CLASS-X

MATHEMATICS

ARITHMETIC PROGRESSION
Learning Objectives
Students will be able to
1. recall some patterns which occur in their day to day life.
2. know the condition when the pattern will be in
arithmetic progression.
3. know various terms associated with AP, i.e 1 st term and
common difference.
4. understand different types of AP, finite & infinite.
5. understand the formula to find nth term and sum of n
terms of an AP.
6. apply the knowledge of finding nth term and sum of n
terms of an AP in solving day to day life problems.
Sequence
A sequence is a set of numbers written in a
particular order
Examples of sequence:
1, 3, 5, 7, 9, . . . .
1, 4, 9, 16, 25, . . . .
1, −1, 1, −1, 1, −1, . . . ,
1, 3, 5, 9
Series
A series is something we obtain from a
sequence by adding all the terms together.
For example, let us consider the sequence of
numbers 1, 2, 3, 4, 5, 6, . . ., n .
 Then S1 = 1, as it is the sum of just the first
term on its own.
The sum of the first two terms is S2 = 1 + 2 = 3.
 Continuing, we get S3 = 1 + 2 + 3 = 6 , S4 = 1
+ 2 + 3 + 4 = 10 ,
Arithmetic progressions
An arithmetic progression, or AP, is a
sequence where each new term after the first
is obtained by adding a constant d, called the
common difference, to the preceding term.
If the first term of the sequence is a then the
arithmetic progression is
a, a + d, a + 2d, a + 3d, . . .
Where d is the common difference
To check that a given terms are in A.P.
or not.
2, 6, 10, 14….
Here first term a = 2,
find differences in the next terms a2-a1
=6 – 2 = 4
a3-a2 = 10 –6 = 4

a4-a3 = 14 – 10 = 4

Since the difference between

the consecutive terms is same
Hence the given terms are in
A.P.
which the given series is in A.P. 4, k –1
, 12
Solution : Given A.P. is 4,k –1 , 12…..
If series is A.P. then the differences will
be common.
d1 = d1
a2 – a1 = a3 – a2
k – 1 – 4 = 12 – (k – 1)
k–5 = 12 – k + 1
k + k = 12 + 1 + 5
2K=18
K=9
 ACTIVITY-1
Objective: To verify that the given
sequence is an arithmetic progression by
paper cutting and pasting method.
Materials required: coloured paper, pair of
scissors, geometry box, fevicol, sketch pens,
one squared paper
 ACTIVITY-1(continued..)
Procedure : Take a given sequence of
numbers say a1 , a2 , a3 …. 2. Cut a
rectangular strip from a colored paper of
width k = 1 cm (say) and length a1 cm. 3.
Repeat this procedure by cutting rectangular
strips of the same width k = 1cm and lengths
a2 , a3 , a4 , … cm. 4. Take 1 cm squared
paper and paste the rectangular strips
adjacent to each other in order.
 ACTIVITY-1(continued..)
Let the sequence be 1, 4, 7, 10, …. Take
strips of lengths 1 cm, 4 cm, 7 cm and 10
cm, all of the same width say 1 cm. Arrange
the strips in order as shown in Fig 2(a).
Observe that the adjoining strips have a
common difference in heights. (In this
example it is 3 cm.)
ACTIVITY-1(continued..)
 ACTIVITY-1(continued..)
Let another sequence be 1, 4, 6, 9, … Take
strips of lengths 1 cm, 4 cm, 6 cm and 9 cm
all of the same width say 1 cm. Arrange
them in an order as shown in Fig 2(b).
Observe that in this case the adjoining strips
do not have the same difference in heights.
ACTIVITY-1(continued..)
 Conclusion
So, from the figures, it is observed that if
the given sequence is an arithmetic
progression, a ladder is formed in which the
difference between the heights of adjoining
steps is constant. If the sequence is not an
arithmetic progression, a ladder is formed in
which the difference between adjoining
steps is not constant.
nth Term of an AP

a, a + d, a + 2d, a + 3d, a + 4d, . . .is an AP.

where a is the first term, and d is the
common difference.

 If we wanted to write down the n-th term,

we would have a + (n − 1)d ,

𝑎𝑛 = 𝑎 + ሺ𝑛 − 1ሻ× 𝑑
 Let’s see an example

EXAMPLE 1: Let a=2, d=2, n=12,find An

An=a+(n-1)d
=2+(12-1)2
=2+(11)2
=2+22

Therefore, An=24

Hence solved.
nth Term of an AP

Example-2: Write down the first five

terms of the AP with first term 8 and
common difference 7.
Ans: 8,15,22,29,36
Example-3: Write down the first five
terms of the AP with first term 2 and
common difference −5.
Ans: 2, -3, -8,-13,-18
nth Term of an AP
Example 4: What is the common
difference of the AP 11, −1, −13,
−25, . . . ?
Ans: common difference = -1-11
=-13-(-1)
=-12
nth Term of an AP
Example-5: Find the 17th term of the
arithmetic progression with first term 5 and
common difference 2.
Ans: Here a= 5 , d=2
𝑎17 = 5 + ሺ17 − 1ሻ× 2 = 5 + 32 = 37
The sum of an arithmetic series
Sometimes we want to add the terms of a sequence.
Let us see the video https://youtu.be/S6F6jeVX-b8
What would we get if we wanted to add the first n
terms of an arithmetic progression?
We would get Sn = a + (a + d) + (a + 2d) + . . . + (ℓ
− 2d) + (ℓ − d) + ℓ .
This is now a series, as we have added together the
n terms of a sequence. This is an arithmetic series,
and we can find its sum by using a trick.
The sum of an arithmetic series
Let us write the series down again, but this
time we shall write it down with the terms in
reverse order.
We get Sn = ℓ + (ℓ − d) + (ℓ − 2d) + . . . +
(a + 2d) + (a + d) + a
2Sn = (a + ℓ) + (a + ℓ) + (a + ℓ) + . . . + (a +
ℓ) + (a + ℓ) + (a + ℓ),
2Sn = n(a + ℓ)
The sum of an arithmetic series
Sn = ½ n(a + ℓ).
ℓ = a + (n − 1)d ,
Sn = ½ n(a + ℓ)
Sn = ½ n(a + a + (n − 1)d)
 Sn= ½ n(2a + (n − 1)d).
The sum of an arithmetic series
Example -1: Find the sum of the first 50
terms of the sequence 1, 3, 5, 7, 9, . . . .
Solution : This is an arithmetic progression,
a = 1 , d = 2 , n = 50 .
Sn = ½ n(2a + (n − 1)d)
S50 = ½ × 50 × (2 × 1 + (50 − 1) × 2)
= 25 × (2 + 49 × 2)
= 25 × (2 + 98) = 2500
The sum of an arithmetic series
Example-2: Find the sum of the series 1 + 3·5
+ 6 + 8·5 + . . . + 101 .
Solution : Given a=1 , d= 3.5 - 1 = 2.5 and ℓ=101
we know ℓ = a + (n − 1)d
101 = 1 + (n − 1) × 2·5 .
100 = (n − 1) × 2·5
40 = n − 1
n = 41
So, Sn = ½ n(a+ ℓ)= ½ ×41(1+101)=2091
The sum of an arithmetic series
Example -3: An arithmetic progression has 3 as its
first term. Also, the sum of the first 8 terms is
twice the sum of the first 5 terms. Find the
common difference.
Solution : Given a = 3.
S8= ½ × 8 × (6 + 7d), S5= ½ × 5 × (6 + 4d)
S8 = 2S5, we see that
½ × 8 × (6 + 7d) = 2 × ½ × 5 × (6 + 4d)
4 × (6 + 7d) = 5 × (6 + 4d)
24 + 28d = 30 + 20d
8d = 6
of
A.P. 100, 105, 110, 115,,………500

Solution : First term is a = 100 , a n = 500

Common difference is d = 105 -100 =

5
nth term a n = a + (n-1)d
is
500 = 100 + (n-
1)5
500 - 100 = 5(n –
1)
400 = 5(n – 1)
5(n – 1) = 400
5(n – 1) =
400 n – 1 =
400/5 n - 1
= 80
n = 80 +
1
n = 81
Hence the
no. of term
is 81.
Example 5 . Find the sum of 30 terms of
given A.P 12 , 20 , 28 , 36
Solution : Given A.P. is 12 , 20, 28 ,
36 Its first term is a = 12
Common difference is d = 20 – 12
=8
The sum to n terms of an arithmetic
progression Sn= n/2 [ 2a + (n - 1)d ]

= ½ x 30 [ 2x 12 + (30-1)x
8]
= 15 [ 24 + 29 x8]
= 15[24 + 232]
= 15 x 246
= 3690

THE SUM OF TERMS IS 3690

 Summary
In this chapter, you have studied the
following points :
An arithmetic progression (AP) is a list of
numbers in which each term is obtained by
adding a fixed number d to the preceding
term, except the first term. The fixed
number d is called the common difference.
The general form of an AP is a, a + d, a +
2d, a + 3d, . . .
 Summary
A given list of numbers a1, a2, a3, . . . is an
AP, if the differences a2 – a1, a3 – a2, a 4 –
a3, . . ., give the same value, i.e., if ak + 1
– ak is the same for different values of k.
 In an AP with first term a and common
difference d, the nth term (or the general
term) is given by
an = a + (n – 1) d.
 Summary
The sum of the first n terms of an AP is
given by :
S n= ½ n[2a+(n-1)d]
If l is the last term of the finite AP, say the
nth term, then the sum of all terms of the AP
is given by :
Sn = ½ n(a+l)
 MIND MAP
Sum of first n positive integers
Let s n = 1 + 2 + 3 + ... n
How many 2-digit numbers are
divisible by 3? a = 1, last term l = n

sn=
n ( a + l ) = n(1+n)
2-digit numbers divisible by 3
12, 15, 18, ... 99 2
n(n+l)
or s n = 2
a = 12, d = 3, a n = 99 n (a+ l)
s =2
n
a n = a + (n–1)d 2
99 = 12 + (n–1)3
87 List of numbers in which each term is
i.e., n–1 = 3 = 29
obtained by adding a fixed number
n = 30
to the preceding term except the first term.
Fixed number is called common difference.

er
a, a+d, a+2d, a+3d, ...
Arithmeti a+(n –1) d
c
Progressio
If a, b, c, are in AP,
n Fixed number in arithmetic
a+ c progression which provides the
b = 2
h Te

to and fro terms by adding/

b is arithmetic mean subtracting from the present
number.
From beginning Can be positive or negative.
a n = a+(n–1)d
When first term of
Here
common differnce is When first & last a – first termterm
given : a- First
terms are given:
d- common
𝑆 = 𝑛{2𝑎 + (𝑛 − 1)𝑑} d di difference
𝑆𝑛 = 𝑛(𝑎 +
𝑛
2
2 –
𝑎𝑛) c
o
m
o