The petals of a flower, the number of steps you climb each day, or even the pocket money you save weekly; they often follow a pattern called sequence. When the terms of a sequence are added together, they form a series. Here in this exercise, we will learn how the numbers are arranged in a certain order and how we calculate their sum.
This Story also Contains
The Solutions of NCERT will provide stepwise calculations with detailed explanations. These NCERT solutions will make tough-looking sums simple and will help you improve your accuracy. Whether you are figuring out the nth term or the logic behind a number pattern, these solutions will help you build a solid grip on the topic.
Question 1: Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:
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:
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:
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:
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:
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:
$a _ n = 4 n - 3 ; a _{17} , a _{24}$
Answer:
$a _ n = 4 n - 3$
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 $a_{17}=65\, \, and\, \, a _ {24 }=93$
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:
$a_1 = 3, a_n = 3a_{n - 1} + 2\: \: for \: \: all \: \: n > 1$
Answer:
Given : $a_1 = 3, a_n = 3a_{n - 1} + 2\: \: for \: \: all \: \: n > 1$
$a_2 = 3a_{2 - 1} + 2=3a_1+2=3(3)+2=11$
$a_3 = 3a_{3 - 1} + 2=3a_2+2=3(11)+2=35$
$a_4 = 3a_{4 - 1} + 2=3a_3+2=3(35)+2=107$
$a_5 = 3a_{5 - 1} + 2=3a_4+2=3(107)+2=323$
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:
$a _ 1 = -1 , a _ n = \frac{a_{n-1}}{n} , n \geq 2$
Answer:
Given : $a _ 1 = -1 , a _ n = \frac{a_{n-1}}{n} , n \geq 2$
$a _ 2 = \frac{a_{2-1}}{2} =\frac{a_1}{2}=\frac{-1}{2}$
$a _ 3 = \frac{a_{3-1}}{3} =\frac{a_2}{3}=\frac{-1}{6}$
$a _ 4 = \frac{a_{4-1}}{4} =\frac{a_3}{4}=\frac{-1}{24}$
$a _ 5 = \frac{a_{5-1}}{5} =\frac{a_4}{5}=\frac{-1}{120}$
Hence, five terms of series are $-1,\frac{-1}{2},\frac{-1}{-6},\frac{-1}{24},\frac {-1}{120}$
Series
$=-1+\frac{-1}{2}+\frac{-1}{-6}+\frac{-1}{24}+\frac {-1}{120}.........................$
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 $1 = a _ 1 = a _2 \: \:and \: \: a _n = a _{n-1} + a _{n-2} , n > 2$
$a _3 = a _{3-1} + a _{3-2} =a_2+a_1=1+1=2$
$a _4 = a _{4-1} + a _{4-2} =a_3+a_2=2+1=3$
$a _5 = a _{5-1} + a _{5-2} =a_4+a_3=3+2=5$
$a _6 = a _{6-1} + a _{6-2} =a_5+a_4=5+3=8$
$For \,\,n=1,\frac{a _{n+1}}{a_n}=\frac {a_{1+1}}{a_1}=\frac{a_2}{a_1}=\frac{1}{1}=1$
$For \,\, n=2,\frac{a _{n+1}}{a_n}=\frac {a_{2+1}}{a_2}=\frac{a_3}{a_2}=\frac{2}{1}=2$
$For \,\, n=3,\frac{a _{n+1}}{a_n}=\frac {a_{3+1}}{a_3}=\frac{a_4}{a_3}=\frac{3}{2}$
$For \,\, n=4,\frac{a _{n+1}}{a_n}=\frac {a_{4+1}}{a_4}=\frac{a_5}{a_4}=\frac{5}{3}$
$For \,\, n=5,\frac{a _{n+1}}{a_n}=\frac {a_{5+1}}{a_5}=\frac{a_6}{a_5}=\frac{8}{5}$
Also read
1. Sequence - A sequence is an ordered list of numbers that follow a specific pattern or rule. We often write it using function notation like a(n) or an to represent the n-th term of the sequence.
For example: 2, 4, 6, 8, 10 is a sequence where each number increases by 2.
2. Finite sequence- A finite sequence is a sequence that has a limited number of terms.
For example: 3, 6, 9, 12 is a finite sequence with 4 terms.
3. Infinite sequence -An infinite sequence has no end.
For example: 1, 2, 3, 4, 5, 6, ... is an infinite sequence because it continues without stopping.
4. Series- A series is what you get when you add the terms of a sequence together.
If a sequence is in the form- $a_1, a_2, a_3, \ldots, a_n$, then the series corresponding to it is given by,
$$
a_1+a_2+a_3+\ldots+a_n+\ldots
$$
Also read
Students can also check out the NCERT solutions for other subjects as well for effective learning.
Use the links below to solve the NCERT exemplar solutions available for all the subjects. This will help you in your exam preparations.
Frequently Asked Questions (FAQs)
An = n (n + 2)
A2 = 2(2+2)
A2 = 8
An = 2^n
A3 = 2^3 = 8
The finite sequence is a sequence that contains a finite number of terms.
The infinite sequence is a sequence in which the number of terms is not finite.
The weightage of sequence and series in JEE Main Maths is 6.6%. Generally, two questions from this chapter is asked in the JEE Main exam.
Take Aakash iACST and get instant scholarship on coaching programs.
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE