Arithmetic Progressions in Mathematics
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ARITHMETIC PROGRESSION
CLASS-X
MATHEMATICS
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
T
o
check
th
a
t
a
g
i
v
e
n
ter
m
s
are
i
n
A
.
P
.
o
r
not
.
2, 6,
10,
14….
Here
first
term
a
=
2,
find
differences
in
the
next
terms
a
2
-a
1
=
6
–
2
=
4
a
3
-a
2
=
10
–6
=
4
a
4
-a
3
=
14
–
10
=
4
Since
the
difference
between
the consecutive terms
is
same
Hence
the
given
terms
are
in
A.P.
EXAMPLE 1
:
Find
the
value
of
k
for
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.
d
1
=
d
1
a
2
–
a
1
=
a
3
–
a
2
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 ,
Let’s
see
an
example
EXAMPLE 1:
Let
a=2,
d=2,
n=12,find
A
n
A
n
=a+(n-1)d
=2+(12-1)2
=2+(11)2
=2+22
Therefore,
A
n
=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:
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)
S
50
= 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, S
n
= 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
d = ¾ .
E
x
a
m
p
l
e
4
:
F
i
n
d
n
u
m
b
e
r
o
f
t
e
r
m
s
o
f
A
.
P
.
1
0
0
,
1
0
5
,
1
1
0
,
1
1
5
,
,
…
…
…
5
0
0
Solution :
First
term
is
a
=
100
,
a
n
=
500
Common
difference
is
d
=
105
-100
=
5
n
t
h
t
er
m
i
s
a
n
=
a
+
(n-1)d
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.
E
x
a
m
p
l
e
5
.
F
i
n
d
t
h
e
s
u
m
o
f
3
0
t
e
r
m
s
o
f
g
i
v
e
n
A
.
P
1
2
,
2
0
,
2
8
,
3
6
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
S
n
=
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
T
H
E
S
U
M
O
F
T
E
R
M
S
I
S
3
6
9
0
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 a
1
, a
2
, a
3
, . . . is an AP,
if the differences a
2
– a
1
, a
3
–
a2,
a
4
– a
3
, . . .,
give the same value, i.e., if a
k
+ 1 – a
k
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
a
n
= 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)
s
n
=
n
(
a
+
l
)
2
If a,
b,
c,
are
in
AP,
2
b is
arithmetic
mean
b
=
a
+
c
M
IND MAP
A
rithmetic
P
r
o
gre
ssi
on
When
first
term
of
common
differnce
is
given
:
When
first
&
last
terms
are
given:
From
beginning
a
n
=
a+(n–1)d
Here
d
–
c
o
m
o
n
a – first
term
di
2
or
s
n
=
n(n+l)
3
n
=
30
i.e.,
n–1
=
87
=
29
2-digit
numbers divisible by
3
12, 15, 18, ...
99
a
=
12,
d
=
3,
a
n
=
99
a
n
=
a
+
(n–1)d
99
=
12
+
(n–1)3
s
n
=
n(a+l)
=
n
(1+n
)
2
2
Let
s
n
=
1
+
2
+
3
+
...
n
a
=
1,
last
term
l
=
n
Sum
of
first
n
positive
integers
How many
2-digit numbers
are
divisible
by
3?
Fixed number
in
arithmetic
progression which provides
the
to
and
fro
terms
by
adding/
subtracting from the
present
number.
Can
be
positive
or
negative.
a,
a+d,
a+2d,
a+3d,
...
a+(n –1)
d
List
of
numbers
in
which each term
is
obtained by
adding a fixed
number
to
the
preceding
term
except
the
first
term.
Fixed
number
is
called
common
difference.
e
r
h
Te
𝑛
𝑆
𝑛
=
2
{2𝑎 +
(𝑛
−
1)𝑑}
𝑛
𝑆
𝑛
=
2
(𝑎
+
𝑎
𝑛
)
a- First
term
d-
common
difference
Explore the concept of arithmetic progressions (AP) in mathematics by understanding sequences, series, and how to identify and work with APs. Learn about common differences, nth terms, sum of terms, and practical applications of AP in daily life. Discover methods to verify if a given sequence is in AP and solve problems related to AP. Engage in activities to reinforce learning and develop problem-solving skills in mathematics.
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Presentation Transcript
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 1stterm 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 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 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 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 thenext 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.
EXAMPLE 1: Find the value of k for which the given series is in A.P. 4, k 1 ,12 Solution : Given A.P.is4, k 1 , 12 .. If series is A.P. then the differenceswill 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 ACTIVITY- -1 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 ACTIVITY- -1(continued..) 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 ACTIVITY- -1(continued..) 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 ACTIVITY- -1(continued..) 1(continued..)
ACTIVITY ACTIVITY- -1(continued..) 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 ACTIVITY- -1(continued..) 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 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 , ??= ? + ? 1 ?
Lets 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 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 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 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 ?17= 5 +17 1 2 = 5 + 32 = 37
The sum of an arithmetic series 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) + . . . + ( 2d) + ( d) + . 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 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 = + ( d) + ( 2d) + . . . + (a + 2d) + (a + d) + a 2Sn = (a + ) + (a + ) + (a + ) + . . . + (a + ) + (a + ) + (a + ), 2Sn = n(a + )
The sum of an arithmetic series The sum of an arithmetic series Sn = n(a + ). = a + (n 1)d , Sn = n(a + ) Sn = n(a + a + (n 1)d) Sn= n(2a + (n 1)d).
The sum of an arithmetic series 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 = n(2a + (n 1)d) S50 = 50 (2 1 + (50 1) 2) = 25 (2 + 49 2) = 25 (2 + 98) = 2500
The sum of an arithmetic series 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 =101 we know = a + (n 1)d 101 = 1 + (n 1) 2 5 . 100 = (n 1) 2 5 40 = n 1 n = 41 So, Sn = n(a+ )= 41(1+101)=2091
The sum of an arithmetic series 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= 8 (6 + 7d), S5= 5 (6 + 4d) S8 = 2S5, we see that 8 (6 + 7d) = 2 5 (6 + 4d) 4 (6 + 7d) = 5 (6 + 4d) 24 + 28d = 30 + 20d 8d = 6 d = .
Example 4: Find number of terms of A.P. 100, 105, 110, 115,, 500 Solution : First term is a = 100 , an =500 Common difference is d = 105-100 =5 nth term is an = a +(n-1)d 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] = x 30 [ 2x 12 + (30-1)x 8] = 15 [ 24 + 29 x8]
= 15[24 +232] = 15 x246 =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= 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 = n(a+l)
MIND MAP Sum of first n positiveintegers Let sn= 1 + 2 + 3 + ... n a = 1, last term l = n How many 2-digit numbersare divisible by 3? sn=n(a+l)=n(1+n) 2 2-digit numbers divisible by 3 12, 15, 18, ... 99 a = 12, d = 3, an= 99 an= a + (n 1)d 99 = 12 + (n 1)3 2 or sn=n(n+l) 2 sn=n(a+l) 2 n 1 =87= 29 i.e., List of numbers in which each term is obtained by adding a fixed number to the precedingtermexceptthe firstterm. Fixednumberiscalledcommon difference. 3 n = 30 er Arithmetic Progression a, a+d, a+2d, a+3d, ... a+(n 1) d If a, b, c, are in AP, b =a + c Fixed number in arithmetic progression which provides the toandfrotermsbyadding/ subtracting from the present number. Can be positive or negative. 2 hTe b is arithmeticmean From beginning an = a+(n 1)d Here a firstterm a- First term d- common difference When first term of common differnceis given : ? ??= 2 {2? + (? 1)?} Whenfirst& last terms aregiven: d co m on di ? ??= 2 (? + ??)
