Prime Optional Mathematics 8

Grade According to new curriculum in compliance with
VIII Curriculum Development Centre (CDC)

PRIME Optional
Mathematics

Pragya Books &
Distributors Pvt. Ltd.

Author Editors

Dirgha Raj Mishra L.N. Upadhyaya
Khem Timsina
J.N. Aryal

Kadambaba Pradhan
Dinesh Silwal

Pragya Books & Distributors Pvt. Ltd.

Lalitpur, Nepal
Tel : 5200575
email : pragyabooks100@gmail.com

© Author

Author Dirgha Raj Mishra

Editors L.N. Upadhyaya
Khem Timsina
J.N. Aryal
Kadambaba Pradhan
Dinesh Silwal

First Edition 2076 B.S. (2019 A.D.)

Price Rs. 296/-

ISBN 978-9937-9170-6-3

Typist Sachin Maharjan
Sujan Thapa

Layout and Design Desktop Team

Printed in Nepal

Preface

Prime Optional Mathematics series is a distinctly outstanding mathematics
series designed according to new curriculum in compliance with Curriculum
Development Centre (CDC) to meet international standard in the school level
additional mathematics. The innovative, lucid and logical arrangement of the
contents make each book in their series coherent. The representation of ideas in
each volume makes the series not only unique but also a pioneer in the evaluation
of activity based mathematics teaching.

The subject is set in an easy and child-friendly pattern so that students will
discover learning mathematics is a fun thing to do even for the harder problems.
A lot of research, experimentation and careful graduation have gone into the
making of the series to ensure that the selection and presentation is systematic,
innovative, and both horizontally and vertically integrated for the students of
different levels.

Prime Optional Mathematics series is based on child-centered teaching
and learning methodologies, so that the teachers can !nd teaching this series
equally enjoyable.

I am optimistic that, this series shall bridge the existing inconsistencies
between the cognitive capacity of children and the subject matter.

I owe an immense dept of gratitude to the publishers (Pragya Books team)
for their creative, thoughtful and inspirational support in bringing about the
series. Similarly, I would like to acknowledge the tremendous support of editors
team, teachers, educationists and well-wishers for their contribution, assistance
and encouragement in making this series a success. I would like to express my
special thanks to Sachin Maharjan (Wonjala Desktop) for his sincere support
of designing part of the book and also Mr. Gopal Krishna Bhattarai to their
memorable support to prepare this series.

I hope this series will be another milestone in the advancement of teaching
and learning Mathematics in Nepal. We solicit feedback and suggestions from
teachers, students and guardians alike so that I can re!ne and improvise the series
in the future editions.

– Author

Algebra : Ordered Pairs Contents Page

S.N. Units 1
2
1. Algebra 11
1.1 Ordered pairs 21
1.2 Surds 28
1.3 Polynomials 37
1.4 Sequence and series 47
63
2. Limits 87
3. Matrices 131
4. Co-ordinate Geometry 149
5. Trigonometry 173
6. Vector Geometry 200
7. Transformation
8. Statistics

Model questions

4 PRIME Opt. Maths Book - VIII

Unit 1 Algebra

1. Algebra

1.1 Ordered pairs
1.2 Surds
1.3 Polynomials
1.4 Sequence and series

Speci•ication•Grid•Table

K(1) U(2) A(4) HA(5) TQ TM Periods

No. of Questions 1 1 1 – 3 7 16
2 4 –
Weight 1

K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks

Objectives : At the end of the lesson
• Students are able to know the ordered pairs and Cartesian products.
• Students are able to represent Cartesian product in arrow diagram and graph.
• Students are able to know the number system including surds.
• Students are able to operate the surds.
• Students are able to know the polynomials, types and degree of polynomials.
• Students are able to identify the sequence and series.
• Student are able to know the difference between arithmetic and geometric

sequence and their nth term.

Materials Required:
• Chart paper
• Chart of number system.
• Chart of types of surds.
• Chart of types polynomials & properties of addition and multiplication.
• Chart of types of sequence and series.

PRIME Opt. Maths Book - VIII 1

Algebra : Ordered Pairs 1.1 Ordered Pair and Cartesian Product
Ordered Pair

Look•at•a sequence of numbers 1, 2, 3, 4, 5, 6, 7, ....., there is a certain rule
between•the•numbers•and•the•next number will be 8 according to the rule.
The•numbers•in•the•above example are in order.

Let•us•consider•the•following•examples:
(2,•3),•(4,•5),•(a,•b),•(Nepal,•Kathmandu)•etc.•
In•above pair of numbers or letters or names are kept inside a round bracket
separated by comma where the position of numbers or objects play very
important•role. This type of presentation of number or letter or objects is
called•an•ordered•pair.

Ordered pair :
The• Pair of numbers or objects in the form of
(x,•y)•with•an•order•is•called•ordered•pair.

For an ordered pair (x, y)
x & y are called the elements of ordered pair, the irst element is called
antecedent• (x–component) and second element is called consequence
(y–component).

(x, y) is ordered pair :
x is antecedent (x – component)
y is consequce (y – component)

There is a vital role of order (position) in ordered pair, If we change the
position of elements, a new ordered pair is formed.

Examples from practical life.
• Ordered pair of country with respect to capital

(Nepal, Kathmandu), (Japan, Tokyo), (China, Beijing)

2 PRIME Opt. Maths Book - VIII

• Order pair of temples with respect to location. Algebra : Ordered Pairs
(Pashupatinath, Kathmandu), (Pathiveradevi, Taplejung),
(Manakamana, Gorkha)

• Order pair of community and their dresses:
(Newari, Haku Patassi), (Sherpa, Bakhu) (Aryan, Gunyu Choli) etc.

Equal Ordered Pairs
Icna•ltlhede••oerqduearle•odrdpearired(1p, a2i)ras.nIdn(th33is, e36x)a,m•thpel•ecoamntpeocendeenntst•saraendeqcuoanlsweqhuicehnaceres
of•both•the•ordered•pairs•are•equal.

Any two ordered pairs are said to be equal ordered
pairs•if•and•only if the corresponding components are
equal.
• If•(x,•y)•and•(2,•3)•are•equal•then•x•=•2•and•y•=•3.

Worked out Examples

1. Write down any ive ordered pairs of currency with respect to country.
Solution:
The•currencies•and•countries•have to be written as x – component and
y – component•respectively as,
(Rupiya, Nepal), (Taka, Bangladesh), (Pound, United Kingdom), (Yen,
China),•(Riyal, Saudi Arebia)

2. If•(x,•3)•and•(2,•y•+•2)•are•two equal ordered pairs, ind x & y.

Solution:

Given (x, 3) = (2, y + 2).

By equating the corresponding elements, we have

x = 2 and 3=y+2

or 3 – 2 = y

or 1 = y

\ x = 2 and y = 1

PRIME Opt. Maths Book - VIII 3

3.• If•(x•+•y,•2x•–•3)•=•(5,•3),• ind•the•value•of•x•and•y.

Solution :

Given, (x + y, 2x – 3) = (5, 3)

By equating the corresponding elements,

2x – 3 = 3 antecedent• = antecedent•
or, 2x = 3 + 3 of•1st pair of•2nd pair
or, 2x = 6

or, x = 3

Again,

x+y=5 [Consequence of 1st pair = Consequence of 2nd pair]

or, 3 + y = 5 [ a x = 3]

• or, y = 5 – 3

y=2

\ x=3

\ y=2

Algebra : Ordered Pairs Cartesian•Product
Let•us•consider•two sets A = {1, 2} and B = {3, 4}. Let us obtain all possible

ordered pairs where antecedent is from set A and consequence from set B

and•shown•below.

1 3 (1,•3)
4 (1,•4)

2 3 (2,•3)
4 (2,•4)

Now, the set of these ordered pairs is {(1, 3), (1, 4), (2, 3), (2, 4)} which is
called•the•Cartesian•product•of•A•and•B•and•we•denote•it•by•A•×•B.

The• set• of• all• possible• ordered pairs (x, y) from the
non–empty• set• A to the non–empty set B is called
Cartesian•product•A•×•B.•where•x•Î A, y Î B.

4 PRIME Opt. Maths Book - VIII

Thus,•the•Cartesian•product of A and B is read as A cross B and denoted by
A × B and•de•ined•as,
A × B = {(x,•y)•:•x•Î A and•y•Î B}
Also,•n(A•× B) = number of ordered pairs of A × B is called the cardinality
of•A•×•B.

Example
If A = {2, 3}, B = {4, 5}
A × B = {(2, 4), (2, 5), (3, 4), (3, 5)}
Here,
n(A × B) = 4 which is Cardinality of A × B.
Also,
B × A = {(4, 2), (4, 3), (5, 2), (5, 3)}

Here, Algebra : Ordered Pairs
n (B × A) = 4 which is Cardinality of B × A.

Then, we conclude that :
A×B¹B×A
But, n(A × B) = n(B × A)

Note : If n(A) = m, n(B) = n then
n(A × B) = n(A) × n(B) = m × n = mn

Representation of Cartesian product:
Taking the sets A = {2, 3}, B = {4, 5}

i) Listing method :
A × B = {(2, 4), (2, 5), (3, 4), (3, 5)}

ii) Arrow diagram : B B•×•A A
A A•×•B B 42
24

35 53

PRIME Opt. Maths Book - VIII 5

iii) Tabular form :

B A×B A B×A
A B
4 5 2 3

2 (2,•4) (2,•5) 4 (4,•2) (4,•3)

3 (3,•4) (3,•5) 5 (5,•2) (5,•3)

Algebra : Ordered Pairs Worked out Examples

4.• If•A = {1, 2, 3}, B = {4, 5, 6}, ind A × B and B × A. Also prove that n (A × B)
= n(B•×•A)
Solution :
A = {1, 2, 3},
B = {4, 5, 6}
A × B = {(x, y) : x Î A and y Î B}
= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
B × A = {(x, y) : x Î B and y Î A}

• = {(4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3)}

Here,
n(A × B) = 3 × 3 = 9
n (B × A) = 3 × 3 = 9
\ n (A × B) = n (B × A)

5.• If•A × B = {(a, x), (a, y), (b, x), (b, y)}, ind the sets A and B. Also ind A × A
and•B•×•B.
Solution:
A × B = {(a, x), (a, y), (b, x), (b, y)}
Set A = {all the Antecedents x of A × B} = {a, b}
Set B = {all the Consequences y of A × B} = {x, y}

Again,
A × A = {a, b} × {a, b}

= {(a, a), (a, b), (b, a), (b, b)}
B × B = {x, y} × {x, y}

= {(x, x), (x, y), (y, x), (y, y)}

6 PRIME Opt. Maths Book - VIII

6. If•A = {a, b, c} and A × B = {(...., x), (...., y), (a, ....), (...., x), (b, ....), (...., z),
(c,•....),•(c,•....),•(c,•....)},• ind•the set B, Complete A × B. Also show A × B in
arrow diagram.
Solution : A = {a, b, c}
A × B = {(...., x), (...., y), (a, ....), (...., x), (b, ....), (...., z), (c, ....), (c, ....), (c, ....)}
By comparing A and A × B, we get B = {x, y, z}
Also, A × B = (a, x), (a, y), (a, z), (b, x), (b, y), (b, z), (c, x), (c, y), (c, z)}

Arrow diagram:
A A•×•B B
ax

by

cz

7. Find•x•and•y•if•(x•+•2y,•1)•=•(3,•2x•–•y).•Also• ind•x•–•y. Algebra : Ordered Pairs

Solution :

Given, (x + 2y, 1) = (3, 2x – y)

From equality of ordered pairs, we have

x + 2y = 3 and 2x – y = 1

or, x = 3 – 2y ................. (i) or, 2(3 – 2y) – y = 1 [\ using (i)]

or, 6 – 4y – y = 1

or, –5y = – 5

or, y = 1

Putting y = 1 in (i) we have

x =3–2×1

=3–2

=1

\ x–y=1–1=0

PRIME Opt. Maths Book - VIII 7

1.•• i)• Exercise 1.1
ii)
Write down any •ive ordered pairs of districts with respect to
headquarter•of•our•country•Nepal.•Also•show•in•arrow diagram.
Write down the countries and their capitals from the given arrow
diagram•in•ordered•pair•form.

India New•Delhi

China Beijing

Japan Tokyo

Algebra : Ordered Pairs iii) Complete the following ordered pairs by •illing the gaps with
suitable•words.
(Bhanubhakta, ....), (...., Aanshukabi), (Lekhnath Poudel, .....)

iv) Taking an order ‘more than by 2’, complete the following ordered
pairs.•(2,•...),•(5,•...),•(...,•9)•(...,•12),•(15,•...)

• v) What is ordered pair? Write down with an example.

2.• Find•the•value•of•‘x’•and•‘y’•from•the•following•equal•ordered•pairs.

• i) (x, y) = (3, 4) ii) (x – 2, y + 1) = (5, 4)

iii) (2x – 1, y + 2) = (5, 6 – y) iv) (x – 1, 2x + y) = (2, 8)

v) (3x – 2, 2y + 5) = (x + 4, 1) iv) (2x – 3y, 13) = (3, 3x + 4y)

3.• If•A•=•{a,•b},•B•=•{1,•2,•3},••ind
• i) A × B and show in arrow diagram.

ii) B × A and show in arrow diagram.
iii) Prove that A × B ¹ B × A
iv) Prove that n(A × B) = n(B × A)

4.• If•P•=•{2,•5},•Q•=•{3,•4},••ind
• i) P × Q and show in tabular form.

ii) Q × P and show in tabular form.
iii) Find the ordered pair of P × P and Q × Q.
iv) Prove that n(P × Q) = n(Q × P)

8 PRIME Opt. Maths Book - VIII

5.• If•A•×•B•=•{(a,•x),•(b,•x),•(a,•y),•(b,•y),•(a,•z),•(b,•z)}••ind,
• i) Set A and Set B.

ii) Prove that A × B ¹ B × A
iii) Prove that n (A × B) = n (B × A)
iv) Find the ordered pair of B × B and A × A.

6.• i)• From the given arrow diagram •ind the set A and set B. Also •ind

B × A & n(B•×•A) A•×•B B
A

25

47

69 Algebra : Ordered Pairs

ii) If A = {x : x Î 1, 2}, B = {y : y Î 3 < N £ 6}, •ind A × B. Also show in
tabular•form.

• iii) If•P = {x : x Î N, 1 £ N < 4}, Q = {x : x Î 3, 4}, •ind P × Q and n(P × Q).
iv) If•A × B = {(....., 3), (5, .....), (....., 4), (....., 4), (5, .....), (2, .....)} and A = {2, 5},
•ind•B.•Also••ind•A•×•B•and•B•×•A•and•prove that A × B ¹ B × A.

• v) If A = {1, 2}, B = {2, 3}, •ind (A È B) × B.

PRIME more creative questions

7.• i)• If•(x•+•y,•6)•=•(6,•2x•–•y),••ind•the•value•of•‘x’•and•‘y’.

ii) If (2x, y + 3) = (y + 3, 3x – 2), •ind the value of ‘x’ & ‘y’.

iii) From the given table, •ind the sets A and B. Also prove that

n(A•×•B)•=•n(B•×•A)

A B×A
B
a b c

x (x,•a) (x,•b) (x,•c)

y (y, a) (y, b) (y, c)

z (z,•a) (z,•b) (z,•c)

iv) If A = {2, 3}, B = {3, 4}, •ind (A È B) × A.
• v) If A = {1, 2}, B = {2, 3}, C = {3, 4, 5}, •ind A × (B È C). Also show

in tabular form.

Project work
8.• Collect•the•cost•of•different vegetables from your local shop and write

down the ordered pair of them with respect to cost.

PRIME Opt. Maths Book - VIII 9

Answer

1.• Show•to•your•subject•teacher

2.• i)• x•=•3,•y•=•4• (ii)• x•=•7,•y•=•3• • (iii) x = 3, y = 2

iv) x = 3, y = 2 (v) x = 3, y = –2

3.• Show•to•your•subject•teacher.

4.• Show•to•your•subject•teacher.

5.• i)• A•=•{a,•b},•B•=•{x,•y,•z}

• ii) n(A × B) = 6, n(B × A) = 6

iii) B × B = {(x, x), (x, y), (x, z), (y, x), (y, y), (y, z), (z, x), (z, y), (z, z)}

iv) Show to your subject teacher.

v) Show to your subject teacher.

6.• i)• A•=•{2,•4,•6},•B•=•{5,•7,•9}

• B × A = {(2, 5), (2, 7), (2, 9), (4, 5), (4, 7), (4, 9), (6, 5), (6, 7), (6, 9)}

Algebra : Ordered Pairs ii) A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6)}

A×B

B 4 5 6
A

1 (1,•4) (1,•5) (1,•6)

2 (2,•4) (2,•5) (2,•6)

iii) P×Q = {(1, 3), (2, 3), (3, 3), (1, 4), (2, 4), (3, 4)}, n(P×Q) = 6
iv) Show to your subject teacher.
v) (A È B) × B = {(1,•2),•(1,•3),•(2,•2),•(2,•3),•(3,•2),•(3,•3)}

7.• i)• x•=•4,•y•=•2•
• ii) x = 2, y = 1

iii) A = {a, b, c}, B = {x, y, z}, n(A × B) = 9, n(B × A) = 9
iv) (A È B) × A = {(2,•2),•(2,•3),•(3,•2),•(3,•3),•(4,•2),•(4,•3)}
• v) A × (B È C) = {(1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5)}

A × (B•È C)

BÈC 2 3 4 5
A

1 (1,•2) (1,•3) (1,•4) (1,•5)

2 (2,•2) (2,•3) (2,•4) (2,•5)

10 PRIME Opt. Maths Book - VIII

1.2• Surds Algebra : Surds

Number•system
• The•set•of•natural•numbers•is•the•set•of•all•counting•numbers.•We denote

the•set•of•natural•numbers•by N. The least natural number is 1 and the
highest•is•not•de•ined.
• N = {1, 2, 3, 4, 5 ....}
• The• set• of• counting• numbers• including• zero is called whole numbers.
The•set•of•whole numbers is denoted by W The least whole number is
zero.

W = {0, 1, 2, 3, 4, 5, ...}
• The• set• of• all• counting• numbers• including• zero and their negatives is

called•integers.•The•set•of•integers•is•denoted•by•Z.
• Z = {....., –4, –3, –2, –1, 0, 1, 2, 3, 4, ....}

In•the•given examples set of natural numbers
is• the• sub–set• of• whole number and set of
whole number is the sub set of integers.

Rational•numbers:

Let• us• consider• the• elements• of• integers• taken• in• different mathematical

operations•as•follows,

4 – 6 = –2 (Integer) 4 + 6 = 10 (Integer)
2
4 × 6 = 24 (Integer) 4÷6= 3 (Not•Integer)
• But 4 ÷ 2 = 2 (Integer)

Thus, the division of any two integers may not be always an integer.
To de•ine such numbers a new number system is introduced which is
called•the•rational•number.

The•set•of•numbers•in•the•form of p/q where p & q are the
integers•and•q•¹ 0 is•called•the•set•of•rational•numbers.
i.e.•Q = {...., –2, –3/2, –1, 0, 1/2, 1, 3/2, 3, ............} is the
set•of•rational•numbers.

PRIME Opt. Maths Book - VIII 11

2 = 4 p
1 2 q

19
4

Algebra : Surds

19
3

12 PRIME Opt. Maths Book - VIII

Try to T•ihnedrethaerereicnu•irnriitnegnduemcibmeralonf uramtiboenraflonrum272bers between any two
Note :

rational numbers.

Irrational•numbers:
Some Examples :

2 = 1.414213562....................
• 3 = 1.732...............
• 5 = 2.236067977..
• 4 =2

9 =3
Here, 2 , 3 , 5 have the non terminating and non recurring
decimal•parts•while 4 & 9 have the integer numbers as their roots

2 , 3 , 5 are•called•the•irrational•numbers.

Note: Algebra : Surds

i.• All•the•roots•which•are•not•rational•are•irrational

ii.• The•non•terminating and non recurring decimal numbers are irrational.

iii.• p= length•of•circumference of a circle is•an•irrational•number.
diameter
iv. Length of a body is either a rational number or an irrational number.

Introduction•of•Surds:
The numbers 2 , 3 , 3 2 etc are the roots of rational numbers where
the•result•is•irrational•are•called•surds.

• 2 , 3 are unlike surds of same order (same orders but different
radicands)
2 , 3 2 are unlike surds of different order (different orders but same
radicands)

3 , 2 3 are•the•like•surds•(same•orders•and•same•radicands)
• 3 5 , 3 10 are•unlike•surds

Any number in the from n a which can not be expressed in the form
p
of• q , where q ¹ 0, p, q Ï z, is called a surd. n is called the order, ‘a’ is

called•the•radicand•and• called•the•sign•of•radical.

PRIME Opt. Maths Book - VIII 13

Like surds are the surds having same
orders and same radicands.

12 is•the•pure•surd•(having•only•irrational•factors)
• 2 3 is the mixed surd (having both irrational factors and rational

factors rather than 1.)
• For the operation of the surds, pure surds should be changed into

mixed•surds.
• Only•like•surds•can•be•operated using plus and minus sign.

Worked out Examples

1.• Insert•any•three rational numbers between 2 and• 3 .
Solution : 3 4

Algebra : Surds The given rational numbers are 2 and• 3
3 4
2 3
The other three rational numbers between 3 and• 4 are•as•follows.

1st rational no. = 2 + 3
3 4
2

= 1 ` 17 j = 17
2 12 24

2nd rational no. = 1 a 2 + 17 k
2 3 24

= 1 ` 33 j = 11
2 24 16

3rd rational no. = 1 ` 17 + 3 j
2 24 4
1 35 35
= 2 a 24 k = 48

So on any other can be found out.

2.• Express the surd 50 into•mixed•surd•and• 2 3 5 into•pure•surd.
Solution :

50 = 52 × 2 = 5 2 (mixed•surd)
2 3 5 = 3 23 × 5 = 3 40 (pure•surd)

14 PRIME Opt. Maths Book - VIII

3. Express the surds 2, 3 5, 4 6 into the surds of same order and arranged
in•ascending•order.
Solution :
The given surds are 2, 3 5, 4 6
L.C.M. of 2, 3, & 4 is 12.
Where,
2 = 2× 6 26 = 12 26 = 12 64

3 5 = 3× 4 54 = 12 54 = 12 625

4 6 = 4 × 3 63 = 12 63 = 12 216

The surds in ascending order
12 64 , 12 216 , 12 625
i.e. 2 , 4 6 , 3 5

Note : If n a and• n b are•two•surds•and•a•>•b•then• n a > n b

4. Simplify : 5 18 + 72 – 3 32 5.• Multiply•the•surds• 3 and• 3 2 . Algebra : Surds
Solution : Solution

5 18 + 72 – 3 32 3 ×3 4
LCM of 2 and 3 is 6.
= 5 32 × 2 + 62 × 2 – 3 42 × 2
= 15 2 + 6 2 – 12 2 = 6 33 × 6 42
= (15 + 6 – 12) 2 = 6 27 × 16
=9 2 = 6 432

6. Rationalise the denominate of 1 .
3+ 2
Solution :
[ a Multiply both numerator
1 = 1 × 3– 2 and•denominator•by conjugate
3+ 2 3+ 2 3– 2
3– 2]

=

= 3– 2
3–2

= 3– 2
1

= 3– 2

PRIME Opt. Maths Book - VIII 15

7.• Simplify : 4 – 3 – 1
Solution : 6 – 2 6+ 3 3+ 2

= 4× 6+ 2– 3 × 6+ 3– 1 × 3+ 2
6– 2 6+ 2 6+ 3 6+ 3 3+ 2 3+ 2

=––

= 4( 6+ 2) – 3( 6– 3) – 3+ 2
4 3 1

= 6+ 2 – 6+ 3 – 3– 2
=0

Algebra : Surds 8.• Solve•for•‘x’•of•:• x – 2 = 3
Solution:

" x –2=3
or x = 3 + 2
or x = 5
Squaring on both sides ^ xh2 = (5)2
\ x = 25

Checking for x = 25 on x – 2 = 3
or, 25 – 2 = 3
or, 5 – 2 = 3

or, 3 = 3 (true)

\ x = 25

9.• Solve•:• x + 9 – x = 1
Solution :

x+9 – x =1
or, x + 9 = x + 1

Squaring on both side

or, ( x + 9 )2 = ( x + 1)2
or, x + 9 = ( x )2 + 2 x – 1 + (1)2
or, x + 9 = x +2 x + 1
or, 8 = 2 x
or, 4 = x

16 PRIME Opt. Maths Book - VIII

Again,
Squaring on both sides,
or, (4)2 = ( x )2
or, 16 = x
\ x = 16

Note : If a + b is a surd then a – b is called the conjugate surd of a + b .

10.• Find•the•square•root•of•4•–•2 3 .
Solution:
Square root of 4 – 2 3 is,

• 4–2 3 =
= ( 3 – 1)2
= 3 –1

Exercise 1.2 Algebra : Surds

1.•• i)• What•is•natural•number?•Write down the set of natural numbers.
ii) How can you say that set of natural numbers is the sub–set of
whole number?
iii) What do you mean by irrational numbers?
iv) What is the set of rational numbers?
v) Differentiate between like surds and unlike surds with
examples.

2.• Pick•out•the•like•surds•from•the•following•with•reason.

• i) 3 2 , 2 , 5 2 ii) 2 3 5 , 5 3 2 , 3 3

iii) 3 24 , 3 3 3 , 3 81 iv) 210, 75, 4 144

v) 20 , 2 45 , 3 250

3.• Convert the followings as indicated in brackets.

i) 98 (into•mixed•surd)• ii)• 3 500 (into•mixed•surd)
iii) 2 5 (into•pure•surd)• iv) 33 2 (into•pure•surd)
v) 25 2 (into•pure•surd)

PRIME Opt. Maths Book - VIII 17

4.• Express•the•followings•as•indicated•in•brackets.

• i) 3 and• 3 2 (into•same•order)
• ii) 3 3, 4 4 (which is greater)

iii) 3, 3 4, 6 20 (in•ascending•order)
• iv) 3 3, 2, 4 5 (in•descending•order)
• v) 3 4, 4 5, 12 50 (in•ascending•order)

5.• Simplify:

• i) 18 + 50 – 3 8 ii) 2 108 – 6 75 + 5 48

iii) 3 40 + 2 3 320 – 3 3 135 iv) 3 700 – 2800 – 2 63

v) 4 3 432 – 5 3 128 + 7 3 2000

6.• Multiply•the•surds.

• i) 2 ×3 3 ii) 3 4 × 6 8
iv) 2 5 × 3 3 4
iii) 3 × 3 5 × 6 4

v) 3 2 × 2 4 8 × 5 6 4

Algebra : Surds 7.• Rationalise the denominator. 1
i) 1 5+ 3
3– 2 ii)
x– a
iii) 4 iv) x+ a
3 + 2 –1

v)

8.• Simplify•the•following.

• i) 2 +1
5+ 3 3– 2

ii) 1 – 3+ 4
3+ 2 6– 3 6– 2

iii) 3 + 2 + 5
5+ 2 7+ 5 2– 7

iv) x+ a + x– a
x– a x+ a

18 PRIME Opt. Maths Book - VIII

v) +

9.• Solve the followings. ii) x – 1 – 2 = 3
iv) x + 6 – 2 = 1
i) x + 3 = 5
iii) x – 2 = 3

v) x2 – 5 + 3 = 5

10.• Solve the followings. ii)• x + 3 – x = 1
iv) x + 7 = x + 1
i) x + 5 + x = 5
iii) x = x + 11 – 1
v) x2 + 9 – x = 3

11.• Find•the•square•root•of: ii)• 42 + 32 + 22 + 12 – 5
• i) 12 + 22 + 22 × 5 iv) 5 + 2. 5 .1•+•1

iii) ( 3 )2 + 2. 3 .2•+•22 Algebra : Surds
• v) 2 + 2. 2 . 3 + 3

12. PRIME more creative questions:

i) Simplify : 3 200 – 2 1250 + 5 4 1024
12
ii) Rationalise : 5+ 3+ 2

iii) Simplify : 5 – 32 + 43
15 + 10 12 – 6 18 – 6

iv) Solve : =2

v) Find the square root of 4 + 2 3
vi) Find the square root of 7 – 4 3 .

PRIME Opt. Maths Book - VIII 19

Answer

1.• Show•to•your•subject•teacher.
2.• Show•to•your•subject•teacher.
3.• Show•to•your•subject•teacher.

4.• i)• 6 27 & 6 4 ii) 3 3 is•greater.
iii) 3 4 < 6 20 < 3 iv) 4 5 > 3 3 > 2
v) 12 50 < 4 5 < 3 4

5.• i)• 2 2 ii) 2 3 iii) 3 5
iv) 4 7 v) 74 3 2

6.• i)• 6 72 ii) 2 6 2 iii) 6 4050
iv) 6 6 2000 v) 20 12 32

Algebra : Surds 7.• i)• ^ 3 + 2h ii) 1 ^ 5– 3h iii) 6 + 2 – 2
2 iii) –2 2
x + a – 2 ax 1 iii) 25
iv) x–a v) 2 ^x – x2 – 4h

8.• i)• 5 + 2 ii) 0

iv) 2^x + ah v) x
x–a

9.• i)• 4 ii) 26
iv) 3 v) ±3

10.• i)• 4 ii) 1 iii) 25
iv) 9 v) 4

11.• i)• 5 ii) 5 iii) 2 + 3
iv) 5 + 1 v)• 3 + 2

12.• i)• 0 3 ii) – 30 + 3 2 + 2 3 iii) 0
2
iv) x = v) 3 + 1 vi)• 2•–• 3

20 PRIME Opt. Maths Book - VIII

1.3• Polynomials

• An• algebraic• term is the product or quotient of number(s) with

variable(s).

Examples : 3x2, 2xy, 3 x3, 2x2 etc.
2 y2

Let us consider an example of algebraic term 2ax3 in•x.

• Here,

2 is called numeral coef icient.

x is called base.

3 is called power (index) or exponent

a is called literal coef icient of x.

• The•combination•of•the•algebraic•terms with (+) or (–) sign is called
algebraic•expression.

• Examples : 2x2 – 3x•+•5,••••••••2x•+•3,•••••••••2x•etc.

The•algebraic•expression having non–negative Algebra : Polynomials
exponents (power) of the variables and having
coef icient•a•real•number•is•called•polynomial.

Standard•form•of•polynomial•in•x•:
A polynomial written in descending or ascending order of the exponent
(power) of the variable x is called the polynomial in standard form. The
highest•value of exponent of the variable is called the degree of the variable.
For example :

p(x) = a0xn + a1xn-1 + a2xn–2 + ..... + anx0, a0 ¹ 0

• Where,•n•is•non–negative integer and
a0, a1, a2, .....•an are•the•real•numbers•(coef icient)•&•a0 ¹ 0.

• n is•called•the•degree•of•the•polynomial•p(x)
• a0xn, a1xn-1•...•are•the•terms•of•the•polynomial.

PRIME Opt. Maths Book - VIII 21

Types of Polynomials
i) According to degree
• The•polynomial•having•degree•‘1’•is•called•a•linear•polynomial.
• p(x) = ax + b, a ¹ 0 (First•degree•polynomial)
• The•polynomial•having•degree•‘2’•is•called•a•quadratic•polynomial.
• p(x) = ax2 + bx•+•c,•a•¹ 0 (Second•degree•polynomial)
• The•polynomial•having•degree•‘3’•is•called•cubic•polynomial.
• p(x) = ax3 + bx2 + cx•+•d,•a•¹ 0 (Third•degree•polynomial)
• The•polynomial•having•degree•‘4’•is•called•biquadratic•polynomial.
• p(x) = ax4 + bx3 + cx2 + dx + e,•a•¹ 0 (Fourth•degree•polynomial)

ii) According to number of terms, polynomials are classi!ied as the

following.

i) Monomial ® Having only one term in the expression.

p(x) = 2x

Algebra : Polynomials ii) Binomial ® Having two terms in the expression.

p(x) = 2x2 + 3

iii) Trinomial ® Having three terms in the expression.

p(x) = 2x3 + 3x•+•2

• iv) Multinomial ® Having more terms in the expression.

or Polynomial p(x) = 2x4 + 3x3 + 5x2 – 3x•+•7

Other•information•on•polynomials
• The•polynomial•may•have more than one variable also.

P(xy) = ax2y + bxy•+•cxy2

p(xyz) = ax2yz•+•bxy2z + cxyz2.

• The•polynomials•having degree and coef icient of the corresponding

terms are same are called equal polynomials.

p(x) = 2x3 + 5x2 – 3x•+•2 8
4 6x 4
• q (x) = 2 x3 + 5x2 – 2 +

= 2x3 + 5x2 – 3x•+•2

• Here, p(x) = q(x)

Note : All polynomial expressions are algebraic expression but all algebraic
expressions are not polynomials.

22 PRIME Opt. Maths Book - VIII

Worked out Examples

1. If•p(x)•= 12x3 + 5x2 – 10x + 7 and q(x) = (2m + 2)x3 + 5x2 – (3n + 1)x + 7

are equal polynomials, ind the value of ‘m’ and ‘n’.

Solution :

p(x) = 12x3 + 5x2 – 10x•+•7

• q(x) = (2m + 2)x3 + 5x2 – (3n•+•1)x•+•7

• Since, the polynomial are equal,

Corresponding coef•icients are also equal,

i.e. 2m + 2 = 12 and –(3n + 1) = –10

or, 2m = 10 or, 3n = 9

\ m=5 \ n=3

2. If•f(x)•=•x3 + 2x2 + 3x•–•2•and•g(x)•3x3 + 5x2 – 7x•–•1,• ind•f(x)•+•g(x). Algebra : Polynomials
Solution :
f(x) = x3 + 2x2 + 3x•–•2

• g(x) = 3x3 + 5x2 – 7x•–•1
• Then,

f(x) + g(x) = (x3 + 2x2 + 3x•–•2)•+•(3x3 + 5x2 – 7x•–•1)
• = x3 + 2x2 + 3x•–•2•+•3x3 + 5x2 – 7x•–•1
• = (x3 + 3x3) + (2x2 + 5x2) + (3x•–•7x)•+•(–•2•–•1)
• = 4x3 + 7x2 – 4x•–•3

3. What•must•be•subtracted•from the polynomial 5x3 – 3x2 + 2x + 5 to get
the polynomial 2x3 – x2 + 3x•–•2.
Solution :
Let, the subtracted polynomial be ‘K’.
Then, by the question,
(5x3 – 3x2 + 2x•+•5)•–•K•=•2x3 – x2 + 3x•–•2

• or, (5x3 – 3x2 + 2x•+•5)•–•(2x3 – x2 + 3x•–•2)•=•K
• or, 5x3 – 3x2 + 2x•+•5•–•2x3 + x2 – 3x•+•2•=•K

\ K = 3x3 – 2x2 – x + 7
\ The subtracted polynomial is, 3x3 – 2x2 – x + 7

Remember : The required polynomial is
= (Subtracted from the polynomial) – (to get the polynomial)
= (5x3 – 3x2 + 2x•+•5)•–•(2x3 – x2 + 3x•–•2)

PRIME Opt. Maths Book - VIII 23

Algebra : Polynomials 4.• What•should•be•added•with•3x3 + 5x2 – 1 – 7x•to•get•x3 + 2x2 + 3x•–•2?
Solution :
Let the required polynomial be K.
According to question
(3x3 + 5x2 – 1 – 7x)•+•k•=•x3 + 2x2 + 3x•–•2

• or, K = (x3 + 2x2 + 3x•–•2)•–•(3x3 + 5x2 – 1 – 7x)
• = x3 + 2x2 + 3x•–•2•–•3x3 – 5x2 + 1 + 7x
• = x3 – 3x3 + 2x2 – 5x2 + 3x•+•7x•–•2•+•1
• = – 2x3 – 3x2 + 10x•–•1
• Remember : The required polynomial is

= (To get the polynomial) – (added with the polynomial)
= (3x3 + 5x2 – 1 – 7x)•–•(x3 + 2x2 + 3x•–•2)
5.• If•P(x)••=•3x3 – 4 and•Q(x)•=•3•–•4x•+•7x2 – 5x3, ind•P(x).Q(x)
Solution :
We have P(x).Q(x) = (3x3 – 4).(3•–•4x•+•7x2 – 5x3)

= 3x3(3•–•4x•+•7x2 – 5x3) – 4(3•–•4x•+•7x2 – 5x3)
= 9x3 – 12x4 + 21x5 – 15x6 – 12 + 16x – 28x2 + 20x3
= –15x6 + 21x5 – 12x4 + 9x3 + 20x3 – 28x2 + 16x – 12
= –15x6 + 21x5 – 12x4 + 29x3 – 28x2 + 16x•–•12

24 PRIME Opt. Maths Book - VIII

Exercise 1.3

1.•• Answer the following questions.

i) What is polynomial? Write down its types according to degree of

polynomial.

• ii) What is algebraic expression? Write down the types of polynomial

according•to•number•of•terms•of•the•polynomials.

• iii) What do you mean by equal polynomials? Explain with an

example.

iv) Write down numerial coef•icient, literal coef•icient, base and

power of the polynomial

(a) 3x3y2. in•y.•• (b)•3x3y2 in•x•.•• • (c) 3ax3y2 in•‘a’.

v) Write down the degree of the polynomial 3x4 – 2x3 + 5x2 – 2x + 7.

Also•write•down•its•type•according•to•degree•of•the•polynomial.

2.• Which• of• the• following algebraic expressions are the polynomials ? Algebra : Polynomials

Write down with reasons.

i) 4x3 + 5x2 – 3x•+•2• ii)• 3 x3 + 4x2 + 2x
1 1 2 3
• iii) x2 – 3 + 4 +x2 iv) x4 ( x3 + x2 + x )+•10
x

• v) 7 x3 + 3x2 – 5 x–7+ 5
3 2 x

3.• Write down the following polynomials in standard form and •ind their
degrees.

• i) 3x – 5 – 3x2 + x4 – 2x3
ii) 7 – 2x2 – 3x•+•5x3
iii) 3 – 5x2 – x3 + 2x4 – 3x•+•x5
iv) 3x – 2 + 5x2 – x5 + 2x3 – 3x4
v) 3x4 + 2x5 – 7 – 3x•–•2x2 + x3

4.• Write down the types of the polynomials according to degree and

number•of•terms.

• i) 3x – 2x2 – 5 + 2x4 – x3 ii) 5 + 3x2 – 2x•+•3x3

iii) –3 + 2x + x2 iv) 7 – 3x

v) 2

PRIME Opt. Maths Book - VIII 25

Algebra : Polynomials 5.• If•the•following•polynomials•are•equal,••ind•the•value•of•‘a’•and•‘b’.
i) p(x) = (2a + 1)x3 – 7x2 + 3x•+•2•and

• q(x) = 7x3 + (2b•–•3)x2 + 3x•+•2
• ii) f(x) = 10x4 – 3x3 + 5x2 – 12x•+•5•and
• g(x) = 10x4 – (3x•–•3)x3 + 5x2 + 3bx•+•5.
• iii) p(x) = 5x2a•–•1•+•(3b•+•2)x2 – 7x•+•3.
• q(x) = 5x3 + 8x2 – 7x•+•3.
• iv) f(x) = 2ax4 – 3x3 + bx2 – 2

g(x) = 4x4 – 3x3 – 2
v) p(x) = (3a – 2)x4 – 2x2b•–•4•–•3x2 + 2x•+•5
• q(x) = 7x4 – 3x2 + 2x•+•5

6.• Find•the•following•polynomials.•
• i) Add the polynomials: p(x) = x3 – 2x2 – x + 5 and q(x) = 3x3 + x2 – 2x – 1

ii) Add the polynomials: f(x) = 2 – 3x + x2 + 2x3 and g(x) = 2x – x2 +3x3 – 3
iii) Find the sum of the polynomials p(x) = 3x2 – 2x + 3x3 – 3 and

q(x)•=•–2x3 – x2 – x + 15.
• iv) Subtract the polynomials p(x) = 2x3 – x2 + 2x + 3 and

q(x)•=•x3 + x2 – x + 5.
• v) Subtract the polynomials x3 + 3x2 – 2x•–•7•from•4x3 – 5 +2x•–•x2.

7.• Find•the•following.
• i) If•p(x)•= 3x3 + 2x2 – 5x + 2 and q(x) = 2x3 – 3x2 – 2x + 3, •ind p(x) +

q(x).
• ii) If•p(x)•+ q(x) = 5x3 – 3x2 + 2x – 5 and p(x) = 3x3 – x2 + 3x – 2, •ind

q(x).
• iii) What•must•be•subtracted•from the polynomial 4x4 – 3x3 + 2x2 – 5x + 1

to get 3x4 – x3 – 5x2 + 2x•+•3•?
• iv) What must be added to the polynomial x3 – 3x2 – 2x + 3 to get

x4 + 3x3 – x2 + 3x•–•2?
• v) What must be subtracted from the sum of x4 + 2x3 – 3x2 + 2x – 5

and•2x4 + x3 – x2 – 3x•+•2•to•get•x4 – x3 + 2x2 – x – 2?

8. PRIME more creative questions
i) If f(x) = 3x3 + 2x2 – x + 2 and g(x) = x4 – x3 + x2 + x – 5, •ind f(x)
+ g(x). Also write down the types of polynomial of the result
according•to•degree•and•number•of•terms.

26 PRIME Opt. Maths Book - VIII

ii) A polynomial x3 + x2 – 3x – 1 is subtracted from p( ) results x4 – x3
+ 2x2 – x + 2,• ind•the•polynomial•p(x).

• iii) If p(x) = (x2 + 2x – 3) and q( ) = (2x2 – x + 2). Find the value of
p(x)•× q(x). Also write down the type of polynomial according to
degree.

• iv) Multiply the polynomial x2 + 2x•–•3•and•3x3 – 2x2 + 3x•–•5.
• v) If f(x) = (x2 + 2), g(x) = 2x2 – x + 3, what must be subtracted from

the•product•of•f(x)•and•g(x)•to•get•x4 – x3 + 2x2 – x – 2?

Answer Algebra : Polynomials

1.• Consult•to•your•subject•teacher.
2.• Consult•to•your•subject•teacher.
3.• Consult•to•your•subject•teacher.
4.• Consult•to•your•subject•teacher.

5.• i)• a•=•3,•b•=•–2• ii)• a•=•2,•b•=•–4• iii)• a•=•2,•b•=•2
• iv) a = 2, b = 0 v) a = 3, b = 2

6.• i)• 4x3 – x2 – 3x•+•4
• ii) 5x3 – x – 1

iii) x3 + 2x2 – 3x•+•12
• iv) x3 – 2x2 + 3x•–•2
• v) 3x3 – 4x2 + 4x•+•2

7.• i)• 5x3 – x2 – 7x•+•5
• ii) 2x3 – 2x2 – x – 3

iii) x4 – 2x3 + 7x2 – 7x•–•2
• iv) x4 + 2x3 + 2x2 + 5x•–•5
• v) 2x4 + 4x3 – 6x2 – 1

8.• i)• x4 + 2x3 + 3x2 ; Biquadratic,•multinomial
• ii) x4 – 2x3 + 4x2 + 2x•+•3
• iii) 2x4 + 3x3 – 6x2 + 7x•–•6•;•Biquadratic.
• iv) 3x5 + 4x4 – 10x3 + 7x2 – 19x•+•15
• v) x4 + 4x2 – x + 8

PRIME Opt. Maths Book - VIII 27

Algebra : Sequence and Series 1.4 Sequence and series

Let•us•take•examples of the set of numbers taken in order.
1,•4,•7,•10,•...,•...,•...,•...,•...,•...,•...,•...
1,•3,•9,•27,••...,•...,•...,•...,•...,•...,•...,•...
40,•30,•20,•10,•0,••...,•...,•...,•...,•...,•...,•...,•...

Here,• the• set• of• numbers• are written in ascending and descending order
under•the•rule•of•increased•by (+3) in •irst, by (×3) in second and decreased
by 10 in third example respectively.

Such• type• of• arrangement• of• the• numbers• as• the• above examples taken
in•ascending•or•in•descending•order under a certain rule with a constant
number•is•called•sequence.

Sequence•: The set of the numbers taken in order
under•a certain rule is called sequence. Eg.: 2, 6,
10,•14,•18,••...,•...,•...,•...,•...,•...,•...,•...

• A sequence having !inite number of terms is called !inite sequence.
For example : 2, 5, 8, 11, 14, 17, 20.

• A sequence having in!inite number of terms is called in!inite
sequence. For example : 3, 7, 11, 15, 19, ...,•...,•...,•...,•...,•...,•...,•...

• Set of numbers of a sequence is also called the progression.

The•sum•of•the•terms of a sequence is called the
series.•
• If•the•no.•of•terms in the sequence is •inite, the

corresponding•series•is•called•the••inite series.
• If•the•number•of•terms is in•inite, the series is

called•in•inite•series.

• Finite series is 2 + 5 + 8 + 11 + 14 + 17
• In!inite series is 3 + 7 + 11 + 15 + 19 + •...,•...,•...,•...,•...,•...,•...,•...

28 PRIME Opt. Maths Book - VIII

Types of sequence Algebra : Sequence and Series
1.• Arithemetic•Sequence•: A sequence is said to be an Arithmetic Sequence

(AS) or Arithmetic Progression (AP), if the difference between any two
consecutive terms remains constant. The constant difference is called
the•common•difference (d) and the next consecutive term is obtained
by adding d with the term.

Arithmetic•Sequence•: The set of number of a sequence
which is written under a certain rule of addition with a
constant•number•is•called•arithmetic•sequence.•
Example : 2, 7, 12, 17, 22, 27.
Here,•the•numbers•taken•in•order•are•increased•by•(+5)
• The•constant•difference is called common difference.

d = t2 – t1

• tte1,rtm2, st3o, tf4s, et5q, u..e.,n..c.,e...o, r...s, .e..r, i.e..,s...., ... etc are taken symbolic form of the

• Symbolically the constant addition number of the terms is ‘d’

called common difference where,

tttt4213 = a, = a + 0 . d = a + (1 – 1)d
= a + d = a + 1.d = a + (2 – 1)d
= a + 2d = a + (3 – 1)d
= a + 3d = a + (4 – 1)d and so on.

nth term of the sequence can be wt3r.i.t..t.e...n...a..s...t..n..=.....a...+.....(.n....–•e1tc).d
Here, d = t2 – t1 = t3 – t2 = t4 –

\ tn = a + (n – 1)d

• More examples of arithmetic sequence.
2, 7, 12, 17, 22, ...,•...,•...,•...,•...,•...,•...,•...
40, 30, 20, 10, 0, –10, –20, ...,•...,•...,•...,•...,•...,•...,•...
a, a + d, a + 2d, a + 3d, a + 4d, ...,•...,•...,•...,•...,•...,•...,•...
a, a – d, a – 2d, a – 3d, a – 4d, ...,•...,•...,•...,•...,•...,•...,•...

PRIME Opt. Maths Book - VIII 29

• Above examples in the form of series.
2 + 7 + 12 + 17 + 22 + ...,•...,•...,•...,•...,•...,•...,•...
40 + 30 + 20 + 10 + 0 + (–10) + (–20) + ...,•...,•...,•...,•...,•...,•...,•...
a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) + ...,•...,•...,•...,•...,•...,•...
a + (a – d) + (a – 2d) + (a – 3d) + (a – 4d) + ...,•...,•...,•...,•...,•...,•...,•...

2.• Geometric•Sequence•: A sequence is said to be a Geometric sequence
(GS)• or• Geometric• progression (GP) if the ratio between any two
consecutive terms remains constant. The constant ratio is called
the• common• ratio (r) and the next consecutive term to any term is
obtained•by•multiplying•the•term•with•common•ratio•‘r’.

Algebra : Sequence and Series Geometric•Sequence•: The set of number of a sequence

which is written under a certain rule of multiplication

with• a constant number is called geometric sequence.

Example : 2, 6, 18, 54, ..............

Here,•the•numbers•are written in ascending order under

the•multiplication•by•a•constant•number•by•(ו3).

• The•constant•ratio•is•called•common•ratio.
t2
r= t1

• tte1,rtm2,st3o, ft4t,hte5, ...,•...,•...,•...,•...,•...,•...,•... etc. are the symbolic form of the
sequence or series.

• The constant multiplicative number is symbolised by ‘r’ which is

called common ratio, where

tttt3421 = a = ar0 = ar1 – 1 1
= ar = ar1 = ar2 –
= ar2 = ar3 – 1
= ar3 = ar4 – 1

Similarly as above by inspection,

nth term of the s=eqttu32en=cetti34s,=tn\.=...t..na..r.=.n..–.a1...r.ne–1tc.
t2
Where, r = t1

30 PRIME Opt. Maths Book - VIII

• More example of geometric sequence.

3, 6, 12, 24, 48, ...,•...,•...,•...,•...,•...,•...,•...

80, 40, 20, 10, 5, ..52.,•..,.,•54...,,•......,,••......,,••......,,••......,,••......,•...,•...,•...
a, ar, ar2, ar3, ar4,
a a a a a
a, r , r2 , r3 , r4 , r5

• Geometric series for the above examples.

3 + 6 + 12 + 24 + 48 + ...•+•...•+•...•+•...•+•...•+•...•+•...

80 + 40 + 20 + 10 + 5 + .52..•++•...•5+4•..+.•+.•....•.+•+•.•....•.+•+•.•....•.+•+•.•....•.+•...•+•...•+•...
a + ar + ar2 + ar3 + ar4 +
a a a a a
a+ r + r2 + r3 + r4 + r5

Worked out Examples Algebra : Sequence and Series

1. What•type of sequence is given below? Also ind the 10th term of 2, 6, 10,

14,•18,•...,•...,•...,•...,•...,•...

Solution:

The given sequence is, 2, 6, 10, 14, 18, .....................

Here,

First term (a) = 2

Common difference (d) = 6t2 – 2t1
= –

=4

Also,

• d = tt43 – tt23 = 10•–•6•=•4
d = – = 14•–•10•=•4

• Hence, It is arithmetic sequence.

We have,

nth term of the sequence is,

ttn10 = a + (n – 1)d
= 2 + (10 – 1)4

=2+9×4

= 38

PRIME Opt. Maths Book - VIII 31

2.• Which sequence represents the given pattern? Also ind the no. of dots
found in 6th diagram.

Solution:

The sequence formed according to the no. of dots present in the given

diagrams•is•3,•5,•7,•...,•...,•...,•...,•...

• Here, ittt32is–– att12ri==th75m––e35tic== 2
Hence, 2
sequence

Algebra : Sequence and Series Where,

First term (a) = 3

Common difference (d) = t52 – 3t1=
= –
2

\ No•of•dots•found•in•6th diagram•is

• tn = a + (n – 1)d
= 3 + (6 – 1)2

=3+5×2

= 13

3.• Find•the•8th term of the sequence 512, 256, 128, 64, ..., ..., ..., ..., ...

Solution :

The given sequence is, 512, 256, 128, 64, ..., ..., ..., ..., ..., ..., ..., ...

Here, t2 – t1 = 256 = 1
512 2
128 1
t3 – t2 = 256 = 2

Hence, It is geometric sequence.

Then, First term (a) = 512
t2
Common ratio (r) = t1 = 256 = 1
8Wthetheramve,(t8) = ? 512 2

• or, tn = arn–1 ` 1 8–1
t8 = (512) 2
j

32 PRIME Opt. Maths Book - VIII

= 512 × ` 1 7
2
j

= 512 × 1
128
\ t8 = 4

4. If• irst•term is 3 of an arithmetic sequence having 9th term 35, ind the

common difference.

Solution:

First term (a) = 3

• c9othmtemrmon(dt9i)ff=er3e9nce (d) = ?

We have,

or, 3tt9n5 = a + (n – 1)d Algebra : Sequence and Series
or, = a + (9 – 1)d
= 3 + 8d

or, 32 = 8d

\d =4

Exercise 1.4

1.• What•types•of•sequences•are given below? Write down with the reason
after•inspection.

• i) 3, 8, 13, 18, 23, ..., ..., ..., ..., ..., ..., ..., ...
ii) 5, 10, 20, 40, ..., ..., ..., ..., ..., ..., ..., ...
iii) 2, 6, 18, 54, ..., ..., ..., ..., ..., ..., ..., ...
iv) a, a + d, a + 2d, a + 3d, ..., ..., ..., ..., ..., ..., ..., ...
v) a, ar, ar2, ar3, ar4, ...,•...,•...,•...,•...,•...,•...,•...

2.• Identify•the•following as sequence and series. Also •ind the common

constant•term•associated•with•them.

• i) 5 + 12 + 19 + 26 + ... + ... + ... + ... + ... + ... + ...

ii) 100, 90, 80, 70, 60, ..., ..., ..., ..., ..., ..., ..., ...

iii) 3 + 6 + 12 + 24 + 48 + 96 + 192 + ... + ... + ... + ... + ... + ... + ...

iv) a + 10d, a + 9d, a + 8d, a + 7d + ..., ..., ..., ..., ..., ..., ..., ...
a a a a
v) a, r , r2 , r3 , r4

PRIME Opt. Maths Book - VIII 33

3.• Identify•the•sequences•or•series•given below as •inite and in•inite.

i) 5 + 12 + 19 + 26 + ... + ... + ... + ... + ... + ... + ...

ii) 100, 90, 80, 70, 60, ..., ..., ..., ..., ..., ..., ..., ...

iii) 3 + 6 + 12 + 24 + 48 + 96 + 192 + ... + ... + ... + ... + ... + ... + ...

iv) a + 10d, a + 9d, a + 8d, a + 7d + ..., ..., ..., ..., ..., ..., ..., ...
aF,inard,•thra2e•,12ra3th t,erra4m of the sequence : 2, 6, 10, 14, 18 ..., ..., ..., ..., ..., ...
4.• v)
i)•

ii) Find the 8th term of the series 40 + 34 + 28 + 22 + 16 ..., ..., ..., ..., ...

iii) If •irst term and common difference of an arithmetic sequence

are 6 and 8 respectively. Find the 15th term.

iv) If •irst term is 50 and common difference is –6, •ind the 11th term

of•the•arithmetic•sequence.

• v) If 10th term of the arithmetic sequence having common difference

4 is•50,••ind•the••irst•term.

Algebra : Sequence and Series 5.• i)• Find•the•8th term of the sequence 3, 6, 12, 24, ..., ..., ..., ..., ..., ..., ..., ...

ii) Find the 6th term of the series 128 + 64 + 32 + ..., ..., ..., ..., ..., ..., ..., ...

iii) Find the 10th term of the sequence 5, 10, 20 40, ..., ..., ..., ..., ..., ..., ...

iv) If •irst term and common ratio of a geometric sequence are 10

and•2•respectively, •ind the 5th term of the sequence.

v) If •irst term and common ratio of a geometric sequence are 80
1
and• 2 respectively, •ind the 6th term of the sequence.

6.• Find•the•followings:
• i) If •irst term and 6th term of an AS are 7 and 27 respectively, •ind

the•common•difference of the sequence.
ii) If common difference and 9th term of an AS are 3 and 44

respectively, •ind the •irst term of the sequence.
iii) If •irst term and common ratio of a GS are 2 and 3 respectively,

•ind•the••ifth•term•of•the•sequence.
• iv) If •irst term and 4th term of a GP are 5 and 135 respectively, •ind

the•common•ratio•of•the•sequence.
• v) If common ratio and 6th term of a GS are 4 and 5120, •ind the •irst

term of the sequence.

34 PRIME Opt. Maths Book - VIII

7. PRIME more creative questions
i) Find the number of dots present in 8th igure in the given pattern
of•triangles.

•
ii) Add two more diagrams for the followings. Also ind the number
of•dots•found•in•10th igure•in•the•given pattern.

iii) Monthly salary of a person in the month of Baishakh is Rs. 4000 Algebra : Sequence and Series
where monthly increment is given for irst 6 months by Rs. 500.
What•will•be•his•salary•in•the•month•of•Aswin?

• iv) If irst term of an arithmetic sequence having 8th term 25 is 4,
ind•the•common•difference of the sequence.

v) Find the 7th term of the series 320 + 160 + 80 + ... + ... + ... + ...

Answer

1.• Show•to•your•teacher.

2.• Show•to•your•teacher.

3.• Show•to•your•teacher.

4.• i)• 46• ii)• –2• iii)• 118• iv) –10 v) 14
iii) 2560 iv) 160 v) 2.5
5.• i)• 384• ii)• 4 iii) 162 iv) 3 v) 5
iii)• Rs. 6,500 iv) 3 v) 5
6.• i)• 4 ii) 20

7.• i)• 24• ii)• 50•

PRIME Opt. Maths Book - VIII 35

Alzebra

Unit Test - 1
Time•:•30•minutes

[1•×•1•+•3•×•2•+•2•×•4•+•1•×•5•=•20]

Attempt all the questions:

1.• Name•the•x-component•and•y-component•of•a•ordered•pair•(x,•y).

2.• a.• If•(3x•– 2, 4) = (x + 4, 3y + 16) are two equal ordered pairs, ind ‘x’

any ‘y’.

b. Which is greater between 3 or•3 5 ?
c. Write down the polynomial 5x3 – 2x2 + x4 – 3x + 2 in standard form

and•mention•its•type•based•on•degree.

3.• a.• Simplify: 4– 1
3– 6– 2 3+ 2
• 6– 3

Algebra : Ordered Pairs b. If A = {1, 2, 3} and B = {4, 5}, ind A × B and show in arrow diagram.
Also• ind•the•cardinality•of•A•×•B.

4.• Find•8th term of the sequence 3, 7, 11, 15, ..............................

Unit Test - 2
Time•:•30•minutes

[1•×•1•+•3•×•2•+•2•×•4•+•1•×•5•=•20]
Attempt all the questions:
1.• Write down one difference between sequence and series.
2.• a.• What• is• ordered pair? Write down any three ordered pairs of

district•with•respect•to•headquarter.

b. Simplify: 18 + 3 50 – 98
c. Find the value of p(x) + q(x) where p(x) = 3x2 – 5x + 2 and q(x) =

x3 – 2x2 + x – 3.
3.• a.• If•A × B = {(a, p), (a, q), (a, r), (b, p), (b, q), (b, r)} ind the sets A and

B.•Also• ind•B•×•A•and•n(B•×•A).

• b. Write down in ascending order of: 2 , 3 4 and• 4 5
4.• Find•the•6th term of the series. 3 + 6 + 12 + ..........................

36 PRIME Opt. Maths Book - VIII

Unit 2 Limits

Speci•ication•Grid•Table

K(1) U(2) A(4) HA(5) TQ TM Periods

No. of Questions – – 1 – 14 4

Weight ––4–

K = Knowledge, U = Understanding, A = Application, HA = Higher ability,
TQ = Total Question, TM = Total Marks

Objectives : At the end of the lesson
• Students are able to know sequence and series in descending and

ascending order.
• Students are able to round up the numbers in decimal and whole number.
• Students are able to know the limiting value of the sequence of numbers.
• Students are able !nd the limiting value of algebraic expressions.

Materials Required:
• Chart paper.
• Chart of numbers taken in order.
• chart of examples of limit values.
• Some of the sequences in diagrams.

PRIME Opt. Maths Book - VIII 37

Limit

Let•us•take example of sequences of in•inite terms. The in•inite term will be
nearly equal to a de•inite number which is called absolute value.
• 0.1,•0.01,•0.001,•0.0001,•0.00001,•...,•...,•...,•...,•...
• 0.9,•0.99,•0.999,•0.9999,•0.99999,•...,•...,•...,•...,•...
• 2.1,•2.01,•2.001,•2.0001,•2.00001,•...,•...,•...,•...,•...
• 2.9,•2.99,•2.999,•2.9999,•2.99999,•...,•...,•...,•...,•...

The•above sequence can be de•ined to their in•inite terms where the term
will•be•nearly equal to zero (0), 1, 2, 3 respectively. They are also called the
absolute values. The absolute value of such sequences are also called the
limiting•value•of•the•sequence•which•is•nearest•whole•number.

Diagrammatic representation of sequence to get absolute value.

i) Let us consider a line segment AB having length 20cm. It is divided

by a point at mid-point and so on as given in diagram.

10cm 20cm
2.5cm5cm
Limits
ASR Q P B

Here, AP = 10cm,
AB = 20cm, AR = 2.5cm,
AQ = 5cm,
AS = 1.25cm

The sequence of the length of the line segments so formed will be,

20, 10, 5, 5 , 5 , ....................•.
2 4

• As the value of the partition ‘n’ increases and for very large ‘n’ the
length• of• the• intersected• line• becomes• very small which is about to
zero•i.e.•as•n ® ¥, l ® 0 and we•write•this•information•as

• n lim l = 0
"3

38 PRIME Opt. Maths Book - VIII

Let us consider the another situation,
Let add the intersected half parts successively as shown in the •ig
which is denoted by S (say).

i.e.• Sn==1B0P+•+5•P+Q2•+.5•Q+R1•+.2•R5S+•+.•....................................
•

As•• n ® an (i.e.•in•initely intersected)
S ® 20 (but•not•equal•to•20)

• We write this information in limit as

n lim Sn = 20
"3

The• sequence• to denote whole line segment AB

from its parts as,

• 10,•5,• 5 , 5 , 5 , ....................•» 0
2 4 8
5 5 5
• 10•+•5•+• 2 + 1 + 8 + .....................•» 0 Limits

ii) When a basketball is dropped on the ground, it is going in
dropping on decreasing height and !inally comes at rest (nearly)
as shown in the diagram after in!inite drops.

• As• the• dropping times increases, the height travelled by the ball
decreases.

• Finally•the•height•travelled by the ball nearly equal to zero.

PRIME Opt. Maths Book - VIII 39

iii) Let a full cup of water of height 16cm is made half in each attempt
as shown in the !igure.

16cm

8cm 4cm

2cm 1cm

The sequence so formed is 16, 8, 4, 2, 1 ,...................................

Discuss the absolute value of the above examples for their sequences
mentioned as above for their in!inite term.

Limits

A function or expression f(x) is said to be have a !inite

limit ‘l’ at x = a, if the value of f(x) approaches towards

the constant ‘l’ as the value of x approaches towards

the number ‘a’.

i.e. As x ® a implies f(x) ® l. We denote it as lim f(x) = l
x"a

i. In a sequence discussed above as 0.1, 0.01, 0.001, 0.0001, ..., ..., ..., the

in!inite term will be nearly zero which is called its limiting value. It can

be symbolized as, x lim f(x) = 0
"3

ii. Lets us take an example given in table.

x 0.9 0.99 0.999 0.9999 0.99999 ....... » 1

f(x) =3x + 2 4.7 4.97 4.997 4.9997 4.99997 ....... » 5

It can be symbolized in limit as,

lim f(x) = lim (3x + 2) =5
x "1 x "1

40 PRIME Opt. Maths Book - VIII

Note•:

1.• For a function y = f(x), f(a) is called the functional value at x = a.

[Put x = a in the expression]

2.• For a polynomial function, the functional value is the limiting value of

any point at x = a

i.e. lim f(x)•=•f(a)
x"a

3.• For a rational function, the functional value is the limiting value at
4.• oIxfr•o=•nnua•mpifu•eittrt•adintooger•sxa•nn=odat•ctiaatnktcaeek•tlehtseh•ef00ocromomr• m0033oonrfof•ar33cmto., rfas.cTtohreinzeptuhtexd=eanotomgientatthoer

limit•at•x•=•a.

Worked out Examples

1. Write down the 10th term of the sequence 1.1, 1.01, 1.001,.................... . Also

write down its limiting value.

Solution:

The given sequence is, Limits

1.1, 1.01, 1.001,.................... .

It’s 10th term can be de ined by inspection as,

• Ht10er=e1, .0000000001.

The in inite term of the sequence is very close to 1 due to which its

limit•value•is•1.

• i.e. lim (for•expression•on•‘x’)•=•1.
x"3

2. Area of a triangle is 32cm2 which is going to be divided by taking the
median of the triangle continuously in a sequence. Show the information
in diagram and examine the limiting value of area.
Solution:
Taking a triangle ABC having area = 32cm2 and dividing as the given
informations•by•taking•median•as.

• 41

PRIME Opt. Maths Book - VIII

The sequence of the area of triangle so formed is 32, 16, 8, 4, 2, 1,

....................•. The limit value of the area so formed is very close to zero

for in•inite diagram.

i.e. lim (Area)•=•0
x"3

3. Complete the given table and examine the limit value.

x 0.1 0.01 0.001 0.0001 0.00001 0.000001

f(x) =2x + 3 ......... ......... ......... ......... ......... .........

Solution:
Sequence of variable ‘x’ is given 0.1, 0.01, 0.001, 0.0001, 0.00001,
0.000001.•The•algebraic•expression•is•f(x)•=•2x•+•3•

Limits • Where,
Taking x = 0.1
f(x) = 2x + 3 = 2(0.1) + 3 = 3.2
Taking x = 0.01,
f(x) = 2x + 3 = 2(0.01) + 3 = 3.02. Then the table is completed as,

x 0.1 0.01 0.001 0.0001 0.00001 0.000001

2x•+•3 3.2 3.02 3.002 3.0002 3.00002 3.000002

• It’s absolute value for the in•inite term is very close to 3 which is taken

as•the•limit.

• i.e. limit = lim (2x•+•3)•=•3
x"0

4. Evaluate : lim (x2 + 2x – 3)
x"2

Solution:

lim (x2 + 2x•–•3)
x"2

• Taking x = 2 (very close to 2)

x2 + 2x•–•3•• =•22 + 2 × 2 – 3

=4+4–3

= 5.

\ lim (x2 + 2x•–•3)•=•5
x"2

42 PRIME Opt. Maths Book - VIII

5. Evaluate: lim x2 – 9
x"3 x–3
Solution:

lim x2 – 9 [when•x•=•3•then•function•becomes• 0 form]
x"3 x–3 0

= lim
x"3

=3+3

=6

\ lim x2 – 9 =6
x"3 x–3

Exercise 2.1

1.• Find•the•next•two•terms•and•limit•of•the•given sequences.

i. 1.9, 1.99, 1.999, 1.9999, ..............................

ii. 3.1, 3.01, 3.001, 3.0001, ..............................

iii. 5, 5 . 5 , 5 , 5 , .............................. Limits
2 4 8 16
• iv. 0.19, 0.199, 0.1999, 0.19999, ..............................

v. 4.101, 4.1001, 4.10001, 4.100001, , ..............................

2.• Complete the following table and diagrams to compute the limit value.

i. x 10 100 1000 10000 100000
f(x)•= 1 x ........... ........... ........... ........... ...........

• ii. x 0.1 0.01 0.001 0.0001 0.00001 0.000001
f(x)•=3x•+•4 ......... ........... ........... ........... ........... ...........

• iii. A BA BA B

P PQ

D CD C D C
upto 6 terms. Where area of rectangle ABCD = 80cm2 and P & Q
are taken as the mid-point of sides.

PRIME Opt. Maths Book - VIII 43

iv.

upto 6 terms where area of circle is 200cm2 & going on dividing
by half.
v. Draw a line segment of length 24cm and going on dividing by half
of•the•line•segment•continuously.

3.• Evaluate the following limits:

i. lim (3x•+•5)• • ii. lim (5x•–•2)
x"0 x"1

• iii. lim (2x2 – x + 3)• • iv. lim (x2 + 2x•–•3
x"2 x"3

• v. lim (x2 + 2ax•+•a2)
x"a

4. PRIME more creative questions:

Limits i. From the given diagram write down the sequence of area of

shaded•region where area of DPQR•= 100cm2 and taking the mid-

point•of•sides•as•shown•in•diagrams. P
PPP

•Q RQ RQ RQ R

• ii. Complete the given table and ind the absolute value.

x 10 100 1000 10000 100000 1000000

4 ........... ........... ........... ........... ........... ...........
x2

• iii. A sphere of radius 14cm is going on cutting down to hemi-sphere,

half•of••hemisphere and so on up to 5 times. Draw the diagram

and•write down the sequence for volume. Also ind its limit value.

iv. Evaluate: lim x2 – 4
x"2 x–2

v. Evaluate : lim x2 – a2 .
x"a x–a

44 PRIME Opt. Maths Book - VIII

Answer

1.• i.• 1.99999,•1.999999;•limit•value•=• x lim (expression•on•x)•=•2
"3

• ii. 3.00001, 3.000001; limit value = lim (expression•on•x)•=•3
x"3

• iii. 5 , 5 ; limit•value•=• x lim (expression•on•x)•=•0
32 64 "3

• iv. 0.199999,•0.1999999;•limit•value = lim (expression on x) = 0.2
x"3

v. 4.1000001,•4.10000001,•limit•value = lim (expression on x) = 4.1
x"3

2.• i.• Limit•value•=• x lim (expression•on•x)•=•0
"3

• ii. Limit value = lim (expression•on•x)•=•4
x"3

• iii. Limit value = lim (area)•=•80cm2 Limits
x"3

iv. Limit value = lim (area)•=•0
x"3

• v. Limit value = lim (length)•=•0
x"3

3.• i.• 5••••• ii.• 3•••••• iii.• 9 iv. 12 v. 4a2

4.• i.• Limit•value•=• lim (area)•=•100cm2
x"3

ii. Limit value = lim (expression•on•x)•=•0.
x"3

• iii. Limit value = lim = (volume)•=•0
x"3

• iv. 4 v. 2a.

PRIME Opt. Maths Book - VIII 45