X Maths Chapter 1 - Arithmetic Sequences

Chapter 1
Arithmetic Sequences

For the arithmetic sequence

x n = an + b

the sum of the first n terms is

x 1 + x 2 + ... + x n = a n(n+1) + bn
2

Prepared by
Cecilia Joseph
St. John De Britto’s, A.I.H.S,
Fortkochi

Arithmetic Sequence 1

Online Class – X − 03 21 / 06 / 2021

1. Arithmetic Sequence – Class 1

To view class

Number Sequences

• Natural Numbers ( Counting Numbers)
1, 2, 3, 4, 5,...............................................

• Even Numbers
2, 4, 6, 8, 10,.................................................

• Odd Numbers
3, 5, 7, 9, 11,................................................

• Multiples of 5
5, 10, 15, 20, 25,...................................................

• Counting numbers ending with 1
1, 11, 21, 31, 41, 51,............................................

• Counting numbers ending with 0
10, 20, 30, 40,....................................

• Numbers which leave remainder 1 when divided by 3
1, 4, 7, 10, 13,..................................................

• Perfect Squares
1, 4, 9, 16, 25, 36,...............................................

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 2

• Consider the following squares

1cm 2cm 3cm

1cm 2cm 3cm

Writing the sides, perimeters and areas of above squares we
have

Side ; 1cm, 2cm, 3cm, 4cm,..................

Perimeter : 4 × 1, 4 × 2, 4 × 3, 4 × 4,...........
Perimeter= 4 × side

4cm, 8cm, 12cm, ...............

Area : 1 × 1, 2 × 2, 3 × 3, 4 × 4,.......................

Area= side × side
1cm2, 4cm2, 9cm2, 16cm2,...........

Analysing above all we can say

A group of numbers written like this as
1st, 2nd,3rd and so on using a particular

condition is called a Number Sequence

********************

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence (W.S) 1

Arithmetic Sequence
Worksheet-1

Q1)Write down sequence of natural numbers which leave
remainder 2 on division by 3.

Ans) 2, 5, 8, 11, 14, .............................

Q2) Consider the sequence 2, 4, 6, 8 · · ·
a) Write two more terms of this sequence
b) Which is he smallest two digit term of this sequence?
c) Which is the largest two digit term of this sequence
d) Can the sum of any two terms of this sequence 75?

Ans) a) 10, 12
b) 10
c) 98
d) No. Here terms are even numbers. The sum of two even
numbers cannot be an odd number. So 75cannot be the
sum.

Q3) Consider the sequence 1, 3, 5, 7 · · ·
a) Write next two more terms of this sequence
b) Which is the smallest two digit term of this sequence?
c) Which is the largest two digit term of this sequence
d) Can the sum of any two terms of this sequence 75?

Ans) a) 9, 11
b) 11
c) 99
d) No . Here terms are odd numbers. The sum of two odd
numbers cannot be an odd number.

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence (W.S) 2

Q4)Look at the picture . Squares are made by joining match
sticks.

a) Write the number of squares in each line as a sequence.
b) Write the number of match sticks in each line as a sequence.
c) How many match sticks are used to make next line.
Ans) a) 1, 2, 3, 4, 5, 6, ......................

b) 4, 7, 10
c) 13

Q5) Consider the sequence 1 , 2 , 3 ........
7 7 7

a) What is the next term of this sequence?

b) What is the position of 1 in this sequence?

c) What is the position of 100 in this sequence?

d) Prove that this sequence contains all natural numbers.
4
Ans) a) 7

b) 7th term = 7 =1
7

So 1 is the seventh term .
700
c) 700th term = 7 = 100

So 100 is the seven hundredth term .

d) For n = 7, 14, 21 · · · we get the counting numbers 1, 2, 3, · ·

·

Q6) Consider the sequence 1, 6, 11, 16 · · ·

a) Describe this sequence in three different ways?

b) Which number just below 100 a term of the sequence?

Ans)a) Sequence of numbers having 1 or 6 comes in one’s place.

Sequence starting from 1 and adding 5 repeatedly

Sequence of numbers 4 less than the multiples of 5

b) 96

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 1

Online Class – X − 04 24 / 06 / 2021

1. Arithmetic Sequence – Class 2

To view class

Algebra of sequences
• Consider sequence of counting numbers

1, 2, 3, 4, 5...............

The numbers forming a sequence are called its terms

Here
1st term = 1, 2nd term= 2 , 3rd term= 3, 4th term = 4 ,.............
10th term = 10, ............. nth term= n

The terms in a sequence are written in algebra as

x1, x2, x3, x4, ..............

∴ x1 = 1

x2 = 2

x3 = 3

x4 = 4, ...............x10 =10 ...................., xn = n

Algebraic form

nth term of a sequence is the general term and hence it
is called the algebraic form of the sequence

Algebraic form of sequence of counting numbers is n

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 2

• Consider sequence of even numbers

2, 4, 6, 8, 10, 12, ...................

Here

This can be written as

1st term (x1) = 2 2=2×1
2nd term (x2) = 4 4=2×2
3rd term (x3) = 6 6=2×3
4th term (x4) = 8 8=2×4
5th term (x5) = 10 10 = 2 × 5

⸽

10th term (x10) = 20 20 = 2 × 10
⸽

20th term ( x20) x20 = 2 × 20 = 40

⸽

50th term ( x50) x50 = 2 × 50 = 100
⸽

nth term ( xn) xn = 2 × n = 2n

Algebraic form of sequence of even numbers is 2n

• Consider sequence of perfect squares
1, 4, 9, 16, 25, 36, ..................

Here
x1 = 1 = 12
x2 = 4 = 22
x3 = 9 = 32
x4 =16 = 42
⸽
x20 = 202 = 400
⸽

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 3

x50 = 502 = 2500
⸽
xn = n2

Algebraic form of sequence of perfect squares is n2

• Consider sequence of reciprocals of counting numbers
Counting numbers : 1, 2, 3, 4, 5 ...............

Reciprocals : 1 , 1 , 1 , 1 ,...............
1 2 3 4

x1 = 1 =1
x2 = 1
x3 = 1
x4 =
⸽ 2
x12 = 1
⸽
xn = 3
1

4

1
12

1
n

Algebraic form of sequence of reciprocal of counting
1
numbers is n

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 4

• Consider sequence of triangles made with dots

Number of 1 3 6 10
dots x1 =1 x2 =1+2 x3 =1+2+3 x4 =1+2+3+4

x5 = Number of dots in 5th triangle = 1 + 2 + 3 + 4 + 5
⸽
x10 = Number of dots in 10th triangle = 1 + 2 + 3 + 4 + ....... + 10
⸽
xn = Number of dots in nth triangle = 1 + 2 + 3 + 4 + ....... + n

•Consider sequence of sum of interior angles of regula polygons
1. Triangle

1 Triangle
Sum of interior angles = 1800 × 1

x1 = 1800 × 1

2. Square

2 Triangles
Sum of interior angles = 1800 × 2

x2 = 1800 × 2

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 5

3. Pentagon

3 Triangles
Sum of interior angles = 1800 × 3

x3 = 1800 × 3

4. Hexagon

4 Triangles
Sum of interior angles = 1800 × 4

x4 = 1800 × 4

Algebraic form of sum of interior angles of polygon is
1800 × n

Discussion: Is it possible to write algebraic form of every
sequence ?

Consider sequence of prime numbers Prime numbers
2, 3, 5, 7, 11, 13, 17, 19, 23, ................ are the numbers
Here we cannot write a general term. that have only
There is no algebraic form exists for the two factors, 1 and
sequence of prime numbers the number itself.
So some sequences have no algebraic form.

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 6

Assignment
T.B Page 18

(1) Write the algebraic expression for each of the sequences
below:

a) Sequence of odd numbers
b) Sequence of natural numbers which leave remainder 1 on

division by 3.
c) The sequence of natural numbers ending in 1.
d) The sequence of natural numbers ending in 1 or 6.

(2) For the sequence of regular polygons starting with an
equilateral triangle, write the algebraic expressions for

a) the sequence of the sums of inner angles
b) the sums of the outer angles
c) the measures of an inner angle
d) the measures of an outer angle.

(3) Look at these pictures:

The first picture is got by removing the small triangle
formed by joining the midpoints of an equilateral triangle.
The second picture is got by removing such a middle triangle
from each of the red triangles of the first picture.
The third picture shows the same thing done on the second.

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 7

i) How many red triangles are there in each picture?
ii) Taking the area of the original uncut triangle as 1, compute

the area of a small triangle in each picture.
iii)What is the total area of all the red triangles in each

picture?
iv) Write the algebraic expressions for these three sequences

obtained by continuing this process.

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 1
Online Class – X − 04

1. Arithmetic Sequence – Class 2

Assignment Answers

T.B Page 18

(1) Write the algebraic expression for each of the sequences

below:

a) Sequence of odd numbers

b) Sequence of natural numbers which leave remainder 1 on

division by 3.

c) The sequence of natural numbers ending in 1.

d) The sequence of natural numbers ending in 1 or 6.

Ans)

a) Sequence of odd numbers is

1, 3, 5, 7, 9,.................

x1 = 1 2×1–1 =1
x2 = 3 2×2–1 =3
x3 = 5 2×3–1 =5
x4 = 7 2×4–1 =7
⸽

xn = 2 × n – 1 = 2n − 1
Algebraic expression is 2n – 1

b) Sequence of natural numbers which leave remainder 1 on

division by 3 is

1, 4, 7, 10, 13,................

x1 = 1 3×1–2 =1
x2 = 4 3×2–2 =4

x3 = 7 3 × 3 – 2 = 7

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 2

x4 = 10 3 × 4 – 2 = 10
⸽

xn = 3 × n – 2 = 3n - 2

Algebraic expression is 3n – 2

c) The sequence of natural numbers ending in 1 is

1, 11, 21, 31, 41, .......................

x1 = 1 10 × 1 – 9 = 1
x2 = 11 10 × 2 – 9 = 11
x3 = 21 10 × 3 – 9 = 21
x4 = 31 10 × 4 – 9 = 31
⸽

xn = 10 × n – 9 = 10n − 9

Algebraic expression is 10n – 9

d) The sequence of natural numbers ending in 1 or 6 is

1, 6, 11, 16, 21, 26, 31, 36, ...................

x1 = 1 5×1–4 =1
x2 = 6 5×2–4 =6
x3 = 11 5 × 3 – 4 = 11
x4 = 16 5 × 4 – 4 = 16
⸽

xn = 5 × n – 4 = 5n− 4

Algebraic expression is 5n – 4

(2) For the sequence of regular polygons starting with an
equilateral triangle, write the algebraic expressions for

a) the sequence of the sums of inner angles
b) the sums of the outer angles
c) the measures of an inner angle
d) the measures of an outer angle.

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 3

Ans)
Sequence of regular polygons

12 3 4

a) Sequence of sums of the inner angles is

1800, 3600, 5400, 7200, ..........................

Term position : 1 234

1 × 1800 , 2 ×1800 , 3 ×1800 , 4 ×1800 ,....................

∴ Algebraic expression is n × 1800 = 1800 n

b) Sequence of sums of the outer angles is
3600 , 3600 , 3600 , 3600 , ..........................................

Term position : 1 2 3 4

∴ Algebraic expression is 3600

c) Sequence of measures of inner angles is

600, 900, 1080, 1200 , ..............................

Term position : 1 2 34

1 x 180 , 2 x 180 , 3 x 180 , 4 x 180 ,............................
3 4 5 6

Algebraic expression is n x 180 = 180 n
n+2 n+ 2

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 4

d) Sequence of measures of outer angles is
1200, 900, 720, ................................

Term position : 1 2 3

360 , , 360 , 360 ..............................
3 4 5

Algebraic expression is 360
n+2

(3) Look at these pictures:

The first picture is got by removing the small triangle
formed by joining the midpoints of an equilateral triangle.
The second picture is got by removing such a middle triangle
from each of the red triangles of the first picture.
The third picture shows the same thing done on the second.

i) How many red triangles are there in each picture?
ii) Taking the area of the original uncut triangle as 1, compute

the area of a small triangle in each picture.
iii) What is the total area of all the red triangles in each

picture?
iv) Write the algebraic expressions for these three sequences

obtained by continuing this process.

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 5

Ans)
i) Number of red triangles in the first figure = 3
Number of red triangles in the second figure = 9
Number of red triangles in the third figure = 27

ii) Area of a small triangle in the first figure = 1
4
1
Area of a small triangle in the second figure = 16

Area of a small triangle in the third figure = 1
64

iii) Total area of all the red triangles in the first figure = 3
4
9
Total area of all the red triangles in the second figure = 16

Total area of all the red triangles in the third figure = 27
64

iv) a) Sequence of the number of red triangles in the each
figure is 3 , 9 , 27 , . . . . . . . . .
Term position : 1 2 3 . . . . . . . . . . n
Terms : 3 9 27 . . . . . . . . . .
31 32 33 . . . . . . . . . 3n
Algebraic expression of the sequence = 3n

b) Sequence of the area of a small triangle in the each figure

is 1 , 1 , 1 ,..........
4 16 64

Term position : 1 2 3 . . . . . . . . . . n
1 1 1
Terms : 4 , 16 , 64 ,..........

( 1 1 , ( 1 2 , ( 1 3 ........ ( 1 n
4 4 4 4
) ) ) )

Algebraic expression of the sequence = ( 1 n
4
)

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 6

c) Sequence of the total area of all the red triangles in the each
3 9 27
figure is 4 , 16 , 64 ,...........

Term position : 1 2 3 . . . . . . . . . .n
3 9 27
Terms : 4 , 16 , 64 ,.........

( 3 1 , ( 3 2 , ( 3 3 ,...... ( 3 n
4 4 4 4
) ) ) )

Algebraic expression of the sequence = ( 3 n
4
)

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 1

Online Class – X − 05 28 / 06 / 2021

1. Arithmetic Sequence – Class 3

To view class

Arithmetic Sequences
Consider following sequences :
• Sequence of natural numbers is

1, 2, 3, 4, 5, 6, . . . . . . . . . Here the sequence start with 1 and
add 1 repeatedly.

• Sequence of even natural numbers is
2, 4, 6, 8, 10, . . . . . . . . . . .Here the sequence start with 2 and
add 2 repeatedly.

• Sequence of multiples of 5 is
5, 10, 15, 20, . . . . . . . . . . Here the sequence start with 5 and
add 5 repeatedly.

• Sequence of natural numbers which leave remainder 2 on
division by 3 is
2, 5, 8, 11, 14, . . . . . . . . Here the sequence start with 2 and
add 3 repeatedly.

• Sequence of perimeters of squares with the length of a side
1, 2, 3, 4, . . . . . . . .is
4, 8, 12, 16, . . . . . . . . . Here the sequence start with 4 and
add 4 repeatedly.

• Sequence of perimeters of squares with the length of a side
1, 1½ ,2 , 2½. . . . . . . . . . . . .is

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 2

4, 6, 8, 10, . . . . . . . . . Here the sequence start with 4 and
add 2 repeatedly.

A sequence got by starting with any number and adding a
fixed number repeatedly is called an arithmetic sequence.

Consider the given sequences :
• Sequence of sums of outer angles of regular polygons is

3600 , 3600 , 3600 , 3600 , . . . . . . . . . . .
Here the sequence start with 360 and add 0 repeatedly.
So this is an arithmetic sequence .

• Sequence of squares with the length of the sides
1, 1½ ,2 , 2½, . . . . . . . . . . .

Here the sequence start with 1 and add ½ repeatedly.

So this is an arithmetic sequence .

• Sequence of length diagonals of squares having side

1, 2, 3, 4 . . . . .is Length of the diagonal of a
√2 , 2√2 , 3√2 , . . . . . . . square = √ 2 × side

Here the sequence start with √2 and add √2 repeatedly.

So this is an arithmetic sequence .

• Consider an object moving along a straight line at 10m/s .
By applying a constant force in the opposite direction the
speed is reduced every second by 2 m/s
Sequence of speed is
10 , 8 , 6 , . . . . . . . . .
Here the sequence start with 10 and subtract 2 repeatedly
or we can say add −2 repeatedly.
So this is an arithmetic sequence .

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 3

An arithmetic sequence is a sequence in which we get the
same number on subtracting from any term, the term

immediately preceding it.

The constant difference got by subtracting from any term the
just previous term, is called the common difference of an
arithmetic sequence.
Common difference usually denoted using the letter ‘d’

• Consider sequence of natural numbers
1, 2, 3, 4, 5, . . . . . . .

This is an arithmetic sequence with common difference 1.

• Multiply each term of the above sequence by 6
Sequence is 6, 12, 18, 24, . . . . . . . . . . . . . . . .

This is an arithmetic sequence with common difference 6.

• Add one to each term of the above sequence
Sequence is 7, 13, 19, 25, . . . . . . . . . . . . . .

This is an arithmetic sequence with common difference 6.

• Subtract two from each term of the above sequence
Sequence is 5, 11, 17, 23, . . . . . . . . . . . . . .

This is an arithmetic sequence with common difference 6.

• Consider sequence of powers of two 22 = 4
Sequence is 4, 8, 16 . . . . . . . . 23 = 8

Since there is no common difference , this is not an 24 =16
arithmetic sequence .

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 4

• Consider sequence of prime numbers
Sequence is 3 , 5, 7, 11, 13, . . . . . . . . . . . . . .

Since there is no common difference , this is not an
arithmetic sequence .

Assignment
T.B Page 18

Ans)

i) Sequence of odd numbers 1, 3, 5, 7,.......
Here the sequence start with 1 and adding 2 repeatedly .
So this is an arithmetic sequence .
Common difference = 2

ii) Sequence of even numbers 2, 4, 6, 8,.......
Here the sequence start with 2 and adding 2 repeatedly .
So this is an arithmetic sequence .
Common difference = 2

•iii) Sequence of fractions got as half the odd numbers

1 , 3 , 5 , 7 ,.....................
2 2 2 2

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 5

Here the sequence start with 1 and
2
3 1 2
adding 1 repeatedly . 2 − 2 = 2

So this is an arithmetic sequence . =1
Common difference = 1

iv) Sequence of powers of 2 is 21, 22, 23, 24,.................
2, 4, 8, 16, ..................

x2 – x1 = 4 – 2 = 2
x3 – x2 = 8 – 4 = 4
Here the difference of two consecutive terms is not a
constant , So this is not an arithmetic sequence .

v) Sequence of reciprocals of natural numbers is
1 1 1
1, 2 , 3 , 4 ,.....................

x2 – x1 = 1 − 1 = −1
2 1 2
1 1 −1
x3 – x2 = 3 − 2 = 6

Here the difference of two consecutive terms is not a

constant , So this is not an arithmetic sequence .

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 6

Ans) Numbers of coloured squares in each picture is
8, 12, 16, ...............
Here the sequence start with 8 and adding 4 repeatedly .
So this is an arithmetic sequence .
Common difference = 4

Ans) i) No., of small squares in the first rectangle = 2
No. of small squares in the second rectangle = 4
No. of small squares in the third rectangle = 6
No. of small squares in the fourth rectangle = 8

ii) No. of large squares in first rectangle = 0
No. of large squares in second rectangle = 1
No. of large squares in third rectangle = 2
No. of large squares in fourth rectangle = 3

iii) Total squares in first rectangle = 2
Total squares in second rectangle = 5
Total squares in third rectangle = 8
Total squares in fourth rectangle = 11

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 7

Sequence of no. of small squares 2, 4, 6, 8.....
Here the sequence start with 2 and adding 2 repeatedly .
So this is an arithmetic sequence .

Sequence of no. of large squares 0, 1, 2, 3.......
Here the sequence start with 0 and adding 1 repeatedly .
So this is an arithmetic sequence .

Sequence of total no. of squares 2, 5, 8, 11.......
Here the sequence start with 2 and adding 3 repeatedly .
So this is an arithmetic sequence .

i) Height from the ground climbing up first step = 10 cm
Height from the ground climbing up second step = 10 + 17.5
= 27.5cm
Height from the ground climbing up third step = 27. 5+ 17.5
= 45 cm
Height from the ground climbing up fourth step = 45 + 17.5
= 62.5 cm

Height from the ground climbing up fourth step = 62.5 + 17.5
= 80 cm

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 8

Height from the ground climbing up fourth step = 80 + 17.5
= 97.5cm

ii) Sequence of height is
10, 27.5, 45, 62.5, 80, 97.5 ...............

Ans) Draw PL, QM, RN, SO parallel to XY

We can see that ∆XAP, ∆PLQ, ∆QMR, ∆RNS, ∆SOT are equal.
(If one side and two angles on it of a triangle are equal to
one side and two angles on it of another triangle , then the
triangles are equal )

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 9

Let AP = x & LQ = d.
Since triangles are equal MR = d, NS = d, OT = d
∴ Lengths of perpendiculars is x, x+d, x+d+d, x+d+d+d,.............

That is x, x+d, x+2d , x+3d ,...................
This is an arithmetic sequence with common difference ‘ d ’

Ans) X3 = 33 – 6 × 32 + 13 × 3– 7
X1 = 13 – 6 × 12 + 13 × 1 – 7 = 27 – 54 + 39 − 7
= 1 – 6 + 13 – 7 = 66 −61
= 14 – 13 =5
=1

X2 = 23 – 6 × 22 + 13 × 2 – 7 X4 = 43 – 6 × 42 + 13 × 4– 7
= 8 – 24 + 26 – 7 = 64 – 96 + 52 −7
= 34 – 31 = 116 −103
=3 = 13

Sequence is 1, 3, 5, 13,.....................
Since there is no common difference this is not an arithmetic
sequence.

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 1

Online Class – X − 06 29 / 06 / 2021

1. Arithmetic Sequence – Class 4

To view class

Position and term
Activity 1
Can you make an arithmetic sequence with 30 and 50 as the
first and second terms?
Ans) First term (x1 ) = 30 , Second term (x2 ) = 50

Common difference (d) = Second term − First term
= 50 − 30
= 20

Third term = Second term + Common difference
= 50 + 20 = 70

∴ Arithmetic sequence is 30, 50, 70, . . . . . . . . . . . .

Note:

Second term =First term + Common difference x2 = x1 + d

Third term =First term + 2 × Common difference x3 = x1 + 2d

Fourth term =First term + 3 × Common difference x4 = x1 + 3d
Fifth term =First term + 4 × Common difference x5 = x1 + 4d

Sixth term =First term + 5 × Common difference x6 = x1 + 5d

⸽⸽

Activity 2
Can you make an arithmetic sequence with 30 and 50 as the
first and third terms?
Ans) 30, ------ , 50

First term = 30 , Third term = 50
Third term = First term + 2 × Common difference
Third term − First term = 2 × Common difference

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 2

50 − 30 = 2 × d

20 = 2 × d
20
d= 2

= 10
∴ Arithmetic sequence is 30, 40, 50, 60, . . . . . . . . . . . .

Activity 3

Can you make an arithmetic sequence with 30 and 50 as the

third and seventh terms?

Ans) ---- , ---- , 30 , ---- , ---- , ----, 50,

Seventh term = Third term + 4 × common difference

Seventh term − Third term = 4 × common difference

50 − 30 = 4 × d

20 = 4 × d
20
d= 4 =5

First term = Third term − 2 × common difference

= 30 – 2 × 5

= 30 – 10

= 20
∴ Arithmetic sequence is 20 , 25 , 30 , 35 , 40 , 45 , 50, . . . . . . . . .

Activity 4
Can you make an arithmetic sequence with 30 and 70 as the
10th and 20th terms?
Ans)

20th term = 10th term + 10 × common difference
20th term − 10th term = 10 × common difference

↑ ↑(20 -10 )

Term difference = Position difference × common difference

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 3

70 − 30 = 10 × d

40 = 10 × d
40
d= 10 =4

First term = 10th term − 9 × common difference
= 30 – 9 × 4 = 30 – 36 = −6

∴ Arithmetic sequence is −6, −2 , 2 , 6 , 10 , . . . . . . . . . . .

Observations

The difference between any two terms of an arithmetic
sequence is the product of the difference of positions and

the common difference

We can put it like this also:

In an arithmetic sequence, term difference is proportional
to position difference; and the constant of proportionality is
the common difference.

In any arithmetic sequence ,

Common difference = Term difference
Position difference

Term difference = Position difference x Common difference

Term difference is a multiple of common difference

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 4

To check whether a given number is a term of a given

arithmetic sequence.

Activity 5

Is 100 a term of the arithmetic sequence 4 , 7 , 10 , . . . ?

Give reasons .

Ans) Common difference = 7 − 4 = 3 96 = 32
Term difference = 100 − 4 = 96 3

96 = 32 × 3

Since 96 is a multiple of common difference 3, 100 is a

term of this sequence.

Note:
When 4 ÷ 3 , remainder = 1
When 7 ÷ 3 , remainder = 1
When 10 ÷ 3 , remainder = 1
⸽
When 100 ÷ 3, remainder = 1

Here we can see that when the terms are divided by common
difference remainder is the same .
So we can say, 100 is a term of this sequence.

Considering an arithmetic sequence with
terms and common difference as natural numbers ,

the terms of this sequence leave same remainder
when they are divided by its common difference

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 5

Assignment
T.B Page 21

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 1

Answers
Q1) Ans
i) Given x1 = 24, x2 = 42

Common difference = 42 – 24 = 18
X3 = 42 + 18 = 60
X4 = 60 + 18 = 78
∴ Arithmetic sequence is 24, 42, 60, 78, . . . . . .

ii) Given x2 = 24, x3 = 42
Common difference = 42 – 24 = 18
X1 = 24 – 18 = 6
X4 = 42 + 18 = 60
∴ Arithmetic sequence is 6, 24, 42, 60, . . . . . . .

iii) Given x3 = 24, x4 = 42
Common difference = 18
X2 = 24 – 18 = 6
X1 = 6 – 18 = –12
∴ Arithmetic sequence is –12, 6, 24, 42, . . . . . .

iv) Given x1 = 24, x3 = 42

Common difference = (42 – 24 ) = 18 =9
3− 1 2

X2 = 24 + 9 = 33

X4 = 42 + 9 = 51
∴ Arithmetic sequence is 24, 33, 42 ,51. . . . . . .

v) Given x2 = 24, x4 = 42

Common difference = (42 – 24 ) = 18 =9
4− 2 2

X1 = 24 – 9 = 15

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 2

X3 = 24 + 9 = 33
∴ Arithmetic sequence is 15, 24, 33, 42, . . . . . . . .

vi) Given x1 = 24, x4 = 42

Common difference = (42 – 24 ) = 18 =6
4−1 3

X2 = 24 + 6 = 30

X3 = 30 + 6 = 36
∴ Arithmetic sequence is 24, 30, 36, 42, . . . . . . . .

Q2)Ans. or
i) Given x3 = 34 and x6 = 67 67− 34
d= 6− 3
x3 + 3d = x6

34 + 3d = 67 = 33
3

3d = 67 – 34 = 11

3d = 33

d =11

So, x2 = 34 – 11= 23 x4 = 34 + 11 = 45

x1 = 23 –11 = 12 x5 = 45 + 11 = 56

∴ First five terms is 12, 23, 34, 45, 56, 67, . . . . .

ii) Given x3 = 43 and x6 = 76

x3 + 3d = x6
43 + 3d = 76

3d = 76 – 43

3d = 33

d =11

So, x2 = 43 – 11= 32 x4= 43 + 11 = 54

x1 = 32 –11 = 21 x5 = 54 +11 = 65

∴ First five terms is 21, 32, 43, 54, 65 ,76, . . . . . . .

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 3

Similar way do other questions also
iii) First 5 terms is 1, 1½ , 2, 2½ , 3 , . . . . . .
iv) First 5 terms is 1, 1⅓, 1⅔, 2, 2 ⅓ , . . . . .
v) First 5 terms is 6, 5, 4, 3, 2 , . . . . . . .

Q3 Ans) or
Given x5 = 38 , x9 = 66 66− 38
x5 + 4d = x9 d= 9− 5

38 + 4d = 66

4d = 66 – 38 = 28
4

4d = 28 =7
28
d= 4

d= 7

x25 = x5 + 20d

= 38 + 20 × 7

= 38 + 140

= 178

Q4 Ans)

a) Common difference = 24 – 13 = 11 88 =8
Term difference = 101 − 13 = 88 11

88 = 8 × 11

Since 88 is a multiple of common difference 11, 101 is a

term of this sequence.

b) Term difference = 1001 – 13 = 988 988 =89.818....
11

Since 988 is not a multiple of common difference 11, 1001

is not a term of this sequence.

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 1

Online Class – X − 07 01 / 07 / 2021

1. Arithmetic Sequences – Class 5

To view class

To the magical world of Arithmetic sequences
through games
Game 1:
In the given 4 × 4 grid , fill each box with numbers such that
each row and each column has to be an arithmetic sequence.

Ans) d=4
Filling each box with natural numbers from 1 to 16d=4
d=4
1 2 3 4 d=1d=4
5 6 7 8 d=1
9 10 11 12 d = 1
13 14 15 16 d = 1

Here ,each row is an arithmetic sequence with common
difference 1 .
Each column is an arithmetic sequence with common
difference 4 .

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 2

Game 2 :
In the given 4 × 4 grid , fill each box with continuous terms of
an arithmetic sequence. The numbers in each row and each
column must be an arithmetic sequence.

2

32 d=8
d=8
Ans) Here two numbers are fixed .d=8
We can fill each box with even natural numbers.d=8

2 4 6 8 d=2
10 12 14 16 d = 2
18 20 22 24 d = 2
26 28 30 32 d = 2

Here ,each row is an arithmetic sequence with common
difference 2 .
Each column is an arithmetic sequence with common
difference 8 .

Game 3 :
In the given 4 × 4 grid , fill each box with continuous terms of
an arithmetic sequence. The numbers in each row and each
column must be an arithmetic sequence.

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 3

1

31d=8
d=8
Ans) d=8
Here two numbers are fixed .
We can fill each box with odd natural numbers.d=8

1 3 5 7 d=2
9 11 13 15 d = 2
17 19 21 23 d = 2
25 27 29 31 d = 2

d = 6 d = 10

Here ,each row is an arithmetic sequence with common
difference 2 .
Each column is an arithmetic sequence with common
difference 8 .
One diagonal is an arithmetic sequence with common
difference 10 .
Other diagonal is an arithmetic sequence with common
difference 6.

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 4

Analysing the above questions it can be seen that
a) If we fix the number in the first box and the common

difference of the arithmetic sequence we are going to write ,
we can fix the number in the last box.

Game Number Common Relation Number in
the last box
Game 1 fixed in the difference
Game 2
Game 3 first box fixed

1 1 1 +15 × 1 =16 16

2 2 2 +15 × 2 =32 32

1 2 1 +15 × 2 =31 31

b) If we fix the number in the first box, the number in the last
box and the common difference of the arithmetic sequence
we are going to write ,we can find the total number of boxes
needed .

Numb Num

Game er Com ber Term Total

fixed mon in Differ Term Difference number of

in the differ the ence Common Difference boxes

first ence last needed

box fixed box

Game 1 1 16 16 − 1 16 − 1 = 15 = 15 15 + 1= 16
1 11

Game 2 2 32 32 − 2 32 − 2 = 30 = 15 15 + 1= 16
2 22

Game 1 2 31 31 −1 31 − 1 = 30 = 15 15 + 1= 16
3 22

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 5

Q1) Suppose we fix the number in the first box as 2, the

number in the last box as 48 and the common difference of

the arithmetic sequence we are going to write as 2 , find

the total number of boxes needed .

Ans) 2

Number fixed in the first box = 2 48

Number fixed in the last box = 48

Common difference fixed = 2

Term difference = 48 − 2

Common difference 2

= 46

2

= 23

ie, Position difference = 23
∴ Total number of boxes needed = 23 + 1 = 24

Q2) Fill up the empty boxes such that the numbers in each row
and column form an arithmetic sequence .

14

7 28

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 6

Ans)

1st 4th Arithmetic
Term Term Common difference Sequence

1st row 1 4 4−1 = 3 =1 1, 2, 3, 4
4th row 7 28 3 3 7, 14, 21, 28
1st column 1 7 1, 3, 5, 7
4th column 4 28 28 − 7 = 21 =7 4, 12, 20, 28
3 3

7−1 = 6 =2
3 3

28 − 4 = 24 =8
3 3

12 34
3 12
5 20
7 14 21 28

Similarly we find the common differences of other rows also
and can fill the empty boxes .

1st 4th Common difference Arithmetic
Term Term Sequence

2nd row 3 12 12 − 3 = 9 =3 3, 6, 9, 12
3rd row 5 20 3 3 5, 10, 15, 20

20 − 5 = 15 =5
3 3

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 7

1 23 4
3 69
5 10 15 12
7 14 21 20
28

In any arithmetic sequence

Common Difference = Term difference
Position difference

Position Difference = Term difference
Common difference

Q3) How many terms are there in the arithmetic sequence

101, 108, 115, . . . . . . . . . . 997 ?

Ans) Common difference = 108 – 101 = 7

Term difference = 997 −101

= 896

Position difference = Term difference
Common difference

= 896
7

= 128

Number of terms = 128 + 1 Term difference = xn – x1

= 129 Position Difference = n – 1
n−1 = 128

n = 128 + 1 = 129

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 8

Q4) How many three-digit numbers are there, which leave

remainder 3 on division by 7 ?

Ans) Sequence is

101, 108, ............................... 997

First-term = x1 = 101

Last term = xn = 997

Common difference = 7

Term difference = xn − x1 = 997 – 101 = 896

Position difference = Term difference
Common difference

= 896
7

= 128

Position difference = n − 1 = 128

n = 128 + 1

= 129

∴ 129 three-digit numbers are there which leave remainder
3 on division by 7 .

Q5) How many two-digit numbers are there, which leave

remainder 2 on division by 3 ?

Ans) Sequence is

11 , 14 , 17 , . . . . . . . . , 98

First-term = x1 = 11

Last term = xn = 98

Common difference = 3

Term difference = xn − x1 = 98 – 11 = 87

Position difference = Term difference
Common difference

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 9

= 87
3

= 29

Position difference = n − 1 = 29

n = 29 + 1

= 30

∴ 30 two-digit numbers are there which leave remainder
2 on division by 3 .

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 1

Online Class – X − 08 05 / 07 / 2021

1. Arithmetic Sequences – Class 6

To view class

Algebra of arithmetic sequences

Activity 1 :

Look at the sequence of natural numbers,

1,2,3,4,...
a) What is the 10th term of the sequence ?
b) What is the 100th term of the sequence ?
c) What is the 199th term of the sequence ?
d) What is the nth term of the sequence ?

Ans)

10th term 10

100th term 100

199th term 199

nth term n

Algebraic form of the sequence of natural numbers is ‘ n ’

Activity 2 :

Look at the sequence of even numbers

2,4,6,8,........
a) What is the 10th term of the sequence ?
b) What is the 100th term of the sequence ?
c) What is the 199th term of the sequence ?
d) What is the nth term of the sequence ?

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 2

Ans)

10th term 2 × 10 = 20
100th term 2 × 100 = 200
199th term 2 × 199 = 398
2 × n = 2n
nth term

Algebraic form of the sequence of even numbers is ‘ 2n ’

Activity 3 :

Look at the sequence of odd numbers

1,3,5,7,.......
a) What is the 10th term of the sequence ?
b) What is the 100th term of the sequence ?
c) What is the nth term of the sequence ?

Ans)

1st term 1 2×1−1 =2 −1 =1
2nd term 3 2×2−1 =4 −1 =3
10th term 2 × 10 − 1 = 20 − 1 = 19
100th term 2 × 100 −1= 200 − 1 =199
nth term 2 × n − 1 = 2n − 1

Algebraic form of the sequence of odd numbers is ‘ 2n − 1 ’

Activity 4 :
Look at the sequence of multiples of 5
5 ,10 , 15 , 20 , . . . . . . .

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 3

a) What is the 10th term of the sequence ?
b) What is the 100th term of the sequence ?
c) What is the nth term of the sequence ?

Ans)

1st term 5 5×1=5

2nd term 10 5 × 2 = 10

10th term 5 × 10 = 50

100th term 5 × 100 = 500

nth term 5 × n = 5n

Algebraic form of sequence of multiples of 5 is ‘ 5n ’

Similar way we can find the following

Sequence Algebraic form

Multiples of 3 3, 6, 9, 12, . . . . . . 3n

Multiples of 4 4, 8, 12, 16, . . . . . 4n

Multiples of 6 6, 12, 18, 20, . . . . 6n

Multiples of 7 7, 14, 21, 28, . . . . 7n

Note:
We can generate a new arithmetic sequence by adding or
subtracting a constant from the multiples of a natural
number.

Consider the following sequences
• Sequence of multiples of 5

Adding 1 to each terms of this sequence we get another
sequence

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 4

Sequence Common Algebraic
difference form

Multiples of 5 5 , 10 , 15 , 20 , . . . . . . 5 5n
5 5n + 1
Adding 1 to 6 , 11 , 16 , 21, . . . . . . .
each term
1st term = 5 × 1 + 1 = 6
2nd term = 5 × 2 + 1 = 11
⸽
10th term = 5 ×10 +1 = 51
100th term= 5 ×100 +1 = 501
nth term = 5 × n +1 = 5n+1

• Sequence of multiples of 3
Adding 2 to each terms of this sequence we get another
sequence .

Sequence Common Algebrai
difference c form

Multiples of 3 3 , 6 , 9 , 12 , . . . . . . 3 3n
3 3n + 2
Adding 2 to 5 , 8 , 11, 14 , . . . . . . .
each term
1st term = 3 × 1 + 2 = 5
2nd term = 3 × 2 + 2 = 8
⸽
10th term = 3 × 10 + 2 = 32
100th term = 3 ×100 +2 = 302
nth term = 3 × n + 2 = 3n + 2

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 5

• Sequence of multiples of 3
Subtracting 1 from each terms of this sequence we get
another sequence .

Sequence Common Algebraic
difference form

Multiples of 3 3 , 6 , 9 , 12 , . . . . . . 3 3n
3 3n − 1
Subtracting 1 2 , 5 , 8 , 11 , . . . . . . .

from each

term 1st term = 3 × 1 − 1 = 2

2nd term = 3 × 2 − 1 = 5

⸽

10th term = 3 × 10 − 1 = 29

100th term = 3 ×100 −1 = 299

nth term = 3 × n −1 = 3n −1

Findings :
To find algebraic form of an arithmetic sequence

* First find the common difference of the sequence.
* Write the multiples of the common difference and its

algebraic form
* See that which constant is added or subtracted to make the

given arithmetic sequence.

The algebraic form of any arithmetic sequence
is of the form a n + b , where a and b are fixed number

and ‘ a ’ is the common difference

The nth term of a sequence is its general form .
The nth term of a sequence is also called its algebraic form

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Arithmetic Sequence 6

nth term or Algebraic form of any arithmetic
sequence is
xn = f + ( n – 1 ) d
or
xn = dn + (f – d)
where, f = first term
d = common difference

Q1) Write the algebraic form of the sequence 2 , 5 , 8 , 11 , . . .

Ans) nth term = dn + (f – d ) f = 2, d = 5 −2 = 3

= 3n − 1 f − d = 2 − 3 =−1

Finding algebraic form without using the formula:
d = 3, here each term of the sequence is one less than
the multiples of 3 . So, algebraic form is 3n − 1

Q2) Write down the sequence of natural numbers which leave

remainder 1 on division by 4 and write its algebraic form.

Ans) Sequence is

1 , 5 , 9 , 13 , . . . . . . .

nth term = dn + (f – d) f = 1, d = 5 − 1 = 4

= 4n − 3 f − d = 1 − 4 =−3

d = 4, here each term of the sequence is 3 less than
the multiples of 4 . So, algebraic form is 4n − 3 .

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi