
Answer
\(a_{1}=3, a_{n}=3 a_{n-1}+2 \text { for all } n > 1\)
\(=a_{2}=3 a_{1}+2=3(3)+2=11\)
\(=a_{3}=3 a_{2}+2=3(11)+2=35\)
\(=a_{4}=3 a_{3}+2=3(35)+2=107\)
\(=a_{5}=3 a_{4}+2=3(107)+2=323\)
Hence, the first five terms of the sequence are \( 3,11,35,107,323 \).
The corresponding series is \( 3+11+35+107+323+\ldots \)
12. Write the first five terms of the following sequence and obtain the corresponding series:
\({a}_{1}=-1, {a}_{{n}}=\frac{a_{n-1}}{n}, {n} \geq 2\)
\({a}_{1}=-1, {a}_{{n}}=\frac{a_{n-1}}{n}, {n} \geq 2\)
Answer
\({a}_{1}=-1, {a}_{{n}}=\frac{a_{n-1}}{n}, {n} \geq 2\)
\(={a}_{2}=\frac{a_{1}}{2}=\frac{-1}{2}\)
\(={a}_{3}=\frac{a_{2}}{3}=\frac{-1}{6}\)
\(={a}_{4}=\frac{a_{3}}{4}=\frac{-1}{24}\)
\(={a}_{5}=\frac{a_{4}}{4}=\frac{-1}{120}\)
Hence, the first terms of the sequence are \( -1, \frac{-1}{2}, \frac{-1}{6}, \frac{-1}{24} \) and \( \frac{-1}{120} \)
The corresponding series is \( (1)+\left(\frac{-1}{2}\right)+\left(\frac{-1}{6}\right)\) \(+\left(\frac{-1}{24}\right)+\left(\frac{-1}{120}\right) \)
13. Write the first five terms of the following sequence and obtain the corresponding series:
\( a_{1}=a_{2}=2 \), an \( =a_{n-1}-1, n > 2 \)
\( a_{1}=a_{2}=2 \), an \( =a_{n-1}-1, n > 2 \)
Answer
\(a_{1}=a_{2}=2, a_{n}=a_{n-1}-1, n > 2\)
\(=a_{3}=a_{2}-1=2-1=1\)
\(=a_{4}=a_{3}-1=1-1=0\)
\(=a_{5}=a_{4}-1=0-1=-1\)
Hence, the first five terms of the sequence are \( 2,2,1,0 \), and \(-1 \).
The corresponding series is \( 2+2+1+0+(-1)+\ldots \)
14. The Fibonacci sequence is defined by \(1 =a_{1}=a_{2} \) and \( a_{n}= {a}_{{n}-1}+{a}_{{n}-2}, {n} > 2 \) find \( \frac{a_{n+1}}{a_{n}} \), for \( {n}=1,2,3,4,5 \)
Answer
\(1=a_{1}=a_{2}\)
\(a_{n}=a_{n-1}+a_{n-2}, n > 2\)
\(=a_{3}=a_{2}+a_{1}=1+1=2\)
\(a_{4}=a_{3}+a_{2}=2+1=3\)
\(a_{5}=a_{4}+a_{3}=3+2=5\)
\(a_{6}=a_{5}+a_{4}=5+3=8\)
For \( {n}=1,=\frac{a_{n+1}}{a_{n}}=\frac{a_{2}}{a_{1}}=\frac{1}{1}=1 \)
For \( {n}=2, \frac{a_{n+1}}{a_{n}}=\frac{a_{3}}{a_{2}}=\frac{2}{1}=2 \)
For \( {n}=3, \frac{a_{n+1}}{a_{n}}=\frac{a_{4}}{a_{3}}=\frac{3}{2} \)
For \( {n}=4, \frac{a_{n+1}}{a_{n}}=\frac{a_{5}}{a_{4}}=\frac{5}{3} \)
For \( {n}=5, \frac{a_{n+1}}{a_{n}}=\frac{a_{6}}{a_{5}}=\frac{8}{5} \)
