O Math 10 2075

Approved by Curriculum Development Centre (CDC) Sanothimi, Napal
as a reference material for schools

PERFECT OPTIONAL

MATHEMATICS

Class 10

N.B. Khatakho

VIDYARTHI PRAKASHAN (P.) LTD.
Publisher and Distributor

Kamalpokhari, Kathmandu

Book : Perfect Optional Mathematics Class 10

Publisher : Vidyarthi Prakashan (P.) Ltd.
Kamalpokhari, Kathmandu
Phone : 01-4423333, 01-4245834

Copyright : Publisher
 

Edition : Revised, 2076

ISBN : 978 – 99937 – 715 – 52 –2

Price : Rs.400/-

Computer : Dynamic Computer Services, Bhaktapur
Printed at : Alliance Printers & Media House Pvt. Ltd.

Contents

Topics Pages

1. Algebra 1 – 78

1.1 Relations and Functions 2 – 12
1.2 Polynomials 13 – 15
1.3 Remainder and Factor Theorem 16 – 21
1.4 Arithmetic Sequence and Series 22 – 28
1.5 Arithmetic Mean 29 – 34
1.6 Sum of Arithmetic Series 34 – 43
1.7 Geometric Sequence and Series 43 – 49
1.8 Geometric Mean 50 – 54
1.9 Sum of Geometric Series 55 – 61
1.10 Linear Programming 62 – 70
1.11 Equations and Graphs 71 – 78

2. Limits and Continuity 79 – 91

3. Matrices 92 – 108

3.1 Determinant 93 – 101
3.2 Solving Equations by Matrix Method 102 – 108

4. Co-ordinate Geometry 109 – 154

4.1 Angle Between Two Straight Lines 110 – 114
4.2 Some Relations between Straight Lines 115 – 124
4.3 Equation of Pair of Straight Lines 125 – 130
4.4 Angle between Pair of Straight Lines 131 – 138
4.5 Conic Section (Circle) 138 – 147
4.6 Equation of Circle on Some Special Cases 147 – 154

5. Trigonometry 155 – 207

5.1 Trigonometric Ratios of Multiple Angles 156 – 167
5.2 Trigonometric Ratios of Sub-Multiple Angles 167 – 174
5.3 Transformation of Product to Sum or Difference 175 – 179
5.4 Transformation of Sum or Difference to Product 179 – 185
5.5 Conditional Trigonometric Identities 186 – 191
5.6 Trigonometric Equations 192 – 198
5.7 Heights and Distances 199 – 207

6. Vectors 208 – 238

6.1 Dot Product of Vectors 209 – 219
6.2 Formulae of Vector Geometry 220 – 227
6.3 Theorems on Vector Geometry 228 – 238

7. Transformation 239 – 285

7.1 Combination of Translations 241 – 244
7.2 Combination of Reflections 245 – 250
7.3 Combination of Rotations 251 – 254
7.4 Combination of Enlargements 255 – 259
7.5 Combination of Different Transformations 260 – 264
7.6 Insversion Transformation 265 – 272
7.7 Transformation Using Matrices 273 – 285

8. Statistics 286 – 303

8.1 Partition Values 287 – 293
8.2 Quartile Deviation and Mean Deviation 294 – 298
8.3 Standard Deviation 299 – 303

Model Questions 304 – 305

Answers 306 – 320

PREFACE

Perfect Optional Mathematics is a series of textbooks on optional mathematics for

the Secondary Level students which has been launched since 2060 BS. Designed specially
to boost the pupils’ creativity, the series is expected to be much more useful and complete
in itself to carter to the needs of the students. As an application of the knowledge and
skills obtained throughout my very long experience in the field of teaching, learning,
training and researches on mathematics, the series incorporates materials in accordance
with the interests of students and teachers. This series not only makes learning easier but
also enables the students to get the real taste of the tests as it includes so many varieties
of sums. For self-learners, the series presents many suitable examples and appropriate
exercises at essential places, which will demostrate more practical ways of learning
and teaching, too. Based on the curriculum developed by CDC, Sanothimi, each book
of the series incorporates exercises for short and long questions in each content given
separately.
Defacto and dejure, as this book is a result of practical researches and knowledge, it is
tuned to the enthusiastic teenaged students’ attitudes. I’m very much hopeful that this text
will also win the learners' heart and then rest in their mind as my other books.
Last but not least, I extend my heartfelt gratitude to Mr. Shanta Gautam, proprietor of
Vidyarthi Prakashan (P ) Ltd., who has been taking an immense responsibility to send the
text to the hands of students and teachers. Likewise, my thanks go to Mr. Binod Kasula,
Mr. Bijendra Bir Karmacharya, Members of Council of Mathematics Education,
Bhaktapur branch and all the wellwishers and generous hands who have helped me in
course of preparing this series in its present shape.
Any constructive suggestions and comments are greatly acknowledged.

Author

1. ALGEBRA Algebra

Curriculum Contents Included Periods

a. Function 35
– Algebraic and trigonometric functions (with graphs) (y = mx + c, y = ax3, y = sin A,

cos A, tan A (– 2p ≤ A ≤ 2p)
– Composite function (only two functions)
– Inverse function (arrow diagrams)
b. Polynomials
– Simple operations of polynomials : Multiplication, Division, Synthetic division method
– Remainder theorem and its use
– Factor theorem and its use
– Use of remainder theorem and factor theorem (solving equation up to 3 degree)
c. Sequence and Series
– Arithmetic sequence and series (Introduction, general term, means, sum)
– Sum of the first n-natural numbers (Even and odd numbers)
– Geometric sequence and series (Introduction, general term, means, sum of finite terms)
d. Linear programing problems
– Introduction
– Linear inequalities (identifying inequality from graph)
– Maxima and minima by linear programing
e. Quadratic equation and graphs
– Graph of quadratic and cubic functions
– Solution of quadratic equations (graph method)
– Solution of simultaneous linear and quadratic equations (graph and substitution method)

Objectives

On completion of this unit, students will be able to
operate the algebraic and trigonometric functions (y = mx + c, y = ax2 + bx + c, y = ax3, y = sin A, y = cos A,
y = tan A) and draw their graphs.
solve the problems of composite and inverse functions.
find the quotient and remainder when a polynomial is divided by monomial or binomial.
state and prove "remainder theorem" and "factor theorem" and use it to solve the poblems of finding
remainder and factor of the given polynomial divided by binomial x – a.
use synthetic division method to find quotient and remainder and factorize the polynomial and solve the
polynomial equations using the synthetic division.
find the genreal term, sum and means of arithmetic and geometric progression.
find the sum of n-natural numbers, even and odd natural numbers.
plot the graph of the given linear inequalities.
maximize/minimize the given objective function defined in the feasible region determined by the given set
of constraints.
draw the graph of quadratic equation of the form y = ax2 + bx + c and find its turning point.
solve the quadratic equation and linear equation by graphical and substitution method.

Specification Grid

Contents Knowledge Understanding Application Higher ability Total Total
Algebra Each of 1 mark Each of 5 marks Quesitons Marks
Each of 2 marks Each of 4 marks
2 1 8 21
32

 /1

1 ALGEBRA

1.1 Relations and Functions

Algebraic Functions

The functions involving algebraic expressions are called algebraic functions. Some examples of

algebraic functions are f(x) = 5, f(x) = 2x + 3, f(x) = x2 + 2x + 1, f(x) = x3 + 2 etc. The algebraic functions

have different degrees. Hence, there are different types of algebraic functions. Following are some special

types of algebraic functions.

Constant Function A B Y

A function f : A→B is said to be a constant f 2 y = f(x) = 2
function if, for every x∈A, there is same image in B 1 4
i.e. for all x∈A, y = f(x) = c, c∈B. For example, f = 2

{(1, 2), (2, 2), (3, 2)} is a constant function. In this case, 3 6 X'-5 4 -3 -2 -1 O 2 3 4 5 X
the range of f will always be a singleton set. Y'

When we discuss the functions over the sets of real numbers, it is difficult to represent them in

various ways except with the help of graph. The nature of the graph of function depends on the sets over

which we define the functions.

Linear Function Y y=x+2

A function f : A→B is said to be a linear function A B

if for all x∈A, y∈B, it can be expressed in the form of y 1 3

= mx + c, m and c being constants. For example, f : A→B 2 4
is defined as y = f(x) = x + 2 where A = {1, 2, 3} and 3
B = {3, 4, 5, 6}. 5
X
6 X' O
Y'

In this case, if we draw the graph of the function, all the ordered pairs lie on a straight line.

Identity Function

A function f : A→A is said to be an identity function if for all

x∈A, y = f(x) = x. Y y=x

For example, A = {1, 2, 3, 4}, Af A X' O X
f : A→A is defined as Y'
f = {(1, 1), (2, 2), (3, 3), (4, 4)} 1 1
2
2 3

3

Here, for x = 1, y = f(1) = 1, 44

for x = 2, y = f(2) = 2,

for x = 3, y = f(3) = 3 and

for x = 4, y = f(4) = 4.

∴ f satisfies y = f(x) = x, so f is an identity function.

2 / Optional Mathematics Class 10

Quadratic Function Y y = x2 – 4 Algebra

A function f : A→B is said to be a quadratic function if, 5
for every x∈A, y∈B, it can be expressed in the form of y = ax2 + 4
bx + c, a ≠ 0, b and c are constants. It represents a curve of conic 3
section named parabola (U shaped). 2
1
For example, y = f(x) = x2 – 4 is a quadratic function.
X' -5 4 -3 -2 --11 O 2 3 4 5 X
We find the value of f(x) for some values of x and
corresponding values of y, then the graph of the function is as -2
shown. -3
-4
x 0 1 2 –1 –2 3 –3 -5

y –4 –3 0 –3 0 5 5 Y'

Cubic Function Y y = x3
A function f : A→B is said to be a cubic function if, for
all x∈A, y∈B, it can be expressed in the form of 40
y = ax3 + bx2 + cx + d, a ≠ 0, b, c and d are constants. 32
For example, y = x3 is a cubic function. 24
16
x 0 1 – 1 2 – 2 3 – 3 8
y 0 1 – 1 8 − 8 27 – 27
X' -5 4 -3 --28 -1 O 2 3 4 5 X
Trigonometric Function
-16
-24
-32
-40

Y'

A function f : A→B is said to be a trigonometric function if the function f involves trigonometric
ratios: sine, cosine, tangent etc. For example, y = sin x, y = cos x, y = sin x + cos x etc. are the trigonometric
functions.

Consider a sine function, y = sin x

x 0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360°
y = sin x 0 0.5 0.87 1 0.87 0.5 0 -0.5 -0.87 -1 -0.87 -0.5 0

Consider a cosine function, y = cos x

x 0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360°
y = cos x 1 0.87 0.5 0 -0.5 -0.87 -1 -0.87 -0.5 0 0.5 0.87 1

Consider a tangent function, y = tan x

x 0° 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360°
y = tan x 0 0.58 1.73 ∞ -1.73 -0.58 0 0.58 1.73 ∞ -1.73 -0.58 0

YYY

1.0 y = sin x 1.0 y = cos x 4.0 y = tan x
3.0
0.5 0.5 2.0 X

X' X X' X X' 1.0 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360°

30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360° 30° 60° 90° 120° 150° 180° 210° 240° 270° 300° 330° 360° -1.0
-0.5 -0.5 -2.0
-3.0
-1.0 -1.0 -4.0

Y' Y' Y'

Relations and Functions / 3

Composite Function AfB Bg C

Let A = {2, 3, 4}, B = {6, 9, 12} and C = {4, 7, 10}. 26 6 4
If f : A→B and g : B→C are defined by 39 7
f = {(2, 6), (3, 9), (4, 12)} and 9 10
g = {(6, 4), (9, 7), (12, 10)} 4 12
The general forms of f and g are f(x) = 3x and g(x) = x – 2. 12
From diagram, it is clear that
f(2) = 6 and g(6) = 4 \ g[f(2)] = 4 Af Bg C
f(3) = 9 and g(9) = 7 \ g[f(3)] = 7
f(4) = 12 and g(12) = 10 \ g[f(4)] = 10 2 64
Then we can draw a single function g⸰f : A→C as 3 97
mapping diagram alongside. 4
g⸰f(x) = g[f(x)] = g[3x] = 3x – 2 12 10
The function from A to C is called the composite function of f and g.
g⸰f C
Definition
A g⸰f 4
7
2 10
3
4

Let f : A→B and g : B→C be two functions. Then the A f B g
new function defined from A to C such that every element of A
C

corresponds with a unique element of C is known as the composite x f(x) g[f(x)]
function of f and g. It is denoted by g⸰f or simply gf. g⸰f

Example 1: Let the function f : R→R be defined by f(x) = 4x2 – 3 for x > 1
2x + 1 for x ≤ 1

find a. f(2) b. f(0) c. f(h) – f(0) for 1 ≥ h
Solution: h

a. As 2 > 1, for f(2)

f(x) = 4x2 – 3

or, f(2) = 4×22 – 3 = 4×4 – 3 = 16 – 3 = 13

b. As 0 < 1, for f(0)

f(x) = 2x + 1

or, f(0) = 2×0 + 1 = 0 + 1 = 1

c. for, f(h) – f(0), f(h) = 2h + 1 as 1 ≥ h
h

or, f(h) – f(0) = [2h + 1] – [2×0 + 1] = 2h + 1 – (0 + 1) = 2h +1 – 1 = 2h = 2
h h h h h

4 / Optional Mathematics Class 10

Example 2: If f(x) = 1 sin x x , find the values of f(30°) and f πc . Algebra
+ cos 4
Solution:

The function given is

f(x) = 1 sin x x
+ cos

f(30°) = 1 sin 30°
+ cos 30°

= 1π4c1+/2=231=+si21nco×π4sc2π4c+2= 3 =1 ×2– 3= 2– 3 = 2 – 3
Again, f 1 2+ 3 2– 3 4– 3

1
2
+1

2

= 1 × 2 = 1 × 2 – 1 = 2 – 1 = 2 – 1
2 2+1 2+1 2–1 2–1

Example 3: If f = {(2, 4), (6, 10), (8, 2)} and g = {(4, 6), (10, 2), (2, 6)}, then show that the function f⸰g

and g⸰f in an arrow diagram and find it in ordered pair form.
Solution:

For f⸰g, gf
f⸰g(4) = f{g(4)} = f(6) = 10
f⸰g(10) = f{g(10)} = f(2) = 4 10 2 4
f⸰g(2) = f{g(2)} = f(6) = 10 4 6 10
Hence, f⸰g = {(4, 10), (10, 4), (2, 10)} 2 82

f⸰g

For g⸰f, f g
g⸰f(2) = g{f(2)} = g(4) = 6
g⸰f(6) = g{f(6)} = g(10) = 2 24 6
g⸰f(8) = g{f(8)} = g(2) = 6 6 10
Hence, g⸰f = {(2, 6), (6, 2), (8, 6)} 82 2

g⸰f

Example 4: If f = {(2, 3), (3, 4), (5, 6)} and g⸰f = {(2, 6), (3, 10), (5, 17)}, find the function g in the form

of set of ordered pairs. fg
Solution:
2 36
Here, f = {(2, 3), (3, 4), (5, 6)} and g⸰f = {(2, 6), (3, 10), (5, 17)} 3 4 10
Showing above information in the mapping diagram 5 6 17

\ g = {(3, 6), (4, 10), (6, 17)} g⸰f

Relations and Functions / 5

Example 5: If f : R→R and g : R→R be the two functions defined by f(x) = 3x + 7 and g(x) = 2(x – 8),

find (g⸰f)(x) and (f⸰g)(x) and test whether (g⸰f)(x) = (f⸰g)(x) or not.
Solution:

Here, f(x) = 3x + 7 and g(x) = 2(x – 8). and (f⸰g)(x) = f(g(x))

Now, (g⸰f)(x) = g(f(x)) = f(2(x – 8))

= g(3x + 7) = f(2x – 16)

= 2{(3x + 7) – 8} = 3(2x – 16) + 7

= 2(3x – 1) = 6x – 48 + 7

= 6x – 2 = 6x – 41

∴ (g⸰f)(x) = 6x – 2 ∴ (fog)(x) = 6x – 41
Hence, (g⸰f)(x) ≠ (f⸰g)(x)

Example 6: If f(x) = 2x – 5 and (f⸰g)x = 4x + 3, then find the equation of linear function g(x) and g(2).

Solution:

(f⸰g)x = 4x + 3
or, f{g(x)} = 4x + 3

or, 2g(x) – 5 = 4x + 3
or, 2g(x) = 4x + 3 + 5

or, 2g(x) = 4x + 8

or, g(x) = 4x + 8
2
 g(x) = 2x + 4.

Now, g(2) = 2×2 + 4 = 4 + 4 = 8.
Example 7: If f(x) = 3x, g(x) = x + 2 and f⸰g(x) = 18, find the value of x.
Solution:

Here, f(x) = 3x, g(x) = x + 2 and f⸰g(x) = 18.
i.e. f[g(x)] = 18

or, f(x + 2) = 18

or, 3(x + 2 ) = 18

or, x + 2 = 6

\ x = 4
Example 8: If f(x) = 2x – 1, g(x) = 3x and h(x) = x + 3, find

a. (f⸰g⸰h)(x) b. (g⸰f⸰h)(2)
Solution:

We have, f(x) = 2x – 1, g(x) = 3x and h(x) = x + 3.

a. (f⸰g⸰h)(x) = f((g⸰h)(x)) = f(g(h(x))) = f(g(x + 3)) = f(3(x + 3))
= f(3x + 9) = 2(3x + 9) – 1 = 6x + 17
∴ (f⸰g⸰h)(x) = 6x + 17

6 / Optional Mathematics Class 10

b. (g⸰f⸰h)(2) = g((f⸰h)(2)) = g(f(h (2))) = g(f(2 + 3)) = g(f(5))

= g(2 × 5 – 1) = g(9) = 3 × 9 = 27 Algebra

∴ (g⸰f⸰h)(2) = 27.

Exercise 1.1A

1. Define the following terms with an example.

a. Inverse function b. Composite function

c. Identity function d. Constant function

2. Write the name of the following functions. Y

a. f g b.

1 23 3 X
3 45
6 78 X' O

c. g⸰f d. Y'
Y
Y
5
5
4 4
3
2 3
1
2

1

X ' -5 -40-3 -2 --11 O1 2 3 4 5 X X ' -5 -40-3 -2 --11 O1 2 3 4 5 X
Y' Y'

3. a. A function f is defined on the set of natural number as follows:
2x – 7 for x is eovdedn,
b. f(x) = 4 – 2x for x is find f(2) and f(3).

4. a. A function f is defined on the set of integers as follows:
b.
5. a. f(x) = 4x + 1 for x < 2, find f(– 3), f(3) and f(8).
1 – 3x for x ≥ 2
b. pc pc
If f(x) = sin 2x + 1, find the values of f 6 and f 4 .
c.
If f(x) = 2 tan x , find the values of f pc and f pc .
1 – tan2x 6 2

If f = {(– 3, 4), (– 1, 5), (– 2, 6), (2, 4), (0, 3)} and g = {(4, 3), (5, 7), (3, 0), (6, 0)}, find the set
of ordered pairs of g⸰f and show in a mapping diagram.

If f = {(3, 9), (2, 4), (1, 1), (0, 0} and g = {(4, 3), (3, 2), (2, 1), (1, 0)}, then show the function
f⸰g in an arrow diagram and find it in ordered pair.

If f : {(2, 1), (3, 2), (4, 3)} and g(x) is an identity function, then find g⸰f and f⸰g. Also show
them in an arrow diagram.

6. a. If the functions f and g are given by f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)},
find (i) f⸰g(2), f⸰g(5) and f⸰g(1) (ii) g⸰f(1), g⸰f(3) and g⸰f(4)

b. If f = {(2, 3), (4, 6), (– 2, 4)} and g = {(4, 5), (3, 7), (6, 8)}, find the value of the following
composite functions.

(i) g⸰f(2) (ii) g⸰f(4) (iii) g⸰f(– 2)

Relations and Functions / 7

7. a. If f = {(4, – 5), (6, – 2), (0, – 1)} and g⸰f = {(0, 7), (4, 0), (6, 4)}, then find the function g by
showing the mapping diagram.

b. If g = {(2, 7), (3, 8), (4, 10)} and g⸰f = {(0, 7), (1, 8), (2, 10)}, then find the function f by
showing the mapping diagram.

8. a. If f(x) = x + 1 and g(x) = 2x, find f⸰g(x) and g⸰f(x).

b. If f(x) = x2 and g(x) = x – 2, find

i. (g⸰f)(x) ii. (f⸰g)(x) iii. (f⸰f)(x)

iv. (g⸰g)(x) v. (g⸰f )(4) vi. (f⸰g)(– 2)
9. a. If f(x) = 2x and g(x) = x2, then prove that f⸰g(x) is an identity function.

b. If f(x) = 2x + 3 and g(x) = 5x – 3, then prove that g⸰f(x) is an identity function.
5 2

10. a. If f(x) = 3x + 4 and g(x) = 2(x + 1), then prove that g⸰f(x) = f⸰g(x).

b. If g(x) = x + 2 and f(x) = 3x – 2, then prove that g⸰f(x) = f⸰g(x).
3

11. a. For the function f(x) = 2x – b, ff(2) = – 7. Find the value of b.

b. Given that function f(x) = 3x + a, if ff(6) = 10, find the value of a.

12. a. If the function f(x) = 2x – 1, g(x) = 3x + a and fg(2) = 9, find the value of a.

b. If the function f(x) = x + a, g(x) = 2x – a and gf(4) = 11, find the value of a.

13. a. If f(x) = 2x, g(x) = x – 4 and fg(x) = 10, find the value of x.

b. If f(x) = 3x – 2, g(x) = 2x – 5 and gf(x) = 15, find the value of x.

14. a. If f(x) = 3x + 1 and fg(x) = x + 3, then find g(x) and g(4).

b. If f(x) = 2x + 1 and f{g(x) + 2} = 4x – 1, find g(x) and g(3).

c. If g(x) = 2(x + 1) and f⸰g(x) = 8x + 13, then find f(x).

15. a. If f(x) = 4x, g(x) = x – 3 and h(x) = 3x – 1, find f⸰g⸰h(x) and (f⸰h⸰g)(x).

b. If f(x) = x + 2, g(x) = x – 4 and h(x) = 3x, find the following

i. (f⸰g⸰h)(x) ii. (g⸰f⸰h)(x)

iii. (h⸰g⸰f)(3) iv. (h⸰f⸰g)(10)

16. Draw the graphs of the following functions.

a. y = f(x) = 3x – 7, x∈N b. y = f(x) = 4 – x2, x∈Z

c. y = f(x) = 4 – x3, x∈Z d. y = f(x) = (x + 1)2, x∈Z

17. Draw the graphs of the following functions.

a. y = f(x) = 3x – 4, x∈R b. y = f(x) = x2 + 4, x∈R
d. y = f(x) = sin x, 0º ≤ x ≤ 360º
c. y = f(x) = x3 – 4, x∈R
e. y = f(x) = tan x, 0º ≤ x ≤ 90º



8 / Optional Mathematics Class 10

Inverse Function

Let A = {2, 3, 4}, B = {a, b, c} and the function f : Algebra

A→B is defined by f = {(2, a), (3, b), (4, c)}. A f B B gA
Also the domain of f = {2, 3, 4} and range of f = {a, b, c}.
2 a a2
b
∴ f is one to one onto function. 3 c b3

Now we define a function g : B→A as c4

g = {(a, 2), (b, 3), (c, 4)}. 4

So, g is also one-to-one onto function.

The function g defined from B to A is said to be the inverse of f and is denoted by f–1.

But for every function there may not exist their inverse. Let us study the following examples.

1. f = {(2, 3), (3, 3), (4, 5)}.

Here f is "many to one onto function." Af B B A

Here, a relation from B to A is not a function as 3∈B 2 3 3 2
3
corresponds with more than one element of A. So, f–1 3 5 4
45
does not exist.

2. f = {(2, 3), (4, 8), (5, 2)}. Af BB A

f is a one to one into function. 2 33 2
4 88 4
Here, a relation from B to A is not a function 99
because 9∈B does not correspond with any element 5 22 5
of A. So f–1 does not exist.

Hence, from the above examples, it is clear that to define an inverse function f must be one to one onto.

Definition

Let f : A→B be a one to one onto function. Then the function from B to A which associates each
element of B with a unique element of A is called the inverse function of f. The inverse function from B to A
is denoted by f–1 : B→A.

Example 1: Let A = {a, b, c} and f : A→A is defined as f = {(a, c), (b, a), (c, b)}, find f .–1

Solution: Af A A f–1 A
Here, f = {(a, c), (b, a), (c, b)}
a a aa
bb
Clearly, the function f is one to one onto, hence f–1 exists and b b cc
c
f–1 = {(c, a), (a, b), (b, c)}. c

Example 2: Let : R→R be defined by f(x) = 2x – 3, find the formula that defines f –1.

Solution:

Here, f : R→R and x is a variable, f is a one to one onto function, f–1 exists.

y = f(x) = 2x – 3,

To find the inverse function, we interchange the role of x and y

i.e. x = 2y – 3

Relations and Functions / 9

or, 2y = x + 3

∴ y = x + 3
2
x + 3.
Hence, inverse function y = f–1(x) = 2

Example 3: If f–1(x) = 2x – 3, find f(x).

Solution:

Let y = f–1(x) = 2x – 3. To find f(x), we interchage the role of x and y.

i.e. x = 2y – 3

or, 2y = x + 3

or, y = x + 3
2
x + 3.
Hence, the required function is f(x) = 2

Example 4: If f(x + 2) = x + 3 , then find f(x) and f –1(x).
x + 2
Solution:

Given f(x + 2) = x + 3
x + 2
x – 2 + 3
or, f(x – 2 + 2) = x – 2 + 2 [as x→x – 2]

or, f(x) = x + 1 = y (let)
x

For f–1(x), interchange the role of x and y.

i.e. x = y + 1
y

or, xy = y + 1

or, xy – y = 1

or, y(x – 1) = 1

or, y = x 1 1
–
1
 f–1(x) = y = x – 1

Example 5: If f(x) = 3 x and f(x) = f­–1(x), find the value of x.
+ 2x
Solution:

Here, let y = f(x) = 3 +x2x. Interchanging the role By question, we get
of x and y, we get
f(x) = f–1(x)
y
x = 3 + 2y or, 3 x = 1 3x
+ 2x – 2x
or, 3x + 2xy = y
or, 9x + 6x2 = x – 2x2

or, y – 2xy = 3x or, 6x2 + 2x2 + 9x – x = 0
or, y(1 – 2x) = 3x or, 8x2 + 8x = 0
or, 8x(x + 1) = 0

\ y = f–1(x) = 3x Either x = 0 or, x + 1 = 0 i.e. x = – 1
– 2x
1

10 / Optional Mathematics Class 10

Example 6: If f(x) = 3x + 4 and g(x) = x + 1. For what value of x, f–1⸰g(x) = 2? Algebra
Solution:

Here, y = f(x) = 3x + 4. Now interchanging the Again, g(x) = x + 1 and f–­1⸰g(x) = 2
role of x and y, we get or, f–1(x + 1) = 2

x = 3y + 4 or, (x + 1) – 4 = 2
3
or, 3y = x – 4
or, x + 1 – 4 = 6
x – 4
or, y = 3 x or, x – 3 = 6

\ f–­1(x) = – 4 or, x = 9
3

Example 7: If f(x) = 3x + 4; and g(x) = 4x – 13. If ff(x) = g–1(x), find the value of x.
2
Solution:

Here, given functions are f(x) = 3x + 4 and g(x) = 4x – 13 Now, ff(x) = g–1(x)
2
2x + 13
For the inverse function g–1(x). or, f(3x + 4) = 4

y = g(x) = 4x – 13 (let) or, 3(3x + 4) + 4 = 2x + 13
2 4

Interchanging the role of x and y, or, 9x + 12 + 4 = 2x + 13
4
x = 4y – 13
2 or, 36x + 64 = 2x + 13

or, 2x = 4y – 13 or, 36x – 2x = 13 – 64

or, 4y = 2x + 13 or, 34x = – 51

∴ y = 2x + 13 or, x = – 51 = –3
4 34 2
2x + 13
Hence, g–1(x) = 4  x = –23.

Exercise 1.1B

1. Find the inverse of the following functions.

a. f = {(1, 2), (3, 4), (5, 6)} b. g⸰f = {(9, 10), (11, 12), (13, 14)}

c. g–1 = {(a, b), (c, d), (e, f)} d. (f⸰g)–1 = {(1, 3), (5, 7), (9, 11)}

2. a. If f : x → 2x + 3 then find f–1(x) and f–1(3).

b. If f = 4x + 5 then find f–1(x) and f–1(1).
x+2
1
c. If f(x + 3) = 3x + 5, then find f–1(x) and f–1 2 .

d. If f x+2 = x + 3, then find f–1(x) and f–1(– 8).
3

3. a. If f–1 : x → 4x – 5, then find f(x) and f (11).

b. If f–1(x) = x + 3 ,then find f(x) and f – 1 .
2 2

c. If f–1(x + 2) = x, then find f(x) and f(a).

Relations and Functions / 11

d. If f–1 x = x + 1, then find f(x) and f 1 .
2 x

4. a. If f(x + 2) = 2x – 3, then find f(x) and f–1­(x).
b. If g(x + 3) = x2+x2, then find g(x) and g–1(x).

5. a. If f–1 = {(1, 5), (2, 4), (3, 6)} and f–1⸰g–1 = {(5, 6), (4, 5), (6, 4)} then find g–1.

b. If f–1 = {(2, 3), (4, 5), (6, 7)} and g–1⸰f–1 = {(2, 4), (4, 6), (6, 8)} then find g.

c. If f–1(x) is an identity function and g–1 = {(2, 5), (3, 6), (4, 7)} then find f–1⸰g–1 and g–1⸰f–1 and
also draw arrow diagram.

Exercise 1.1C

1. a. If f(x) = 3x + 4 and g(x) = x + 2, find (f⸰g)–1(x) and (g⸰f)–1(x).
3
x – 5,
b. If f(x) = 2x + 1 and g(x) = 2 then find the value of f–1⸰g–1(3).

c. If f(x) = 6x and g(x) = 3x + 4, find the value of f–1⸰g(x) and f⸰g–1(x).

2. a. If f(x) = x 3, find the value of x for which f(x) = f–1(x).
2x –
4x
b. If g(x) = x+3 and g–1(x) = g(x), find the value of x.

3. a. If f(x) = 2x + 3 and f–­1(1) = k, then what is the value of k.

b. If g(x) = x + 3 and g–1(x) = 25, then what is the value of x.
4

c. If f(x) = 2x – a and f(4) + f–1(3) = 7, find the value of a.

4. a. If g(x) = 2x + 4, f(x) = ax + 5 and f–1(6) = g–­1(6), then find the value of a.

b. If g(x) = bx + 9, f(x) = 3x + 8 and f–1(11) = g–1(10), then find the value of b.

5. a. If g(x) = 2 + 3x, then prove that gg–1(x) = x.

b. If f(x) = 8 – 7x, then prove that ff–1 is an identity function.

6. a. If g(x) = 2x – 3, then prove that g–1g = gg–1.

b. If f(x) = x + 2 and g(x) = 3x – 2, then prove that fg = gf.
3

7. a. It is given that the function f(x) = 4x + 7 and g(x) = 3x – 5. If fg–1(x) = 15, find the value of x.

b. It is given that f(x) = x2 – 2x and g(x) = 2x + 3. If fg–1(x) = 3, calculate the value of x.

8. a. If f(x) = 2x + 1 and g(x) = 3x + 1 are two functions. If ff(x) = g–1(x), find the value of x.
2
2x + 8.
b. It is given that f(x) = 4x – 17 and g(x) = 5 If ff–1(x) = g–1(x), find the value of x.

c. If functions f(x) = x + 3 and g(x) = 2x + 5 are given, then find the value of g –1(x) and fg –1(3).
2 x+2
x + 5
9. a. It is given that f = (x, 2x + 3) and g = x, 2 . If g–1(x) = ff(x), find the value of x.

b. If f(x) = m + 2x, g(x) = x – 3 and fg–1(x) = 2g(x), then find the value of m.

10. a. If f(x) = 4x – 2, g(x) = 1 and {f–1(6)} × {gf(x)} = 1, find the value of x.
x

b. If f(m) = 2m – 3, g(m) = 1 + m and f(m) + g–1(m) = 0, then find the value of m.



12 / Optional Mathematics Class 10

1.2 Polynomials Algebra

Multiplication of Polynomials

When we multiply a polynomial by a constant, each term of the polynomial is multiplied by the
constant. In case of two polynomials, each term of a polynomial is multiplied by every term of another
polynomial and the result is simplified.

Example 1: Multiply the polynomial p(x) = x3 – 3x + 4 by polynomial g(x) = x2 + 4x + 1.
Solution:

Here, f(x). g(x)

= (x3 – 3x + 4) (x2 + 4x + 1)

= x3(x2 + 4x + 1) – 3x(x2 + 4x + 1) + 4(x2 + 4x + 1)

= x5 + 4x4 + x3 – 3x3 – 12x2 – 3x + 4x2 + 16x + 4

= x5 + 4x4 – 2x3 – 8x2 + 13x + 4

Division of Polynomials

When we divide one polynomial by another, we perform the following steps:
a. Divide the first term of the dividend by the first term of the divisor and get the first term of the quotient.

b. Multiply each term of the divisor by the first term of the quotient and keep the product below the

dividend. Dividend


Divisor → x – 1 ) x3 + 7x2 – 3x + 2 ( x2 + 8x + 5 ← Quotient
– x3 +– x2
8x2 – 3x

–8x2 –+ 8x
5x + 2

–5x +– 5
7 ← Remainder

c. Subtract the product from the dividend.

d. Take this remainder as the new dividend and repeat the process until the degree of the remainder is
less than that of the divisor.

Note that in division process, we get

Dividend = Divisor × Quotient + Remainder

In the above example,

Divisor × Quotient + Remainder

= (x – 1) (x2 + 8x + 5) + 7

= x3 + 8x2 + 5x – x2 – 8x – 5 + 7

= x3 + 7x2 – 3x + 2 (Dividend)

Polynomials / 13

Example 2: Divide the polynomial x3 + 7x2 – 3 by x + 2.

Solution:

x + 2 ) x3 + 7x2 – 3 ( x2 + 5x – 10
– x3 –+ 2x2
5x2 – 3

–5x2 +– 10x
– 10x – 3

+– 10x –+ 20
17

Quotient : x2 + 5x – 10, Remainder : 17.
Example 3: Find the quotient Q(x) and remainder (R) in the following process of division

2x3 – 9x2 + 4x – 6 = (2x – 3) Q(x) + R.
Solution:

Here, 2x3 – 9x2 + 4x – 6 is to be divided by 2x – 3

2x – 3 ) 2x3 – 9x2 + 4x – 6 ( x2 – 3x – 5
–2x3 +– 3x2 2

– 6x2 + 4x

+– 6x2 +– 9x
– 5x – 6
15
+– 5x +– 2

– 6 – 15 = – 27
2 2

or, 2x3 – 9x2 + 4x – 6 = (2x – 3) (x2 – 3x – 52) – 27
2
52) 27
 Q(x) = (x2 – 3x – and R = – 2

Note: If the remainder of the division of polynomials is zero, the divisor is a factor of the dividend.

Synthetic Division

The method of dividing a polynomial by a polynomial x – a is lengthy. Such division of polynomials
can be performed by a short method called synthetic division. Study the following example of synthetic
division. (3x3 – 4x2 + 7x – 3) ÷ (x – 2)

The constant term in divisor with sign changed = 2.
Writing coefficients in order, we get

2 3 – 4 + 7 – 3
6 4 22
3 2 11 19

Quotient = 3.x2 + 2.x + 11 = 3x2 + 2x + 11 and Remainder = 19.

14 / Optional Mathematics Class 10

Process Algebra
a. List the coefficients of the terms of the dividend in standard form and write the coefficient 0

for any missing term.

b. Write the constant term of the divisor with sign changed.

c. Bring down the first coefficient of the dividend.

d. Multiply the coefficient by the constant term of the divisor with sign changed.

e. Write the product under the next term and add.

f. Multiply the result obtained by the same constant and write the result under the another
coefficient and add.

g. Repeat the process to get the last sum which is the remainder of the division process.

Example 4: Use the synthetic division to divide x4 – 7x2 + 3 by x + 2.

Solution:

Dividend = x4 – 7x2 + 3 = x4 + 0.x3 – 7x2 + 0.x + 3 and Divisor = x + 2 = x – (– 2)
Writing the coefficient in order:.

– 2 1 0 – 7 0 3
– 2 4 6 – 12

1 – 2 – 3 6 – 9
Quotient = 1.x3 + (– 2).x2 + (– 3).x + 6 = x3 – 2x2 – 3x + 6

Remainder = – 9.

Exercise 1.2

1. Find the product of the following polynomials :

a. p(x) = 2x – 3, q(x) = x + 4 b. p(x) = x2 + 3x + 1, q(x) = x2 + 3x + 2

2. Divide p(x) by d(x) in each of the following :

a. p(x) = x4 + 3x3 – 2x + 3, d(x) = x + 4 b. p(x) = x3 – 3x2 + 4, d(x) = x – 4

3. Find the value of Q(x) and R in the each of the following cases :

a. 6x2 – 7x + 12 = (x – 2) Q(x) + R b. 2x3 – x2 – 6x – 6 = (2x – 3) Q(x) + R

c. 2x3 – 9x2 + 5x – 5 = (2x – 3) Q(x) + R d. 2x3 + 3x2 – 4x – 5 = (x – 21) Q(x) + R
e. x3 – 19x2 – 30 =  (x + 2) Q(x) + R e. 3x4 – 9x2 – 4 = (x + 2) Q(x) + R

4. Using synthetic division, find the quotient and the remainder in the following cases :

a. (x3 – 6x2 + 11x – 6) ÷ (x – 2) b. (6x3 – 2x2 + 8) ÷ (x + 1)

c. (6x3 + 5x2 – 10x + 4) ÷ (x + 2) d. (x4 – 3x3 – 2x2 + x + 5) ÷ (x + 1)

5. a. A polynomial was divided by x – 1 and the quotient was x2 + 2x + 1 with the remainder 0. What
was the original polynomial ?

b. When a polynomial is divided by x + 2, the quotient is x2 + 2x – 1 and the remainder is 5. What
is the original polynomial ?

c. When a polynomial is divided by (x – 1) quotient is (x2 – 8x + 16) and leaves remainder – 4.
Find the original polynomial.

Polynomials / 15

1.3 Remainder and Factor Theorem

Introduction

In this section, we study two main theorems on polynomials:
a. Remainder Theorem and
b. Factor Theorem.
Remainder theorem helps us to find the remainder of the division of a polynomial and factor theo-
rem helps us to decide whether x – a is a factor of the given polynomial or not.

Remainder Theorem

A remainder of division of a polynomial can be obtained either by actual division or by synthetic
division. The remainder can also be obtained directly without actual division by using Remainder Theorem.

Theorem

If the polynomial f(x) is divided by x – a, the remainder is f(a).
Proof:

Let Q(x) be the quotient and R the remainder when f(x) is divided by x – a, then

x – a ) f(x) ( Q(x)
... ...

R

f(x) = (x – a) Q(x) + R

This result is true for all values of x.

So, when x = a, f(a) = (a – a).Q(x) + R

∴ f(a) = R

Note: If we divide f(x) by ax – b, the remainder is f b .
Example 1: Find the remainder when x3 – 4x2 + a

8x – 5 is divided by x – 2.

Solution:

Here, f(x) = x3 – 4x2 + 8x – 5.

Comparing x – 2 with x – a, a = 2
Remainder R = f(a) = 23 – 4.22 + 8.2 – 5

= 8 – 16 + 16 – 5 = 3.
Example 2: Find the remainder when 4x3 + 2x2– 4x + 3 is divided by 2x + 3.

Solution:

Here, f(x) = 4x3 + 2x2 – 4x + 3 and

Divisor 2x + 3 can be written as 2 x + 3 and
2

Comparing with x – a, a = – 3
2

16 / Optional Mathematics Class 10

3 3 3 2 3
2 2 2
∴ Remainder, R = f(a) = 4 – + 2 – – 4 – + 3

= 4 × – 27 + 2 × 9 + 4 × 3 + 3 Algebra
8 4 2

= – 27 + 9 + 6 + 3
2 2

= – 27 + 9 + 9 = – 9 + 9 = 0.
2
Example 3: When a polynomial x3 – 3x2 + mx + m is divided by x + 2, a remainder – 2m is obtained.

Find the value of m.

Solution:

Here, f(x) = x3 – 3x2 + mx + m
Comparing x + 2 with x – a, a = –  2

Remainder, f(– 2) = R
or, (– 2)3 – 3(– 2)2 + m(– 2) + m = – 2m

or, – 8 – 12 – 2m + m = – 2m
or, – 20 – m = – 2m

or, 2m – m = 20
 m = 20
The value of m = 20

Factor Theorem

If the polynomial f(x) is divided by x – a and R = f(a) = 0, then x – a is a factor of f(x).

Proof:

Let Q(x) be the quotient when f(x) is divided by x – a, then

f(x) = (x – a) Q(x) + R

or, f(x) = (x – a) Q(x) + 0

or, f(x) = (x – a) Q(x)

Which shows that (x – a) is a factor of f(x).

Note: We use factor theorem to identify whether the expression x – a is a factor of a polynomial f(x) or not.
Example 4: Show that x + 1 is a factor of f(x) = x3 + 4x + 5.
Solution:

Here, f(x) = x3 + 4x + 5.
Comparing x + 1 with x – a, we have a = – 1
Remainder, R = f(– 1) = (– 1)3 + 4(– 1) + 5
= – 1 – 4 + 5
= 0.
∴ x + 1 is a factor of x3 + 4x + 5.

Remainder and Factor Theorem / 17

Example 5: Find the values of the polynomial f(x) = 2x3 – 3x2 – 11x + 6 at x = 1, –  2, 3, –  3, 1 . Find also
2
the factors of f(x).

Solution:

Here, f(x) = 2x3 – 3x2 – 11x + 6,

When x = 1, f(1) = 2 × 13 – 3 × 12 – 11 × 1 + 6

= 2 – 3 – 11 + 6 = – 6

∴ (x – 1) is not a factor of f(x).

When x = – 2, f(– 2) = 2 × (– 2)3 – 3 × (–  2)2 – 11 × (– 2) + 6

= – 16 – 12 + 22 + 6 = – 28 + 28 = 0

∴ (x + 2) is a factor of f(x).

When x = 3, f(3) = 2 × 33 – 3 × 32 – 11 × 3 + 6

= 54 – 27 – 33 + 6

= 60 – 60 = 0.

∴ (x – 3) is a factor of f(x)

When x = – 3, f(– 3) = 2 × (– 3)3 – 3 × (– 3)2 – 11 × (– 3) + 6

= – 54 – 27 + 33 + 6

= – 81 + 39 = – 42

∴ (x + 3) is not a factor of f(x).

When x = 12, 1 1 3 1 2 1
2 2 2 2
f =2× – 3 × – 11 × +6

= 1 – 3 – 11 + 6
4 4 2

= 1 – 3 – 22 + 24 = 0
4

∴ (2x – 1) is a factor of f(x).

As f(x) = 2x3 – 3x2 – 11x + 6 is a polynomial of degree three, it will have at most three factors of first
degree, hence, f(x) = 2x3 – 3x2 – 11x + 6

= (x + 2) (x – 3) (2x – 1).
Example 6: If (x – k) is a factor of x3 – kx2 – 2x + 4, find the value of k.
Solution:

Here, f(x) = x3 – kx2 – 2x + 4
Comparing x – k with x – a, we have, a = k
Remainder, R = f(k)
or, 0 = k3 – k(k)2 – 2k + 4
or, 0 = k3 – k3 – 2k + 4
or, 0 = – 2k + 4
or, 2k = 4
 k = 2.

18 / Optional Mathematics Class 10

Example 7: Factorise: x3 – 6x2 + 11x – 6 Algebra
Solution:

Since the constant term of polynomial is 6, the factors of 6 may be the constant term of the factors.

When x = 1, f(1) = 13 – 6 × 12 + 11 × 1 – 6

= 1 – 6 + 11 – 6 = 0

∴ (x – 1) is a factor of x3 – 6x2 + 11x – 6.

Now, by using synthetic division.
1 1 – 6 11 – 6

1 – 5 6

1 – 5 6 0

∴ Quotient, Q(x) = x2 – 5x + 6
Hence, f(x) = (x – 1) Q(x)
x3 – 6x2 + 11x – 6
= (x – 1) (x2 – 5x + 6)
= (x – 1) (x2 – 3x – 2x + 6)
= (x – 1) {x(x – 3) – 2(x – 3)}
= (x – 1) (x – 3) (x – 2).
Example 8: Solve : x3 + 3x2 – 4x – 12 = 0.
Solution:
Given equation is x3 + 3x2 – 4x – 12 = 0.
We can factorise the expression f(x) = x3 + 3x2 – 4x – 12 by using factor theorem.
When x = 2, f(2) = x3 + 3x2 – 4x – 12
= 8 + 12 – 8 – 12 = 0.
Hence, (x – 2) is a factor of f(x).

By using synthetic division,
2 1 3 – 4 – 12
2 10 12

1 5 6 0

Quotient, Q(x) = x2 + 5x + 6

So, f(x) = (x – 2) Q(x)

= (x – 2) (x2 + 5x + 6)

= (x – 2) (x2 + 3x + 2x + 6)

= (x – 2) {x(x + 3) + 2(x + 3)}

= (x – 2) (x + 3) (x + 2)

Hence, the given equation is

x3 + 3x2 – 4x – 12 = 0

Remainder and Factor Theorem / 19

or, (x – 2) (x + 3) (x + 2) = 0

Either, x – 2 = 0 ⇒ x = 2

or, x + 3 = 0 ⇒ x = – 3

or, x + 2 = 0 ⇒ x = – 2

∴ x = 2, – 3, – 2.
Example 9: If (x – 2) is a factor of x3 + mx2 – 4x + 12, then prove that x – m is the factor of x2 + 4x + 3.

Solution:

Here, f(x) = x3 + mx2 – 4x + 12 and comparing x – 2 with x – a , a = 2

Remainder f(2) = 0 Again, let h(x) = x2 + 4x + 3
or, (2)3 + m(2)2 – 4 × 2 + 12 = 0 Comparing x + 3 with x – a, a = – 3
or, 8 + 4m – 8 + 12 = 0 Remainder is h(– 3)
or, 4m = – 12 = (– 3)2 + 4×(– 3) + 3
 m = – 3 = 9 – 12 + 3
Now, x – m = x – (– 3) = x + 3 = 12 – 12 = 0

Hence, x – m is a factor of x2 + 4x + 3.

Exercise 1.3A

1. Using remainder theorem, find the remainder when f(x) is divided by g(x):
a. f(x) = x3 + 3x2 – 2x + 6, g(x) = x – 3 b. f(x) = 2x3 + 3x2 + 2x + 3, g(x) = x + 2
c. f(x) = x3 – x + 20, g(x) = x + 1 d. f(x) = 4x3 – 2x2 – 4x + 3, g(x) = 2x – 3
2. a. Find the value of m if x3 + 3x2 – 2x + m leaves a remainder 4 when divided by x – 2.
b. Find the value of a if 2x3 – 3x2 + ax – 8 leaves a remainder 10 when divided by x – 3.
c. Find the value of k if 4x3 – 6x2 – 3x + k leaves a remainder – 5 when divided by x + 3.
d. Find the value of m, if x3 + 2x2 – mx + 4 leaves a remainder 2m when divided by x – 2.
e. Find the value of a, if x3 – ax2 – 2x + 2a + 6 leaves a remainder a + 2 when divided by x – a.
f. Find the value of p, if x3 – (p – 2)x2 – px + 27 leaves a remainder 10 when divided by x + 3.
3. a. If f(x) = x3 – 9x2 + 24x – k and f(2) = 0, find the value of k.
b. If p(x) = x3 – (a – 1)x2 + ax + 54 and p(– 3) = 0, find the value of a.
c. If f(x) = 3x3 – (a + 1)x2 + x + 5 and f(– 1) = 6, find the value of a.
4. a. If x – m is a factor of x3 – mx2 – 4x + m + 12, calculate the value of m.
b. If (x – 1) is a factor of 2x3 – 3x2 + 6x + k, find the value of k.
c. If (x – 3) is a factor of x3 – 9x2 + mx – 30, find the value of m.
d. If (x – a) is a factor of x3 – ax2 – 2x + a – 2, find the value of a.
e. If x – a is a factor of x3 – ax2 – 2x + a + 4, find the value of a.
5. a. If (3x + 2) is a factor of 3x3 + 8x2 + x – b, find the value of b.
b. If (3x + 1) is a factor of 3x2 + mx – 2, find the other factor.
c. If (2x – 3) is a factor of mx2 – 11x – 6, find the other factor.

20 / Optional Mathematics Class 10

6. Show that (x + 1) is a factor of each of the following polynomials:

a. x3 – 4x – 3 b. x3 – 7x2 – 4x + 4 Algebra

c. 4x3 + 6x2 – 3x – 5 d. x3 + 3x2 + 6x + 4

7. a. Prove that x – 1 is the factor of 24x3 – 26x2 + 9x – 1.
2
3
b. Prove that x – 2 is a factor of (x – 1) (2x2 + 15x + 15) – 21.

8. a. If x – 3 is a factor of 2xm – 3xm–1 + 4x – 39, find the value of m.

b. If x – 2 is a factor of xm – 3xm – 1 + 4x – 4, find the value of m.

9. a. What must be added to x3 – 3x2 + 4x – 15 to make it exactly divisible be (x – 3)?

b. What must be subtracted from 2x3 + x2 – 2x + 5 to make it exactly divisible by (x + 1)?

Exercise 1.3B

1. a. Find the value of the polynomial f(x) = 2x3 – 3x2 – 11x + 6 at x = 1, 2, – 2, 3, 1 and 1. Also find
23
the factors of f(x).

b. Find the values of the polynomial f(x) = x3 + 4x2 + x – 6 at x = 1, 2, – 2, – 3, 3. Also find the
factors of f(x).

2. a. Find the value of a and b, if (x – 1), (x – 2) are the factors of x2(x – a) + bx – 8.

b. Find the value of a and b, if (x + a) and (2x – 7) are the factors of 2x2 – bx – 21.

3. a. Find the value of a and b if x3 + ax2 + bx – 6 has a factor x + 3 and leaves a remainder – 8 when
divided by x – 1.

b. Find the value of a and b, if 2x3 + ax2 + bx – 2 has a factor (x + 2) and leaves a remainder 7
when divided by 2x – 3.

4. a. When the polynomial x3 + mx2 + nx + 1 is divided by x – 2 and x + 3 the remainders are 9 and 19
respectively, find the values of m and n.

b. When the polynomial ax3 + bx2 – 11x – 6 is divisible by x + 1 and x – 2, the remainder are 6

and – 12 respectively. Find the values of a and b.

c. The expression ax3 – 8x2 + bx + 6 is exactly divisible by x2 – 2x – 3. Find the values of a and b.

5. a. If (x – 1) is a factor of x2 + mx + 1, then show that x – m is a factor of x2 + 3x + 2.

b. One factor of x2 – mx + 6 is x – 3. Prove that x – m is a factor of x2 – 7x + 12. m = s – pq.
6. a. If x2 – px + q and x2 – rx + s both have a common factor x – m, then prove that r –

b. If a common factor of polynomial x2 + 2ax + b and x2 + cx + d is x + m, then prove that

m(2a – c) = (b – d).

7. Solve the following equations : b. 2x3 + 3x2 – 11x – 6 = 0
a. 2x3 + x2 – 2x – 1 = 0 d. 2x3 + 6 – 3x2 – 11x = 0
c. 2x3 – 4x2 – 7x + 14 = 0

e. 6x3 + 2 = 5x2 + 3x f. 6x3 = 4 – 13x2

8. Factorise the polynomial p(x), then solve the equation p(x) = 0 :

a. p(x) = x3 + 2x2 – 5x – 6 b. p(x) = x3 – 9x2 + 26x – 2

c. p(x) = 2x3 + x2 – 5x + 2 d. p(x) = 2x3 – 11x2 + 18x – 9

e. p(x) = x4 – 4x3 – x2 + 16x – 12 f. p(x) = x3 – 2ax2 – a2x + 2a3



Remainder and Factor Theorem / 21

1.4 Arithmetic Sequence and Series

Introduction

Consider the following sequences.

a. 1, 6, 11, 16, ... ... ...

b. 6, 2, – 2, – 6, ... ... ...

c. 3, 431, 523, 7, ... ... ...
In the first sequence, each term is increased by 5 more than the preceding term. In the second se-

quence, each term is decreased by 4 than the preceding term. Each term in the third sequece is increased
4
by 3 more than the preceding term. In each of the above sequences, the difference between a term and its

preceding term is equal or constant. Such a sequence is said to be an Arithmetic Sequence or Arithmetic

Progression.

Definition

A sequence is said to be an arithmetic sequence or arithmetic progression if the difference between
a term and its preceding term is equal or constant throughout the whole sequence. It is denoted by A.P.

The constant difference obtained by subtracting a term from its succeeding term is called the common

difference. In the above example 3, 8, 13, 18, ... ... ..., the common difference = 8 – 3 = 13 – 8 = 18 – 13 =
5. Similarly, the common difference of 40, 36, 32, 28, ... ... ... is – 4 and that of 3, 413, 532, 7, ... ... ... is 43.
The common difference of an arithmetic sequence is denoted by d.

A series corresponding to any arithmetic sequence is known as the arithmetic series associated with
the given arithmetic sequence. Hence, 3 + 8 + 13 + 18 + ... ... is an arithmetic series associated with the
arithmetic sequence 3, 8, 13, 18, ... ... .

General term of an A.P.

If a be the first term and d be the common difference of an arithmetic progression, then the terms of
the progression are a, a + d, a + 2d, a + 3d, ... ... ...

If t1, t2, t3, t4, ... ... ..., tn be the first, second, third, fourth, ... ... ... , nth term of an A.P., then
t1 = a = a + (1 – 1)d
t2 = a + d = a + (2 – 1)d
t3 = a + 2d = a + (3 – 1)d
t4 = a + 3d = a + (4 – 1)d
... ... ... ... ... ...

tn = a + (n – 1)d

Thus if the first term and the common difference of an A.P. are known, we can find any term.
Example 1: What is the 7th term of an A.P., when its first term and the common difference are 7 and 3

respectively.
Solution:

Here, first term (a) = 7

22 / Optional Mathematics Class 10

Common difference (d) = 3 Algebra

7th term (t7­) = ?
We have, tn = a + (n – 1)d
or, t7 = 7 + (7 – 1)3
or, t7 = 7 + 6 × 3
∴ t7 = 25
Example 2: If the 15th term of an A.P. with the first term 8 is 92, find the common difference.

Solution:

Here, 15th term (t15) = 92

First term (a) = 8

Common difference (d) = ?

By using formula, we have

tn = a + (n – 1) d
or, t15 = 8 + (15 – 1)d

or, 92 = 8 + 14d

or, d = 92 – 8
14
∴ d = 6

The required common difference = 6.
Example 3: If three numbers 2k – 3, 3k and 5k – 2 are in A.P.,

a. find the value of k.

b. find the three numbers in A.P.
Solution:

a. As the three numbers 2k – 3, 3k and 5k – 2 are in A.P.

3k – (2k – 3) = 5k – 2 – 3k

or, 3k – 2k + 3 = 2k – 2

or, k + 3 = 2k – 2
∴ k = 5
b. The three numbers are 2k – 3, 3k and 5k – 2

or, 2 × 5 – 3, 3 × 5 and 5 × 5 – 2

∴ 7, 15 and 23.

Example 4: Is – 4 a term of the arithmetic sequence, 18, 229, 11, ... ... ...
Solution:

Here, first term (a) = 18

Common difference (d) = 29 – 18 = – 7
2 2
Last term (tn) = – 4

Number of terms (n) = ?

Arithmetic Sequence and Series / 23

By using formula, we have

tn = a + ( n – 1)d

or, – 4 = 18 + (n – 1) × – 7
2

or, – 4 – 18 = – 7 n + 7
2 2

or, – 22 – 7 = – 7 n
2 2

or, 27 n = 51
2

∴ n = 571.

As the number of terms (n) is a fraction, – 4 is not a term of the given series.

Example 5: Show that, tn = 4 – 7n is a general term of an AP. Find its common difference.
Solution:

Here, tn = 4 – 7n when n = 3
when n = 1

tn = 4 – 7n tn = 4 – 7n
or, t1 = 4 – 7×1 or, t3 = 4 – 7×3
i.e t1 = 4 – 7 = – 3. i.e t3 = 4 – 21 = – 17.
when n = 2 when n = 4

tn = 4 – 7n tn = 4 – 7n
or, t2 = 4 – 7×2 or, t4 = 4 – 7×4
i.e. t2 = – 10 i.e. t4 = – 24
Hence, it gives the terms

t1, t2, t3, t4, ... ... ... ... tn
i.e. – 3, – 10, – 17, – 24, .. .. .. , 4 – 7n

Now common difference = t2 – t1 = t3 – t2

– 10 – (– 3) = – 17 – (– 10)

or, – 10 + 3 = – 17 + 10

or, – 7 = – 7

Hence, tn = 4 – 7n is a general term of A.P.
Example 6: If the 3rd term and 9th term of an A.P. are 20 and 5 respectively, find the 19th term of the

series.

Solution:

Here, 3rd term (t3) = 20, 9th term (t9) = 5, first term (a) = ?
By using formula, we have

tn = a + (n – 1)d
or, t3 = a + (3 – 1)d
or, 20 = a + 2d ... ... ... (i)

24 / Optional Mathematics Class 10

Again, t9 = a + (9 – 1)d Algebra
or, 5 = a + 8d ... ... ... (ii)

Solving equation (i) and (ii),

20 = a + 2d
– 5 =–a +–8d

15 = 0 – 6d

∴ d = – 5
2

Substituting the value of d in equation (i),

20 = a + 2 × – 5
2

or, 20 = a – 5

∴ a = 25
Hence, the first term is 25 and the common difference is – 52.
For the 19th term, number of terms (n) = 19

So, tn = a + (n – 1)d
or, t19
= 25 + (19 – 1) × – 5
5 2
= 25 – 18 × 2

= 25 – 45 = – 20

Example 7: An arithmetic series is given as 2 + 8 + 14 + 20 + ... ... ... + 80

a. Find the number of terms. b. Find the 8th term of the series.
Solution:

Here, the first term (a) = 2

Common difference (d) = 8 – 2 = 6

Last term (tn) = 80
Number of terms (n) = ?

By using formula, b. 8th term (t8) = ?
tn = a + (n – 1)d Again, by using formula,

or, 80 = 2 + (n – 1)6 tn = a + (n – 1)d
or, 78 = 6n – 6 or, t8 = 2 + (8 – 1)6
or, 6n = 84 ∴ t8 = 44
∴ n = 14 Hence, the 8th term is 44.

Hence, the number of terms = 14 t7
t11
Example 8: In an arithmetic series = 171. Find the value of t18 .
Solution:

Here given t7 = 11
t11 7

or, aa++160dd = 11
7

Arithmetic Sequence and Series / 25

or, 7a + 42d = 11a + 110d

or, 0 = 11a – 7a + 110d – 42d

or, 0 = 4a + 68d

or, 0 = a + 17d.

Now, t18 = a + (18 – 1)d

or, t18 = a + 17d

i.e. t18 = 0.
Example 9: If the nth terms of two AP is – 19 – 12 – 5 + 2 ... ... and 1 + 6 + 11 + ... ... are equal then find

the value of n.
Solution:

Here, in 1st AP, – 19 – 12 – 5 + 2 ... ... ... In 2nd AP, 1 + 6 + 11 + ... ... ...

First term (a) = – 19 First term (a) = 1

Commond difference (d) = – 12 + 19 = 7 Commond difference (d) = 6 – 1 = 5

Now nth term = tn = a + (n – 1)d tn = a + (n – 1)d
or, tn = 1 + (n – 1)5
or, tn = – 19 + (n – 1)7 or, tn = 1 + 5n – 5
i.e. tn = 5n – 4
or, tn = – 19 + 7n – 7 ... ... ... (2)

i.e. tn = 7n – 26 ... ... ... (1)

Equating equation (1) and equation (2),

7n – 26 = 5n – 4

or, 7n – 5n = – 4 + 26

or, 2n = 22

i.e. n = 11
Example 10: A taxi meter reads Rs.7:00 at the time of starting and Rs.9:00 for each additional kilometer.

If the distance covered is 12km, find the charge read by the taxi meter.
Solution:

The sequence of charges read by the taxi meter is Rs.7, Rs.(7 + 9) (in 1km), Rs.(7 + 9 + 9) (in 2km), ...

i.e. Rs.7, Rs.16, Rs.25, ... ...

First term (a) = 7

Common difference (d) = 9

Number of terms (n) = 13

tn = a + (n – 1)d
or, tn = 7 + 12 × 9
= 7 + 108

= 115

Total charge read by the taxi meter = Rs.115.

26 / Optional Mathematics Class 10

Exercise 1.4A Algebra

1. a. Define sequence and series with an example.

b. Define arithmetic progression with an example

c. What do you mean by common difference?

2. Examine whether the following sequences form arithmetic sequence. Also find their common difference:

a. 8, 13, 18, 23, ... ... ... b. 12, 321, 19, 425, ... ... ...

c. 7, 631, 531 , 423, ... ... ... d. 4a, 5a + b, a + b, 3b – a, ... ...
2 2
3. From the following arithmetic sequences, find the first term, common difference, next two terms and

the nth term: b. 58, 2, 152, 154, ... ... ...
a. 30, 27, 24, 21, ... ... ...

c. 22, 1912, 17, 1421, ... ... ... d. – 3a – b, b – a, 3b + a, 3b + 3a, ... ... ...
2 2 2 2

4. a. If the first term and third term of AS are (a + b) and (a – b), then find the second term.

b. If the first term and second term of AS are (a + b)2 and a2 + b2, then prove that the third term is (a – b)2.

5. From the following first terms and the common differences of the arithmetic sequences, write the

first three terms and the indicated terms: 9
2
a. First term (a) = 18, common difference (d) = – , t9 = ?

b. First term (a) = 25, common difference (d) = – 5 , t7­­ = ?
2
1031, 2
c. First term (a) = common difference (d) = 3 , t10 = ?

6. a. If there are 35 terms in an A.P. whose first term is 8 and the last term 59, find its common dif-

ference.

b. If the first term of an A.P. is 21, the common difference is – 5 and the last term – 14, find the

number of terms.
c. If the 17th term of an A.P. is 6 and its common difference 45, find its first term.
7. a. If x + 1, x + 5 and 3x + 1 are the first three terms of an A.P., find the value of x and its first three terms.

b. If 8m + 4, 6m – 2 and 2m + 7 are the first three terms of an A.P., find the value of m and its first
three terms.

c. If 2k + 8, 6k and 8k – 1 are in A.P., find the value of k and the three numbers. — 1.

8. a. Is 44 a term of the arithmetic sequence 17, 20, 23, ... ... ... ?
aseriqthumenecteic3s4e,q3u2e21n,c3e12,529+122,2..12.
b. Which term of the ... ... will be 16? be – 10 ?
c. Which term of the + 20 + ... ... will

9. a. If the mt­h term of 25, 50, 75 is 1000, find the value of m.

b. If the kth term of 84 + 78 + 72 + ... ... ... is 0, find the value of k.

10. a. Find the 15th term of the sequence 3, 6, 9, 12, ... ... ... .

b. What is the 13th term of the series 3 + 11 + 19 + ... ... ... ?

11. Find the number of terms and the value of 10th term in each of the following series:
b. 7 + 512 + 4 + 221 + ... ... ... – 23
a. 4 + 11 + 18 + 25 + ... ... ... + 74

Arithmetic Sequence and Series / 27

12. a. Show that tn = 4n – 3 is an general term of AP.

b. Show that 49 – 5n is an general term of AP.
2
13. a. In an arithmetic sequence, 5t5 = 9t9, then prove that t14 = 0.

b. In an arithmetic sequence, 3t3 = 7t7, then prove tha t10 = 0.

Exercise 1.4B

1. a. If the first term of an A.P. is 7 and its 12th term 40, find its common difference and 31st term.

b. If the common difference of an A.P. is 3 and its 20th term 3521, find its first term and 15th term.
2
6
c. There are 22 terms of an A.P. whose common difference is 7 and the last term 36. Find its 15th term.

2. If the 10th and 17th terms of an A.P. are 7 and – 7 respectively,

a. find the first term and common difference.

b. write the first four terms.

c. find the 21st term of the sequence.

3. If the 7th and 25th terms of an A.P. are 2212 and 36 respectively,

a. find the first term and the common difference.

b. write the first four terms.

c. find the 17th term of the sequence.

4. a. If the 5th and 13th terms of an A.P. are 17 and – 15 respectively, find its 19th term.

b. If the 9th term of an A.P. is thrice its 3rd term and the 13th term is 52, find its 20th term.

c. If the sum of the 5th and 11th term of an A.P. is 48 and the 7th term 21. Find the 14th term.

5. a. If the nth term of the A.P. – 5, – 2, 1, 4, ... ... ... is equal to the nth term of the A.P. 30, 28, 26,
24, ... ... ..., find the number of terms.

b. If the (n – 2)th term of the A.P. – 13, – 10, – 7, – 4, .... is equal to the nth 8erm of the A.P. 1, 3,
5, 7, ... ... ..., find the value of n.

6. a. If p, q and r are in arithmetic sequence, prove that : (p – r)2 = 4(q2 – pr).
b. The three numbers a, b, c are in AP. If y2 = xz, prove that : xb – c . yc – a . za – b = 1

c. If a, b, c are pth, qth and rth terms of an AP, prove that : p(b – c) + q(c – a) + r(a – b) = 0

7. a. A taxi meter reads Rs.9.00 at the time of starting and Rs.12 for each additional kilometer.
Find the charge read by the taxi meter when the distance covered is 13 kilometers.

b. The cost of 20 units of electricity per month is Rs.90 and Rs.8 for each additional unit. If the
total electricity consumed in a particular month is 55 units, find the total cost of electricity of
that particular month.

c. A man takes a job of Rs.40000 a year. He receives an annual increase of Rs.2000. What is his
salary during the 11th year ?



28 / Optional Mathematics Class 10

1.5 Arithmetic Mean

Introduction Algebra

If three numbers a, b and c are in A.P. then the middle term b is said to be the arithmetic mean (A.M.)
between the other two. For example, 10, 14 and 18 are in A.P. so 14 is the arithmetic mean between 10 and
18. If any number of terms is in A.P. then the terms between the first and the last terms are known as the
arithmetic means between the first and the last. For example, 11, 14, 17, 20, 23, 26, 29, 32 are in A.P., so 14,
17, 20, 23, 26, 29 are the arithmetic means between 11 and 32.

Arithmetic Mean between Two Numbers

Let M be the arithmetic mean between two numbers a and b, then a, M, b are in A.P.

Now by the definition of A.P.,

M–a=b–M

or, 2M = b + a

∴ M = a + b
2

Example 1: Find the arithmetic mean between 2 and 19.

Solution:

Here, the first term (a) = 2

The last term (b) = 19

Arithmetic mean (A.M.) = ?

Then, A.M. = a + b = 2 + 19
2 2

∴ A.M. = 21 = 1012
2

Example 2: If a, 30, b, 50, c are in an A.P., find the values of a, b and c.

Solution:

t1 = a, t2 = 30, t3 = b, t4 = 50, t5 = c Similarly, t4 = t3 + t5 [as t4 is the mean of t3 and t5]
2
Now, t3 = t2 + t4 [as t3 is the mean of t2 and t4]
2 b + c
or, 50 = 2
30 + 50
or, b = 2 40 + c
or, 50 = 2
∴ b = 40
or, 100 = 40 + c
t1 + t3
Again, t2 = 2 [as t2 is the mean of t1 and t3] or, c = 100 – 40
∴ c = 60
or, 30 = a + b = a + 40
2 2

or, 60 = a + 40

or, a = 60 – 40
∴ a = 20

Hence, values of a, b and c are 20, 40 and 60 respectively.

Arithmetic Mean / 29

Example 3: If 10, m + 5, ... ... ... 3m, 35 are the terms of A.P., find the value of m.
Solution:

First term (a) = 10 = t1
Second term (t2) = m + 5
Second last term (tn – 1) = 3m
Last term (tn) = 35
Now, we know that,

t2 – t1 = tn – tn–1
or, m + 5 – 10 = 35 – 3m
or, m – 5 = 35 – 3m
or, m + 3m = 35 + 5
or, 4m = 40
i.e. m = 10
The value of m = 10.

Arithmetic Means between Two Numbers

Let m1, m2, ... ... ... mn1 be the arithmetic means between a and b, then a, m1, m2, m3, ........., mn1, b
are in A.P.

If d be the common difference, then Now, the required arithmetic means are,

tn = a + (n – 1)d m1 = a + d = a + b – a
n1 + 1
As there are n1 + 2 terms in all i.e. n = n1 + 2
m2 = a + 2d = a + 2×nb1 – a
+ 1
b = a + (n1 + 2 – 1)d
3×nb1 – a
or, b – a = (n1 + 1)d m3 = a + 3d = a + + 1

∴ d = b – a ... ... ... ... ... ...
n1 + 1
mn1 = a + n1d = a + n1×nb1 – a
+ 1

Note: The second term of the sequence is the first mean and the third term is the second mean and so on.

Example 4: If there are 6 A.M. between 4 and 11, then find the common difference.
Solution:

Here, a = 4 and b = 11

Number of means (n1) = 6

We have, d = b – a
n1 + 1

or, d = 11 – 4 = 7 = 1
6+1 7

30 / Optional Mathematics Class 10

Example 5: Insert 5 arithmetic means between 2 and 18. Algebra
Solution:

Here, the first term (a) = 2 Hence, the required arithmetic means are,

The last term (b) = 18 m1 = a + d = 2 + 8 = 423
3
Number of means (n1) = 5
Hence, the common difference, m2 = a + 2d = 2 + 2×38 = 713

d = b – a m3 = a + 3d = 2 + 3×38 = 10
n1 + 1 m4 = a + 4d = 2 + 4×38 = 1232

or, d = 18 – 2
5+1

or, d = 16 = 8 m5 = a + 5d = 2 + 5×38 = 1513
Example 6: 6 3

There are n1 arithmetic means between 12 and 33. If the fourth arithmetic mean is 24, find

the value of n1.

Solution:

Here, the first term (a) = 12 Now, the fourth mean, m4 = a + 4d

The last term (b) = 33 or, 24 = 12 + 4 × 21 1
n1 +
The fourth mean (m4) = 24
or, 12 = 84
Number of means (n) = ? n1 + 1

Common difference, d = b – a or, n1 + 1 = 84
n + 1 12

or, d = 33 – 12 = 21 1 ∴ n1 = 6
n1 +1 n1 +

Example 7: There are n A.M.'s between 3 and 39. Find n so that the third mean : the last mean = 3:7.

Solution:

Here, the first term (a) = 3

The last term (b) = 39

Number of means = n

Third mean: last mean (m3:mn) = 3:7

Now, common difference (d) = b – a
n + 1

or, d = 39 – 3
n+1

or, d = 36
n+ 1

And, m3 = a + 3d
mn a + n.d

or, 37 = 3 + 3 × 36
3 + n × n+ 1
36
n+ 1

Arithmetic Mean / 31

3n + 3 + 3 × 36

or, 73 = n+ 1
3n + 3 + 36n

n+ 1

or, 73 = 3n + 111
39n +3

or, 117n + 9 = 21n + 777

or, 96n = 768

∴ n = 8.
Example 8: If the product of two numbers is 375 and their arithmetic mean is 20, find the

numbers.
Solution:

Let the numbers be a and b the product of two numbers is 375

i.e. a × b = 375 Now putting the value of a, we get
3b75 + b = 40
∴ a = 375 or, 375 + b2 = 40b
b or, b2 – 40b + 375 = 0
or, b2 – 25b – 15b + 375 = 0
their A.M. is 20, i.e., a + b = 20
2 or, b(b – 25) – 15(b – 25) = 0

or, a + b = 20 × 2 or, (b – 25) (b – 15) = 0

or, a + b = 40

Either, b – 25 = 0 i.e. b = 25.

or, b – 15 = 0 i.e. b = 15.

When b = 25, a = 375 = 375 = 15
b 25

and when b = 15, a = 375 = 375 = 25
b 15

Required numbers are 15 and 25 or 25 and 15.

Example 9: One number is reciprocal to another and the arithmetic mean between them is 54. Find those
numbers.

Solution:

Let one of the number be x then other number is x1.
Now, 5
A.M. = 4

x + 1 = 5
or, 2 x 4

or, x22+x 1 = 5
4

or, 4x2 + 4 = 10x

or, 2x2 + 2 = 5x

32 / Optional Mathematics Class 10

or, 2x2 – 5x + 2 = 0 Algebra
or, 2x2 – 4x – x + 2 = 0
or, 2x(x – 2) – 1(x – 2) = 0
or, (x – 2) (2x – 1) = 0
Either x – 2 = 0
∴ x = 2

or, 2x – 1 = 0

∴ x = 1
2
1 1
Hence, required numbers be 2 and 2 or 2 and 2.

Exercise 1.5A

1. a. Define arithmetic mean.

b. Write the formula of A.M. between any two numbers a and b.

2. Find the arithmetic mean between the following two numbers :

a. 20 and 28 b. 20  1 and 10  1
2 3

c. 2k + 3 and k + 1 d. (a + b)2 and (a – b)2

3. a. If the arithmetic mean between k – 1 and 2k is 7, find the value of k.
b. If 3, 6, m are in AP, then find the value of m.
c. If 3, m, n, – 9 are in AP, find the values of m and n.
d. If – 17, a, b, c, 3 are in A.P., find a, b and c.
e. If m, 30, n, 20, p are in an AP, find the values of m, n and p.
4. a. If 3rd term = 9 and 5th term = 15, find the 4th term.
b. If 6th term = 54 and 8th term = 38, then find the7th term.
c. If 4th term = 16 and 5th term = 25, then find 3rd term.
d. If 6th term = 19 and 7th term = 24, then find 8th term.
e. If 7th term = 16 and 9th term = 20, then find 10th and 6th term.
5. a. If there are 4 A.M. between 3 and 28, then find the common difference.
b. If there are 6 A.M. between 8 and 85, find the common difference.
6. a. One number is 5 times of the other and their A.M. is 12. Find the numbers.
b. Two numbers are in the ratio of 5:3. if their A.M. is 80, find the numbers.
c. If one number in greater than other by 10 and their A.M. is 15, then find them.

Exercise 1.5B

1. a. Insert 3 arithmetic means between 12 and – 4.
b. Insert 4 arithmetic means between 112 and 9.

c. Insert 5 arithmetic means between 17 and 4.

Arithmetic Mean / 33

2. a. If the difference of two numbers is 2 and their A.M. is 13, then find them.
b. If the product of two numbers is 12 and their A.M. is 72, find the numbers.
3. a. 7 A.M. are inserted between 2 and b. If the fourth mean is zero, find the value of b.

b. Insert n A.M. between 7 and 71 in such away that the 5th A.M. is 27. Find the value of n.

4. a. n arithmetic means are inserted between 7 and 28. If the second mean is 14, find the value of n.

b. n arithmetic means are inserted between 20 and 50. If the third mean is 35, find the value of n.

5. a. 4 arithmetic means are inserted between a and b. If the second and the fourth means are respec-
tively 15 and 31, find the values of a and b and the remaining means.

b. 8 arithmetic means are inserted between two numbers a and b. If the second and the sixth
means are 6 and 12 respectively, find the values of a and b.

6. a. There are n arithmetic means between 4 and 24. If the ratio of the third mean to the last mean
is 4:5 then find the number of terms in the series.

b. n arithmetic means are inserted between 20 and 5. If the ratio of the third mean and the last
mean is 31:13, find the value of n.

c. There are n arithmetic means between 3 and 17 and the ratio of last mean is to first mean is 3:1.
Find the value of n.

d. n arithmetic means are inserted between 2 and 29. If the ratio of the second and the last mean
is 4:13, find the value of n.


1.6 Sum of Arithmetic Series

Sum of Arithmetic Series

The series associated with an arithmetic sequence is known as an arithmetic series. For example,
3 + 7 + 11 + 15 + 19 + 23 + 27 + 31 is the arithmetic series associated with the arithmetic sequence, 3, 7,

11, 15, 19, 23, 27, 31.

Let a be the first term, d be the common difference, n be the number of terms,  be the last term and
Sn be the sum of the n terms of an A.S., then
Sn = a + (a + d) + (a + 2d) + ... ... ... + ( – 2d) + ( – d) +  ... ... ... (i)

Writing in the reverse order, we have,

Sn =  + ( – d) + ( – 2d) + ... ... ... + (a + 2d) + (a + d) + a ... ... ... (ii)
Adding relations (i) and (ii), we get

2Sn = (a + ) + (a + ) + (a + ) + ... ... ... + (a + ) + (a + ) + (a + ).

or, 2Sn = n(a + )

∴ Sn = n (a + l)
2

Again, the last term,  = a + (n – 1)d

Sn = 2n{a + a + (n – 1)d}

∴ Sn = n {2a + (n – 1)d}
2

34 / Optional Mathematics Class 10

Example 1: Find the sum of the series 2 + 8 + 14 + 20 + ... ... ... 20 terms. Algebra
Solution:

First term (a) = 2 Common difference (d) = 8 – 2 = 6
Number of terms (n) = 20 Sum (Sn) = ?
By using formula, we have

Sn = 2n{2a + (n – 1)d}
or, Sn = 220{2 × 2 + (20 – 1) × 6}

or, Sn = 10(4 + 19 × 6)

or, Sn = 10(4 + 114)

∴ Sn = 1180

Hence, the sum of the 20 terms is 1180.

Example 2: Find the sum of the series : 22 + 27 1 + 32 2 + ... ... ... + 54
3 3
Solution:

First term (a) = 22 2731 531 16
3
Common difference (d) = – 22 = =

Last term () = 54, Number of terms (n) = ?, Sum (Sn) = ?

Now, we have  = a + (n – 1)d

or, 54 = 22 + (n – 1) × 16
3
or, 54 – 22 = 136(n – 1)
3
or, 32 × 16 = n – 1

or, n – 1 = 6

∴ n = 7

Again, Sn = n (a + ) = 7 (22 + 54) = 7 × 76 = 266
2 2 2

Hence, the required sum is 266.

Example 3: 7

In an arithmetic series ∑ (2n – m) = 58. What is the value of m?
n=4
Solution:

Sn = 7
Given, ∑ (3n – m) = 58

n=4

or, (3 × 4 – m) + (3 × 5 – m) + (3 × 6 – m) + (3 × 7 – m) = 58

or, (12 – m) + (15 – m) + (18 – m) + (21 – m) = 58
or, 12 – m + 15 – m + 18 – m + 21 – m = 58
or, 66 – 4m = 58
or, 4m = 66 – 58
or, 4m = 8
∴ m = 2

Sum of Arithmetic Series / 35

Example 4: If the sum of the first n terms of an arithmetic series is Sn = n2 + n , then find the nth term tn.
2
Solution:

Given Sn = n2 + n
2

same as Sn–1 = (n – 1)2 + (n – 1) [as n → n – 1]
2
n2
or, Sn–1 = – 2n +1 + n – 1
2
n2
∴ Sn–1 = – n
2

We know that, tn = Sn – Sn – 1

or, tn = n2 + n – n2 – n
2 2
n2 n2
or, tn = + n – + n
2

or, tn = 2n = n.
2

Example 5: How many terms of an A.S. 17 + 15 + 13 + ... ... are needed to give the sum 80?

Solution:

First term (a) = 17, Common difference (d) = – 2

Sum (Sn) = 80 Number of terms (n) = ?
By using formula,

Sn = n2 [2a + (n – 1)d]
or, 80 = 2n [2 × 17 + (n – 1) × – 2]
or, 160 = n [34 – 2n + 2]

or, 2n2 – 36n + 160 = 0
or, n2 – 18n + 80 = 0
or, n2 – 10n – 8n + 80 = 0

or, (n – 10) (n – 8) = 0

∴ n = 8 or 10.
Example 6: If the sum of the first 16 terms of an A.S. is 280 and the 6th term of the series is 14, find the

sum of the first 21 terms.
Solution:

Let a be the first term and d be the common difference of the given series.

Sum of 16 terms (S16) = 280

6th term (t6) = 14
By using formula, we have Sn = n2 [2a + (n – 1)d]

36 / Optional Mathematics Class 10

or, S16 = 126{2a + (16 – 1)d} Algebra
or, 280 = 8(2a + 15d)

∴ 2a + 15d = 35 ... ... ... (i)

Again, tn = a + (n – 1)d ... ... ... (ii)
or, t6 = a + (6 – 1)d
∴ a + 5d = 14

Solving equations (i) and (ii), we get

2a + 15d = 35

–2a +– 10d =–28

5d = 7

∴ d = 7
5
Substituting the value in equation (ii),

a + 5 × 7 = 14, i.e. a = 7
5
And Sn = 2n [2a + (n – 1)d]
21 7
S21 = 2 2 × 7 + (21 – 1) × 5

or, S21 = 21 14 + 20 × 7
2 5

or, S21 = 21 × 42
2

∴ S21 = 441

Hence, the sum of 21 terms of the series is 441.

Example 7: Find the sum of all natural numbers less than 100 which are exactly divisible by 7.
Solution:

The natural numbers less than 100 and exactly divisible by 7 are 7, 14, 21, ... ... 98.
Here, first term (a) = 7, common difference (d) = 7, last term () = 98, sum (Sn) = ?
Number of terms (n) = ?

By using formula,
 = a + (n – 1)d

or, 98 = 7 + (n – 1)7

or, 98 = 7 + 7n – 7

∴ n = 14.

Hence, Sn = n (a + )
2

or, S14 = 14 (7 + 98)
2

or, S14 = 7 × 105

∴ S14 = 735
Hence, the required sum is 735.

Sum of Arithmetic Series / 37

Sum of the First n Natural Numbers

The first natural numbers are 1, 2, 3, 4, ... ... n.
Let Sn = 1 + 2 + 3 ... ... ... + n
First term (a) = 1
Common difference (d) = 2 – 1 = 1
Number of terms (n) = n
Sum of n terms (Sn) = ?
By using formula,

Sn = n2{2a + (n – 1)d}

or, Sn = n2{2 × 1 + (n – 1)1}

or, Sn = n (2 + n – 1)
2

∴ Sn = n(n + 1)
2

Sum of the First n Odd Natural Numbers

The first n odd natural number are 1, 3, 5, ... ..., (2n – 1)

Let Sn = 1 + 3 + 5 + ... ... + (2n – 1)
Here, first term (a) = 1

Common difference (d) = 3 – 1 = 2

Number of terms (n) = n

Sum of n terms (Sn) = ?

By using formula,

Sn = 2n{2a + (n – 1)d}
or, Sn = n2{2 × 1 + (n – 1) × 2}
or, Sn = 2n(2 + 2n – 2)
∴ Sn = n2

The Sum of the First n Even Natural Numbers

The first n even natural numbers are 2, 4, 6, ... ..., 2n.

Let Sn = 2 + 4 + 6 + ... ... + 2n

Here, first term (a) = 2,

Common difference (d) = 4 – 2 = 2

Number of terms (n) = n
Sum of n terms (Sn) = ?

38 / Optional Mathematics Class 10

By using formula,

Sn = n [2a + (n – 1)d] Algebra
2
n
or, Sn = 2 [2 × 2 + (n – 1)2]

or, Sn = n [4 + 2n – 2]
2
n
or, Sn = 2 (2n + 2)

∴ Sn = n(n + 1)

Sum of the Squares of the First n Natural Numbers

The squares of the first natural numbers are 12, 22, 32, ... ..., n2
Let Sn = 12 + 22 + 32 + ... ... + n2

Consider the identity,

r3 – (r – 1)3 = r3 – (r3 – 3r2 + 3r – 1)
∴ r3 – (r – 1)3 = 3r2 – 3r + 1
Putting r = 1, 2, 3, 4, ... ..., (n – 1), n

13 – 03 = 3.12 – 3.1 + 1

23 – 13 = 3.22 – 3.2 + 1

33 – 23 = 3.32 – 3.3 + 1

43 – 33 = 3.42 – 3.4 + 1
... ... ... ... ... ... ... ... ...

(n – 1)3 – (n – 2)3 = 3(n – 1)2 – 3.(n – 1) + 1
n3 – (n – 1)3 = 3.n2 – 3.n + 1

Adding all above identities, we have

n3 = 3(12 + 22 + 32 + ... ... + n2) – 3(1 + 2 + 3 + ... ... + n) + n

or, n3 = 3Sn – 3n(n + 1) + n
2

or, 3Sn = n3 – n + 3n(n + 1)
2

or, 3Sn = n(n + 1)(n – 1) + 3n(n + 1)
2
3
or, 3Sn = n(n + 1) (n – 1) + 2

or, 3Sn = n(n + 1) 2n – 2 + 3
2

∴ Sn = n(n + 1)(2n + 1)
6

Example 8: Find the sum of the following series:

a. 2 + 4 + 6 + ... ... ... 12 terms b. 1 + 3 + 5 + ... ... 15 terms
Solution:

a. The given series is 2 + 4 + 6 + ... ... 12 terms

Sum of Arithmetic Series / 39

This is the sum of the first 12 even natural numbers.

Here, n = 12 and sum (Sn) = ?
∴ Sn = n(n + 1) = 12(12 + 1) = 156.
Alternately,

First term (a) = 2, Common difference (d) = 2, Number of terms (n) = 12, Sum (Sn) = ?

We have, Sn = n [2a + (n – 1)d]
2

or, Sn = 12 [2 × 2 + (12 – 1) × 2]
2

or, Sn = 6[4 + 22]

∴ Sn = 156

b. The given series 1 + 3 + 5 + ... ... 15 terms

First term (a) = 1, Common difference (d) = 2, Number of terms (n) = 15, Sum (Sn) = ?

We have, Sn = n [2a + (n – 1)d]
2

or, Sn = 15 [2 × 1 + (15 – 1) × 2]
2

or, Sn = 15 [2 + 28] = 15 [30]
2 2

∴ Sn = 15 × 15 = 225

Example 9: The sum of three numbers in A.P. is 27 and the sum of their squares is 293. Find the
numbers.

Solution:
Let the three numbers in A.P. be a – d, a and a + d.
Then (a – d) + a + (a + d) = 27 i.e. a = 9
and (a – d)2 + a2 + (a + d)2 = 293
or, a2 – 2ad + d2 + a2 + a2 + 2ad + d2 = 293
or, 3a2 + 2d2 = 293
or, 3 × 81 + 2d2 = 293
or, d2 = 25
∴ d = ±5.
When a = 9 and d = 5, the required three numbers are a – d, a and a + d
i.e. 4, 9 and 14.
When a = 9 and d = – 5, the required three numbers are a – d, a and a + d
i.e. 14, 9 and 4.
Hence, the numbers are 4, 9 and 14 or 14, 9 and 4.

40 / Optional Mathematics Class 10

Example 10: Find the nth term and the sum to n terms of the series 2 × 3 + 3 × 4 + 4 × 5 + ... n terms. Algebra
Solution:

The given series is 2 × 3 + 3 × 4 + 4 × 5 + ... ... n terms.

nth term of 2, 3, 4, ... ... = 2 + (n – 1)1 = n + 1

nth term of 3, 4, 5, ... ... = 3 + (n – 1)1 = n + 2

The nth term of the given series = (n + 1) (n + 2) = n2 + 3n + 2

Sum of the series to n terms (Sn) = (n2 + 3n + 2)
Sn = ∑n2 + ∑3n + ∑2

= ∑n2 + 3∑n + 2n

= n(n + 1)(2n + 1) + 3 × n(n + 1) + 2n
62

= 6n {(n + 1) (2n + 1) + 9(n + 1) + 12}

= n (2n2 + 3n + 1 + 9n + 9 + 12)
6

= n(2n2 + 12n + 22)
6

= 3n (n2 + 6n + 11).

Exercise 1.6A

1. Write the formula for the sum of the first n terms of the following series:

a. 1 + 2 + 3 + 4 + ... ... b. 2 + 4 + 6 + 8 + ... ...

c. 1 + 3 + 5 + 7 + ... ... d. 12 + 22 + 32 + 42 + ... ...

2. Find the sum of the following series :

a. 7 + 9 + 11 + ... ... 11 terms. b. 21 + 17 + 13 + ... ... 12 terms.

c. 212 + 313 + 416 + ... ... 24 terms. d. 1 + 313 + 532 + ... ... 10 terms.

3. a. Find the common difference of an A.S. whose first term is – 1 and the sum of the first 9 terms is 207.

b. Find the first term of an A.P. whose common difference is – 3 and the sum of the first 10 terms 110.

4. Find the sum of the following:

68 10 12
a. ∑ (2k – 1) b. ∑ (5n + 2) c. ∑ (4k – 1)
k=1 n=2 k=2 d. ∑ (7r + 3)
r=5

3

5. a. In an arithmetic series ∑ (3n + b) = 36, find b.
n=1

10

b. In an arithmetic series ∑ (nb + 6) = 100, find the value of b.
n=3

6. a. In an arithmetic series tn = 5(n + 1) – 7, then find Sn.

b. If tn = 5(n + 1) + 3 in an arithmetic series then find Sn.

Sum of Arithmetic Series / 41

7. a. If t16 = 59 in an arithmetic sequence, then find S31.
b. If t7 = 70 in arithmetic sequence, find S13.
8. a. Find the sum of all natural numbers less than 100 exactly divisible by 5.

b. Find the sum of the first 20 natural numbers which are exactly divisible by 6.

c. Find the sum of all positive integers less than 500 which are exactly divisible by 3 and 7.

9. a. If the sum of three numbers in A.P. is 51 and the product of their extremes is 273, find the
numbers.

b. Find the numbers whose product is 91 and whose arithmetic mean is 10.

10. a. If Sn = n(n + 1) in an arithmetic sequence, then find tn.

b. If Sn = n(n + 1) in an arithmetic sequence, then find tn.
2

c. If Sn = n2 in an arithmetic sequence, then find tn.

Exercise 1.6B

1. Find the sum of the following arithmetic series :

a. 2 + 5 + 8 + ... ... + 44 b. 47 + 45 + 43 + ... ... + 29

c. 6 + 5 23 + 5 13 + ... ... + 23 d. 7 + 8 41 + 9 21 + ... ... + 17

2. a. How many terms of the A.P. – 6 – 11 – 5 ... ... are needed to give the sum – 25 ?
2
b. How many terms of the series 24 + 20 + 16 + ... ... are needed to give the sum 72 ?

c. If the first term of an A.P. is 22, the common difference is – 4 and the sum to n terms is 64, find n.

3. a. If the third term of an A.P. is 1 and its fifth term is 7, find the sum of first ten terms of the series.

b. Find the sum of the first 32 terms of an AP whose 3rd term is 1 and 6th term – 11.

c. The last term of an arithmetic series of 20 terms is 195 and the common difference 5. Calculate

the sum of the series.
4. a. If the fourth term of an A.P. is 1 and the sum of its first eight terms is 18, find the tenth term of

the series.
b. If the twelfth term of an A.P. is 23 and the sum of the first 22 terms 484, find the sum of the first

30 terms.
5. a. If the sum of the first three terms of an arithmetic series is 42 and that of the first five terms 80,

find the twentieth term of the series.

b. The sum of first eight terms of an arithmetic series is 180 and its fifth term is five times of the
first term, then find the first term and common difference.

c. In an arithmetic sequence, the sixth term is equal to three times the fourth term and the sum of
the first three terms is – 12. Find the sum of the first ten terms.

d. If the sum of the first 12 terms of an AS is 168 and that of the first 15 terms 255, find the sum
of the first 21 terms of the series.

6. a. In an arithmetic series, the sum of the first ten terms is 520. If its seventh term is double of its
third term, calculate the first term and the common difference of the series.

42 / Optional Mathematics Class 10

b. The sum of first three terms of an arithmetic series is 21. If the sum of the first two terms is Algebra
subtracted from the third term then it would be 9, find the three terms of the series.

c. The first and last terms of an arithmetic series are – 24 and 72 respectively. If sum of all terms
of the series is 600, find the number of terms and the common difference.

7. a. The sum of the first six terms of an arithmetic progression is 42. The ratio of the 10th terms to
the 30th term of the A.P. is 13. Calculate the first term and the 13th term.

b. The third term of an arithmetic progression is 7 and the seventh term is 2 more than 3 times the
third term. Find the sum of the first 20 terms.

8. a. A sum of Rs.6240 is paid off in 30 instalments, such that each instalment is Rs.10 more than
the proceeding instalment. Calculate the value of the first instalment.

b. The interior angles of a polygon are in arithmetic progression. The smallest angle is 52° and
the common difference is 8°. Find the number of sides of the polygon.

9. a. The sum of the three numbers in A.P. is 33 and the sum of their squares is 461, find the numbers.

b. The sum of the four numbers in A.P. is 28 and the sum of their squares is 216, find the numbers.
10. Find the nth term and the sum to n terms of the following series:
a. 1 × 3 + 2 × 4 + 3 × 5 + ... ... ... n terms.
b. 1 × 3 + 3 × 5 + 5 × 7 + ... ... ... n terms.
c. 12 + (12 + 22) + (12 + 22 + 32) + ... ... ... n terms.



1.7 Geometric Sequence and Series

Introduction

Consider the following sequences.

a. 2, 4, 8, 16, ... ... ... ...

b. 27, 9, 3, 1, ... ... ... ...

c. 16, – 24, 36, – 54, ... ... ... ...

In the first sequence, each term is double of its previous term. In the second sequence, each term is

one-third of its previous term. Each term of the third sequence is – 3 times its previous term. In each of the
2
above sequences, the ratio of a term and its previous term is equal or constant. Such a sequence is known

as a Geometric Sequence or Geometric Progression.

Definition

A sequence is said to be a geometric sequence or geometric progression if the ratio between any term

and its previous term is equal or constant throughout the whole sequence. It is denoted by G.P.

The constant ratio obtained by dividing a term by its previous term is called common ratio of the
4 8 16
sequence. In the above example 2, 4, 8, 16, ... ... the common ratio is 2 = 4 = 8 = 2. Similarly, the common

ratio of 27, 9, 3, 1, ... ... is 1 and that of 16, – 24, 36, – 54, ... ... is – 32. The common ratio of a geometric
3
sequence is denoted by r.

Geometric Sequence and Series / 43

A series corresponding to any geometric sequence is known as the geometric series associated with
the given geometric sequence. Hence, 2 + 8 + 32 + 64 + ... ... ... is a geometric series associated with the
geometric sequence 2, 8, 32, 64, ... ... ... .

General Term of a G.P.

If a be the first term and r be the common ratio of a geometric progression, then the terms of the
progression are: a, ar, ar2, ar3 ... ... ... ...

If t1, t2, t3, t4, ... ... ..., tn be the first, second, third, fourth, ... ... ... , nth terms of a G.P. respectively,
then, t1 = a = ar1 – 1
t2 = ar = ar2 – 1
t3 = ar2 = ar3 – 1
t4 = ar3 = ar4 – 1
... ... ... ... ... ...

tn = ar n – 1
Thus, if the first term and the common ratio of a G.P. are known, we can find any term.

Example 1: If the first term and the common ratio of a G.P. are 1 and – 2 respectively, find the seventh
8
term of the G.P.

Solution:

Here, first term (a) = 1
8

Common ratio (r) = – 2

Seventh term (t7) = ?
By using formula, we have,

tn = arn – 1

or, t7 = 1 × (– 2)7 – 1
8

or, t7 = 1 × 64
8

∴ t7 = 8. 1 1 1 81
18 6 2 2
Example 2: For a geometric series + + + ... ... ... +

a. Find the number of terms

b. Find the 9th term of the series

Solution:

a. Here, the first term (a) = 1
18
1/6
Common ratio (r) = 1/18 = 3

Last term (tn) = 81
2

Number of terms (n) = ?

44 / Optional Mathematics Class 10

By using formula, b. 9th term (t9) = ?
By using formula,
tn = arn – 1 Algebra
tn = arn – 1
or, 821 1
= 18 × 3n – 1 or, t9 = 1 × 39 – 1
18
or, 81 × 9 = 3n – 1 1
or, t9 = 18 × 38
or, 3n – 1 = 36
1
or, n – 1 = 6 or, t9 = 18 × 81 × 81
∴ n = 7.
∴ t9 = 729
2

Example 3: Is 1458 a term of the series 2 + 6 + 18 + 54 + ... ... ... ?
Solution:

Here, the series is 2 + 6 + 18 + 54 + ... ... ...

The first term (a) = 2, common ratio (r) = 6 = 3
2
Suppose, tn = 1458

or, arn – 1 = 1458

or, 2(3)n – 1 = 1458

or, 3n – 1 = 1458
2

or, 3n – 1 = 729

or, 3n – 1 = 36

or, n – 1 = 6
∴ n = 7

Yes, 1458 is the term of given series.

Example 4: Find the value of x for which x + 9, x – 6, 4 are the three terms of a geometric progression

and calculate the fourth term of geometric progression in each case.
Solution:

The given three numbers in G.P. are x + 9, x – 6, 4.

From the definition of G.P., Case I, when x = 0, the given numbers are 9, – 6, 4

xx – 6 = x 4 6 First term (a) = 9
+ 9 –
Common ratio (r) = – 2
or, (x – 6)2 = 4(x + 9) 3

or, x2 – 12x + 36 = 4x + 36 Fourth term (t4) = ?

or, x2 – 16x = 0 We have, tn = arn – 1

or x(x – 16) = 0 or, t4 = 9 × – 2 4 – 1
3
∴ x = 0 or 16.
or, t4 = 9 × –8
27

∴ t4 = – 38.

Geometric Sequence and Series / 45

Case II, when x = 16, the given numbers are 25, 10, 4.

First term (a) = 25

Common ration (r) = 2
3

Fourth term (t4) = ?

We have, tn = arn – 1

2 4 – 1 8
5 125
or, t4 = 25 × = 25 ×

∴ t4 = 58. and the tenth terms are 32 and 1 respectively, find the G.P. Also, find its 15th
Example 5: If the fourth 2

term.

Solution:

Let a be the first term and r be the common ratio of the G.P.

Fourth term (t4) = 32, Tenth term (t10) = 1 Dividing equation (ii) by equation (i),
2

We have, tn = arn – 1 aarr93 = 1/2
or, t4 = ar4 – 1 32
or, ar3 = 32
... ... ... (i) or, r6 = 1
64

and t10 = ar10 – 1 +– 1 6
2
1 or, r6 =
2
or, ar9 = ... ... ... (ii)

∴ r = ± 1
2

Case I, when r = 21,

ar3 = 32

or, a × 1 = 32
8

or, a = 256

The given G.P. is 256, 128, 64 ... ... and the 15th term is

t15 = ar15 – 1

1 14
2
or, t15 = 256 ×

or, t15 = 256 × 28 1 26
×

or, t15 = 256 × 1 64
256 ×

∴ t15 = 1
64

Case II, when r = – 12,

ar3 = 32

46 / Optional Mathematics Class 10