The document defines an arithmetic progression (AP) as a sequence of numbers where each term is calculated by adding a constant value, called the common difference, to the preceding term. The general formula for calculating any term in an AP is: tn = a + (n - 1)d, where a is the first term, d is the common difference, and n is the term number. Several examples of AP sequences are provided to illustrate the definition and formula. Methods for calculating the nth term, sum of terms, and finding consecutive terms in an AP are also explained.

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B
C
A
SERIES
E
SERIES
F
D
SERIES
G
10,20,30,40,50, . . . A . P
1,5,9,13,17,21, . .A.P
1,4,9,16,25,36,. s e q u e n c e .
.
1,8,27,64,125, . . .
12,8,4,0,-4,-8, . . .
1,8,10,14,24,27, . . .
2 , 4 , 8,1 6 ,3 2,6 4,1 28 , . . s e quence
SEQUENCE ARITHMETIC P.
A set of numbers is
arranged in a definite
order according to any
rule is called sequence.

2. Definition
A. PROGRESSION
It is a list of numbers in which each term is
obtained by adding a fixed number to the
preceding term except the first term is as an
arithmetic progression .
This fixed number is called the common difference (d) of the AP.
Remember that ‘d’ can be positive, negative or zero.
9,11,13,15,17, . . . Common difference (d)= t2 – t1 = 2.

3. 8,12,16,20,24, . . .
7,17,27,37,47, . . .
20,16,10,6,1,-6, . . .
𝟏
𝟐
, 𝟏,
𝟑
𝟐
, 2,
𝟓
𝟐
, 3,
𝟕
𝟐
, . . .
𝑡1=a= 8, 𝑡2 =12, 𝑡3 =16,𝑡4 =20,
Consider the following sequences
= 4
d = 𝒕𝟐 − 𝒕𝟏 = 12 - 8
d = 𝒕𝟑 − 𝒕𝟐 = 16 - 12
= 4
d = 𝒕𝟒 − 𝒕𝟑 = 20 - 16
= 4
Common difference(d) is constant.
Given sequence is an A.P.
𝟏
𝟐
,
𝟏
𝟑
,
𝟏
𝟒
,
𝟏
𝟓
,
𝟏
𝟔
,
𝟏
𝟕
, . . .
ARITHMETIC PROGRESSION

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General formula of Arithmetic progression
𝒕𝒏 = 𝒂 + (𝒏 − 𝟏)𝒅
In an A.P. the difference between two consecutive terms is
constant and is denoted by d. and ‘a’ = t1 = first term of an A.P
In an A.P. if the first term is a, and common difference is d
Terms in the A.P. are a, (a + d), (a + 2d), (a + 3d), . . .
Or
Terms in the A.P. are a, t1 + d, t2 + d, t3 + d + . . .

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To find nth term of an A. P : keep value of ‘a’ and ‘d’
𝒕𝒏 = 𝒂 + (𝒏 − 𝟏)𝒅
e.g. find nth term of an A. P 5, 8, 11, 14, 17, . . . .
𝒕𝒏 = 𝒂 + (𝒏 − 𝟏)𝒅
𝒕𝒏 = 𝟓 + 𝒏 − 𝟏 𝟑
𝒂 = 𝟓, 𝒅 = 𝟑, 𝒕𝒏 = ?
𝒕𝒏 = 𝟓 + 𝟑𝒏 − 𝟑
𝒕𝒏 = 𝟐 + 𝟑𝒏
nth term of an A. P is 𝟐 + 𝟑𝒏

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Q. How many two digit numbers are divisible by 4 ?
Solution: List of two digit numbers divisible by 4 is 12, 16, 20, 24, . . . , 96.
Let’s find how many such numbers are there.
tn = 96, a = 12, d = 4
From this we willfindthe value of n. tn= 96,
By formula, 𝒕𝒏 = 𝒂 + (𝒏−𝟏)𝒅
96 = 12 + (n - 1) 4
96 = 12 + 4n - 4
96 = 8 + 4n
96 - 8 = 4n
4n = 88
n = 88/4
n = 22
There are 22 two digit numbers are divisible by 4.

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General formula ( General term ) of Arithmetic
progression 𝒕𝒏 = 𝒂 +
(𝒏 − 𝟏)𝒅
All terms in an A.P. will be as follows if first term is ‘a’ and
common difference ‘d’
a, (a + d), (a + 2d), (a + 3d), (a + 4d), . . . , 𝒂 + (𝒏 − 𝟑)𝒅,
𝒂 + (𝒏 − 𝟐)𝒅, 𝒂 + (𝒏 − 𝟏)𝒅,
Sum of all terms of an A.P. if we know first term, last term and
total number of terms.
Sn =
𝒏
𝟐
( t1 + tn ), t1 = a first term, tn = last term, n = total number of terms

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Sum of all terms of an A.P. if we know first term, last
term and common difference.
Three consecutive terms of an A.P. are as follows
Sn =
𝒏
𝟐
[2a + (n-1) d ] t1 = a = first term, tn = last term, d = common difference
(a – d ), a, (a + d )
(a - 3d ), (a – d ), (a + d ), (a + 3d )
Four consecutiveterms of anA.P. are as follows
(a - 2d ), (a - d ), a, (a + d ), (a + 2d )
Five consecutivetermsof anA.P. are as follows
Eg. In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms.
(Assume that three consecutive terms in A.P. are a - d , a, a + d.)
eg. Find four consecutive terms in an A.P. whose
sum is 12 and sum of 3rd and 4th term is 14.

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ARITHMETIC PROGRESSION
MATHS & SCIENCE TEACHER
Compiled by SANJAY J. GANGAN.M.Sc.,B.Ed.
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