Arithmetic Progression (AP) Concepts
If various terms of a sequence are
formed by adding a fixed number to
the previous term or the difference
between two successive terms is a
fixed number, then the sequence is
called AP.
For example : 5, 10, 15, 20, 25…..
In this each term is obtained by adding
5 to the preceding term except first
term
The general form of an Arithmetic
Progression is
a , a +d , a + 2d , a + 3d ………………, a
+ (n-1)d
Where,
‘a’ is first term and
‘d’ is called common difference.
Common Difference
- The fixed
number which is obtained by
subtracting any term of AP from its
succeeding term.
If we take
first term of an AP
as
a
and
Common Difference
as
d
,
Then,
n
th
th
term of that AP will be
a
n
= a + (n-1)d
To check that a given term is in
A.P. or not.
2, 6, 10, 14….
Here first term a = 2,
Now, 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 differences are common.
Hence the given terms are in A.P.
Problem : 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.
a
2
– a
1
=
a
3
– a
2
k – 1 – 4 = 12 – (k – 1)
k – 5 = 12 – k + 1
k + k = 12 + 1 + 5
2 k = 18 or k = 9
Its formula is
S
U
M
O
F
n
T
E
R
M
S
O
F
A
N
A
R
I
T
H
M
E
T
I
C
P
R
O
G
R
E
S
S
I
O
N
It can also be written as
S
n
= ½ n [ 2a + (n - 1)d ]
S
n
= ½ n [ a + a
n
]
DERIVATION
T
h
e
s
u
m
t
o
n
t
e
r
m
s
i
s
g
i
v
e
n
b
y
:
S
n
=
a
+
(
a
+
d
)
+
(
a
+
2
d
)
+
…
+
[
a
+
(
n
–
1
)
d
]
(
1
)
I
f
w
e
w
r
i
t
e
t
h
i
s
o
u
t
b
a
c
k
w
a
r
d
s
,
w
e
g
e
t
:
S
n
=
[
a
+
(
n
–
1
)
d
]
+
(
a
+
(
n
–
2
)
d
)
+
…
+
a
(
2
)
N
o
w
l
e
t
’
s
a
d
d
(
1
)
a
n
d
(
2
)
:
2
S
n
=
[
2
a
+
(
n
–
1
)
d
]
+
[
2
a
+
(
n
–
1
)
d
]
+
…
…
…
…
+
[
2
a
+
(
n
–
1
)
d
]
S
o
,
S
n
=
½
n
[
2
a
+
(
n
–
1
)
d
]
P
r
o
b
l
e
m
.
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
S
o
l
u
t
i
o
n
.
F
i
r
s
t
t
e
r
m
i
s
a
=
1
0
0
,
a
n
=
5
0
0
C
o
m
m
o
n
d
i
f
f
e
r
e
n
c
e
i
s
d
=
1
0
5
-
1
0
0
=
5
n
t
h
t
e
r
m
i
s
a
n
=
a
+
(
n
-
1
)
d
5
0
0
=
1
0
0
+
(
n
-
1
)
5
5
0
0
-
1
0
0
=
5
(
n
–
1
)
4
0
0
=
5
(
n
–
1
)
5
(
n
–
1
)
=
4
0
0
5
(
n
–
1
)
=
4
0
0
n
–
1
=
4
0
0
/
5
n
-
1
=
8
0
n
=
8
0
+
1
n
=
8
1
H
e
n
c
e
t
h
e
n
o
.
o
f
t
e
r
m
s
i
n
t
h
e
A
P
a
r
e
8
1
.
P
r
o
b
l
e
m
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
…
…
…
S
o
l
u
t
i
o
n
:
G
i
v
e
n
A
.
P
.
i
s
1
2
,
2
0
,
2
8
,
3
6
I
t
s
f
i
r
s
t
t
e
r
m
i
s
a
=
1
2
C
o
m
m
o
n
d
i
f
f
e
r
e
n
c
e
i
s
d
=
2
0
–
1
2
=
8
T
h
e
s
u
m
t
o
n
t
e
r
m
s
o
f
a
n
A
P
S
n
=
½
n
[
2
a
+
(
n
-
1
)
d
]
=
½
x
3
0
[
2
x
1
2
+
(
3
0
-
1
)
x
8
]
=
1
5
[
2
4
+
2
9
x
8
]
=
1
5
[
2
4
+
2
3
2
]
=
1
5
x
2
4
6
=
3
6
9
0
THE SUM OF TERMS IS 3690
P
r
o
b
l
e
m
.
F
i
n
d
t
h
e
s
u
m
o
f
t
e
r
m
s
i
n
g
i
v
e
n
A
.
P
.
2
,
4
,
6
,
8
,
…
…
…
…
…
…
2
0
0
S
o
l
u
t
i
o
n
:
I
t
s
f
i
r
s
t
t
e
r
m
i
s
a
=
2
C
o
m
m
o
n
d
i
f
f
e
r
e
n
c
e
i
s
d
=
4
–
2
=
2
n
t
h
t
e
r
m
i
s
a
n
=
a
+
(
n
-
1
)
d
2
0
0
=
2
+
(
n
-
1
)
2
2
0
0
-
2
=
2
(
n
–
1
)
2
(
n
–
1
)
=
1
9
8
n
–
1
=
9
9
,
n
=
1
0
0
N
o
w
,
t
h
e
s
u
m
t
o
n
t
e
r
m
s
o
f
a
n
a
r
i
t
h
m
e
t
i
c
p
r
o
g
r
e
s
s
i
o
n
S
n
=
½
n
[
2
a
+
(
n
-
1
)
d
]
S
1
0
0
=
½
x
1
0
0
[
2
x
2
+
(
1
0
0
-
1
)
x
2
]
=
5
0
[
4
+
1
9
8
]
=
5
0
[
2
0
2
]
=
1
0
1
0
0
Arithmetic Progression (AP) is a sequence where each term is derived by adding a fixed number to the previous term. This concept is explored through common difference, identifying terms in an AP, finding value of unknowns, sum of terms formula, and derivations. Examples and problems demonstrate the application of AP concepts in determining the number of terms.
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Presentation Transcript
If various terms of a sequence are formed by adding a fixed number to the previous term or the difference between two successive terms is a fixed number, then the sequence is called AP. For example : 5, 10, 15, 20, 25 .. In this each term is obtained by adding 5 to the preceding term except first term
The general form of an Arithmetic Progression is a , a +d , a + 2d , a + 3d , a + (n-1)d Where, a is first term and d is called common difference.
Common Difference - The fixed number which is obtained by subtracting any term of AP from its succeeding term. If we take first term of an AP as a and Common Difference as d, Then, nth term of that AP will be an = a + (n-1)d
To check that a given term is in A.P. or not. 2, 6, 10, 14 . Here first term a = 2, Now, find differences in the next terms a2-a1 = 6 2 = 4 a3-a2 = 10 6 = 4 a4-a3 = 14 10 = 4 Since the differences are common. Hence the given terms are in A.P.
Problem : 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. a2 a1 =a3 a2 k 1 4 = 12 (k 1) k 5 = 12 k + 1 k + k = 12 + 1 + 5 2 k = 18 or k = 9
SUM OF n TERMS OF AN ARITHMETIC PROGRESSION Its formula is Sn = n [ 2a + (n - 1)d ] It can also be written as Sn = n [ a + an ]
DERIVATION The sum to n terms is given by: Sn= a + (a + d) + (a + 2d) + + [a + (n 1)d] (1) If we write this out backwards, we get: Sn = [a + (n 1)d] + (a + (n 2)d) + +a (2) Now let s add (1) and (2): 2Sn = [2a + (n 1)d] + [2a + (n 1)d] + + [2a + (n 1)d] So, Sn = n [2a + (n 1)d]
Problem . 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 terms in the AP are 81.
Problem 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 AP Sn = n [ 2a + (n - 1)d ] = x 30 [ 2 x 12 + (30-1)x 8]
= 15 [ 24 + 29 x 8] = 15[24 + 232] = 15 x 246 = 3690 THE SUM OF TERMS IS 3690
Problem . Find the sum of terms in given A.P. 2 , 4 , 6 , 8 , 200 Solution: Its first term is a = 2 Common difference is d = 4 2 = 2 nth term is an = a + (n-1)d 200 = 2 + (n-1)2 200 - 2 = 2(n 1) 2(n 1) = 198 n 1 = 99, n = 100
Now, the sum to n terms of an arithmetic progression Sn = n [ 2a + (n - 1)d ] S100 = x 100 [ 2x 2 + (100-1)x 2] = 50 [ 4 + 198] = 50[202] = 10100
