Exercise 8.1 (Revised) - Chapter 9 - Sequences & Series - Ncert Solutions class 11 - Maths

You can Download the Exercise 8.1 (Revised) - Chapter 9 - Sequences & Series - Ncert Solutions class 11 - Maths with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

Chapter 8 - Sequences & Series | NCERT Solutions for Class 11 Maths

Write the first five terms of each of the sequences in Exercises 1 to 6 whose $n^n$ terms are:
Ex 8.1 Question 1.

$a_n=n(n+2)$

Answer.

Given: $a_n=n(n+2)$
Putting $n=1,2,3,4$ and 5 , we get,
$
\begin{aligned}
& a_1=1(1+2)=1 \times 3=3 \\
& a_2=2(2+2)=2 \times 4=8 \\
& a_3=3(3+2)=3 \times 5=15 \\
& a_4=4(4+2)=4 \times 6=24 \\
& a_5=5(5+2)=5 \times 7=35
\end{aligned}
$

Therefore, the first five terms are 3, 8, 15, 24 and 35 .
Ex 8.1 Question 2.

$a_n=\frac{n}{n+1}$

Answer.

Given: $a_n=\frac{n}{n+1}$ Putting $n=1,2,3,4$ and 5, we get,

$
\begin{aligned}
& 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}
\end{aligned}
$

Therefore, the first five terms are $\frac{1}{2}=\frac{2}{3}, \frac{3}{4}=\frac{4}{5}$ and $\frac{5}{6}$.
Ex 8.1 Question 3.

$a_n=2^n$

Answer.

Given: $a_n=2^n$
Putting $n=1,2,3,4$ and 5 , we get,
$
\begin{aligned}
& a_1=2^1=2 \\
& a_2=2^2=4
\end{aligned}
$

$
\begin{aligned}
& a_3=2^3=8 \\
& a_4=2^4=16 \\
& a_5=2^5=32
\end{aligned}
$

Therefore, the first five terms are $2,4,8,16$ and 32 .

Ex 8.1 Question 4.

$a_n=\frac{2 n-3}{6}$

Answer.

Given: $a_n=\frac{2 n-3}{6}$
Putting $n=1,2,3,4$ and 5 , we get,
$
\begin{aligned}
& a_1=\frac{2 \times 1-3}{6}=\frac{2-3}{6}=\frac{-1}{6} \\
& a_2=\frac{2 \times 2-3}{6}=\frac{4-3}{6}=\frac{1}{6} \\
& a_3=\frac{2 \times 3-3}{6}=\frac{6-3}{6}=\frac{3}{6}=\frac{1}{2} \\
& a_4=\frac{2 \times 4-3}{6}=\frac{8-3}{6}=\frac{5}{6} \\
& a_5=\frac{2 \times 5-3}{6}=\frac{10-3}{6}=\frac{7}{6}
\end{aligned}
$

Therefore, the first five terms are $\frac{-1}{6}=\frac{1}{6}, \frac{1}{2}, \frac{5}{6}$ and $\frac{7}{6}$.

Ex 8.1 Question 5.

$a_n=(-1)^{n-1} \cdot 5^{n+1}$

Answer.

Given: $a_n=(-1)^{n-1} \cdot 5^{n+1}$
Putting $n=1,2,3,4$ and 5 , we get,
$
\begin{aligned}
& a_1=(-1)^{1-1} \cdot 5^{1+1}=(-1)^0 \cdot 5^2=1 \times 25=25 \\
& a_2=(-1)^{2-1} 5^{2+1}=(-1)^1 \cdot 5^3=-1 \times 125=-125 \\
& a_3=(-1)^{3-1} \cdot 5^{3+1}=(-1)^2 \cdot 5^4=1 \times 625=625
\end{aligned}
$

$
\begin{aligned}
& a_4=(-1)^{4-1} \cdot 5^{4+1}=(-1)^3 \cdot 5^5=-1 \times 3125=-3125 \\
& a_5=(-1)^{5-1} \cdot 5^{5+1}=(-1)^4 \cdot 5^6=1 \times 15625=15625
\end{aligned}
$

Therefore, the first five terms are $25,-125,625,-3125$ and 15625 .
Ex 8.1 Question 6.

$a_n=n \cdot \frac{n^2+5}{4}$

Answer.

Given: $a_n=n \frac{n^2+5}{4}$
Putting $n=1,2,3,4$ and 5 , we get,
$
\begin{aligned}
& a_1=1 \cdot \frac{1^2+5}{4}=1 \cdot \frac{1+5}{4}=\frac{6}{4}=\frac{3}{2} \\
& a_2=2 \cdot \frac{2^2+5}{4}=2 \cdot \frac{4+5}{4}=\frac{18}{4}=\frac{9}{2} \\
& a_3=3 \cdot \frac{3^2+5}{4}=3 \cdot \frac{9+5}{4}=3 \times \frac{14}{4}=\frac{42}{4}=\frac{21}{2} \\
& a_4=4 \cdot \frac{4^2+5}{4}=4 \cdot \frac{16+5}{4}=\frac{84}{4}=21
\end{aligned}
$

$
a_5=5 \cdot \frac{5^2+5}{4}=5 \cdot \frac{25+5}{4}=5 \times \frac{30}{4}=\frac{150}{4}=\frac{75}{2}
$

Therefore, the first five terms are $\frac{3}{2}, \frac{9}{2}, \frac{21}{2}, 21$ and $\frac{75}{2}$.

Find the indicated terms in each of the sequences in Exercises 7 to 10 where $n^n$ terms are:
Ex 8.1 Question 7.

$a_n=4 n-3 ; \quad a_{17}, a_{24}$

Answer.

Given: $a_n=4 n-3$
$
\begin{aligned}
& \therefore a_{17}=4 \times 17-3=68-3=65 \\
& a_{24}=4 \times 24-3=96-3=93
\end{aligned}
$

Therefore, $17^{\text {th }}$ and $24^{\text {th }}$ terms are 65 and 93 respectively.
Ex 8.1 Question 8.

$a_n=\frac{n^2}{2^n} \div a_7$

Answer.

Given: $a_n=\frac{n^2}{2^n}$
$
\therefore a_7=\frac{7^2}{2}=\frac{49}{128}
$

Therefore, $7^{\text {th }}$ term is $\frac{49}{128}$.

Ex 8.1 Question 9.

$a_n=(-1)^{n-1} n^3 ; \quad a_0$

Answer.

Given: $a_n=(-1)^{n-1} n^3$
$
\therefore a_9=(-1)^{9-1} \times(9)^3=(-1)^3 \times 729=729
$

Therefore, $9^{\text {th }}$ term is 729 .
Ex 8.1 Question 10.

$a_n=\frac{n(n-2)}{n+3} ; a_{20}$

Answer.

Given: $a_n=\frac{n(n-2)}{n+3}$

$
\therefore a_{20}=\frac{20(20-2)}{20+3}=\frac{20 \times 18}{23}=\frac{360}{23}
$

Therefore, $20^{\text {th }}$ term is $\frac{360}{23}$.

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:
Ex 8.1 Question 11.

$a_1=3, a_n=3 a_{n-1}+2$ for all $n>1$

Answer.

Given: $a_1=3, a_n=3 a_{n-1}+2$ for all $n>1$
Putting $n=2,3,4$ and 5 , we get
$
\begin{aligned}
& a_2=3 a_{2-1}+2=3 a_1+2=3 \times 3+2=9+2=11 \\
& a_3=3 a_{3-1}+2=3 a_2+2=3 \times 11+2=33+2=35 \\
& a_4=3 a_{4-1}+2=3 a_3+2=3 \times 35+2=105+2=107 \\
& a_5=3 a_{5-1}+2=3 a_4+2=3 \times 107+2=321+2=323
\end{aligned}
$

Hence the first five terms are $3,11,35,107,323$.
Therefore, corresponding series is $3+11+35+107+323+$ $\qquad$

Ex 8.1 Question 12.

$a_1=-1, \quad a_n=\frac{a_{n-1}}{n}, n \geq 2$

Answer.

Given: $a_1=-1, \quad a_n=\frac{a_{n-1}}{n}, n \geq 2$
Putting $n=2,3,4$ and 5 , we get
$
a_2=\frac{a_{2-1}}{2}=\frac{a_1}{2}=\frac{-1}{2}
$

$
\begin{aligned}
& a_3=\frac{a_{3-1}}{3}=\frac{a_2}{3}=\frac{-1 / 2}{3}=\frac{-1}{6} \\
& a_4=\frac{a_{4-1}}{4}=\frac{a_3}{4}=\frac{-1 / 6}{4}=\frac{-1}{24} \\
& a_5=\frac{a_{5-1}}{5}=\frac{a_4}{5}=\frac{-1 / 24}{5}=\frac{-1}{120}
\end{aligned}
$

Hence the first five terms are $-1, \frac{-1}{2}, \frac{-1}{6}, \frac{-1}{24}, \frac{-1}{120}$
$\therefore$ Corresponding series is $-1+\left(\frac{-1}{2}\right)+\left(\frac{-1}{6}\right)+\left(\frac{-1}{24}\right)+\left(\frac{-1}{120}\right) \ldots \ldots \ldots \ldots$.

Ex 8.1 Question 13.

$a_1=a_2=2, a_n=a_{n-1}-1, n>2$

Answer.

Given: $a_1=a_2=2, a_n=a_{n-1}-1, n>2$
Putting $n=3,4$ and 5 , we get
$
\begin{aligned}
& a_3=a_{3-1}-1=a_2-1=2-1=1 \\
& a_4=a_{4-1}-1=a_3-1=1-1=0 \\
& a_5=a_{5-1}-1=a_4-1=0-1=-1
\end{aligned}
$

Hence the first five terms are 2,2,1,0,-1.
Therefore, corresponding series is $2+2+1+0+(-1)+\ldots . .$.
Ex 8.1 Question 14.

The Fibonacci sequence is defined by $1=a_1=a_2$ and $a_{n-1}+a_{n-2} n>2$. Find $\frac{a_{n+1}}{a_n}$, for $n=1,2,3,4,5$.

Answer.

Given: $a_1=a_2=1$ and $a_{n-1}+a_{n-2}, n>2$

Putting $n=3,4,5$ and 6 , we have
$
\begin{aligned}
& a_3=a_{3-1}+a_{3-2}=a_2+a_1=1+1=2 \\
& a_4=a_{4-1}+a_{4-2}=a_3+a_2=2+1=3 \\
& a_5=a_{5-1}+a_{5-2}=a_4+a_3=3+2=5 \\
& a_6=a_{6-1}+a_{6-2}=a_5+a_4=5+3=8
\end{aligned}
$

Now, $\frac{a_{n+1}}{a_n}$
For $n=1, \frac{a_{1+1}}{a_1}=\frac{a_2}{a_1}=\frac{1}{1}=1$
For $n=2, \frac{a_{2+1}}{a_2}=\frac{a_3}{a_2}=\frac{2}{1}=2$
For $n=3, \frac{a_{3+1}}{a_3}=\frac{a_4}{a_3}=\frac{3}{2}$
For $n=4: \frac{a_{4+1}}{a_4}=\frac{a_5}{a_4}=\frac{5}{3}$
For $n=5, \frac{a_{3+1}}{a_5}=\frac{a_6}{a_5}=\frac{8}{5}$