NCERT Solutions for Class 11 Maths Chapter 8 Exercise 8.1 - Sequences and Series

Question 1: Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

$a_n=n(n+2)$

Answer:

Given : $a_n=n(n+2)$

$a_1=1(1+2)=3$

$a_2=2(2+2)=8$

$a_3=3(3+2)=15$

$a_4=4(4+2)=24$

$a_5=5(5+2)=35$

Therefore, the required number of terms =3, 8, 15, 24, 35

Question 2: Write the first five terms of each of the sequences in Exercises 1 to 6 whose nthterms are:

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

Answer:

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

$a_1=\frac{1}{1+1}=\frac{1}{2}$

$a_2=\frac{2}{2+1}=\frac{2}{3}$

$a_3=\frac{3}{3+1}=\frac{3}{4}$

$a_4=\frac{4}{4+1}=\frac{4}{5}$

$a_5=\frac{5}{5+1}=\frac{5}{6}$

Therefore, the required number of terms $\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6}$

Question 3: Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

$a_n=2^n$

Answer:

Given : $a_n=2^n$

$a_1=2^1=2$

$a_2=2^2=4$

$a_3=2^3=8$

$a_4=2^4=16$

$a_5=2^5=32$

Therefore, required number of terms $=2,4,8,16,32.$

Question 4: Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

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

Answer:

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

$a_1=\frac{2\times 1-3}{6}=\frac{-1}{6}$

$a_2=\frac{2\times 2-3}{6}=\frac{1}{6}$

$a_3=\frac{2\times 3-3}{6}=\frac{3}{6}=\frac{1}{2}$

$a_4=\frac{2\times 4-3}{6}=\frac{5}{6}$

$a_5=\frac{2\times 5-3}{6}=\frac{7}{6}$

Therefore, the required number of terms $=\frac{-1}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6},\frac{7}{6}$

Question 5: Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

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

Answer:

Given : $a_n=(-1{)}^{n-1}{5}^{n+1}$

$a_1=(-1{)}^{1-1}{5}^{1+1}=(-1{)}^0{.5}^2=25$

$a_2=(-1{)}^{2-1}{5}^{2+1}=(-1{)}^1{.5}^3=-125$

$a_3=(-1{)}^{3-1}{5}^{3+1}=(-1{)}^2{.5}^4=625$

$a_4=(-1{)}^{4-1}{5}^{4+1}=(-1{)}^3{.5}^5=-3125$

$a_5=(-1{)}^{5-1}{5}^{5+1}=(-1{)}^4{.5}^6=15625$

Therefore, the required number of terms $=25,-125,625,-3125,15625$

Question 6: Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

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

Answer:

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

$a_1=1.\frac{1^2+5}{4}=\frac{6}{4}=\frac{3}{2}$

$a_2=2.\frac{2^2+5}{4}=\frac{18}{4}=\frac{9}{2}$

$a_3=3.\frac{3^2+5}{4}=\frac{42}{4}=\frac{21}{2}$

$a_4=4.\frac{4^2+5}{4}=\frac{84}{4}=21$

$a_5=5.\frac{5^2+5}{4}=\frac{150}{4}=\frac{75}{2}$

Therefore, the required number of terms $=\frac{3}{2},\frac{9}{2},\frac{21}{2},21,\frac{75}{2}$

Question 7: Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

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

Answer:

$a_n=4n-3$

Put $n=17,$

$a{1}_7=4(17)-3=68-3=65$

Put n=24,

$a_{24}=4(24)-3=96-3=93$

Hence, we have $a_{17}=65and{a}_{24}=93$

Question 8: Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are

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

Answer:

Given : $a_n=\frac{n^2}{2^n}$

Put n=7,

$a_7=\frac{7^2}{2^7}=\frac{49}{128}$

Heence, we have $a_7=\frac{49}{128}$

Question 9: Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

$a_n=(-1{)}^{n-1}{n}^3,a_9$

Answer:

Given : $a_n=(-1{)}^{n-1}{n}^3$

Put n =9,

$a_9=(-1{)}^{9-1}{9}^3=(1).(729)=729$

The value of $a_9=729$

Question 10: Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:

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

Answer:

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

Put n=20,

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

Hence, value of $a_{20}=\frac{360}{23}$

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

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

Answer:

Given : $a_1=3,a_n=3a_{n-1}+2foralln>1$

$a_2=3a_{2-1}+2=3a_1+2=3(3)+2=11$

$a_3=3a_{3-1}+2=3a_2+2=3(11)+2=35$

$a_4=3a_{4-1}+2=3a_3+2=3(35)+2=107$

$a_5=3a_{5-1}+2=3a_4+2=3(107)+2=323$

Hence, five terms of series are $3,11,35,107,323$

Series $=3+11+35+107+323+...............$

Question 12: Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:

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

Answer:

Given : $a_1=-1,a_n=\frac{a_{n-1}}{n},n\ge 2$

$a_2=\frac{a_{2-1}}{2}=\frac{a_1}{2}=\frac{-1}{2}$

$a_3=\frac{a_{3-1}}{3}=\frac{a_2}{3}=\frac{-1}{6}$

$a_4=\frac{a_{4-1}}{4}=\frac{a_3}{4}=\frac{-1}{24}$

$a_5=\frac{a_{5-1}}{5}=\frac{a_4}{5}=\frac{-1}{120}$

Hence, five terms of series are $-1,\frac{-1}{2},\frac{-1}{-6},\frac{-1}{24},\frac{-1}{120}$

Series

$=-1+\frac{-1}{2}+\frac{-1}{-6}+\frac{-1}{24}+\frac{-1}{120}.........................$

Question 13: Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series: $a_1=a_2=2,a_n=a_{n-1}-1,n>2$

Answer:

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

$a_3=a_{3-1}-1=a_2-1=2-1=1$

$a_4=a_{4-1}-1=a_3-1=1-1=0$

$a_5=a_{5-1}-1=a_4-1=0-1=-1$

Hence, five terms of series are $2,2,1,0,-1$

Series $=2+2+1+0+(-1)+..................$

Question 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:

Given : The Fibonacci sequence is defined by $1=a_1=a_{2}and{a}_n=a_{n-1}+a_{n-2},n>2$

$a_3=a_{3-1}+a_{3-2}=a_2+a_1=1+1=2$

$a_4=a_{4-1}+a_{4-2}=a_3+a_2=2+1=3$

$a_5=a_{5-1}+a_{5-2}=a_4+a_3=3+2=5$

$a_6=a_{6-1}+a_{6-2}=a_5+a_4=5+3=8$

$Forn=1,\frac{a_{n+1}}{a_n}=\frac{a_{1+1}}{a_1}=\frac{a_2}{a_1}=\frac{1}{1}=1$

$Forn=2,\frac{a_{n+1}}{a_n}=\frac{a_{2+1}}{a_2}=\frac{a_3}{a_2}=\frac{2}{1}=2$

$Forn=3,\frac{a_{n+1}}{a_n}=\frac{a_{3+1}}{a_3}=\frac{a_4}{a_3}=\frac{3}{2}$

$Forn=4,\frac{a_{n+1}}{a_n}=\frac{a_{4+1}}{a_4}=\frac{a_5}{a_4}=\frac{5}{3}$

$Forn=5,\frac{a_{n+1}}{a_n}=\frac{a_{5+1}}{a_5}=\frac{a_6}{a_5}=\frac{8}{5}$