Arithmetic Progression for SSC Maths 2014, How to Prepare for SSC Maths, SSC Maths Preparation.

1. Arithmetic progression
Sequence
A sequence is a collection of numbers arranged in a definite order
according to some definite rule.
Each number in the sequence is called a term of the sequence.
The number in the nth position is called the nth term and is denoted by ‘t n’.
A sequence is usually denoted by t n  or t n and is read as sequence t n.
Sum of the first n terms of a sequence
If t n  is a sequence then we denote the sum of the first n terms of this
sequence by
Thus we see that
Sn  t1  t 2  t 3  ....  t n
S1  t1
S2  t1  t 2
S3  t1  t 2  t 3

  
Sn  t1  t 2  t 3    t n
From the above we have

2. S1  t1
S2  S1  t 2
S3  S2  t 3
Sn  Sn 1  t n
 t n  Sn  Sn 1
Thus if S n is given we can find nth term and therefore any term of the
sequence.
If the number of terms in a sequence is finite then it is called a finite
sequence, otherwise it is an infinite sequence.
e.g. i) 1,3,5,7…. is an infinite sequence.
ii) 4,8,12,….,400 is a finite sequence.
Progressions
A progression is a special type of sequence in which the relationship
between any two consecutive terms is the same.
There are three types of progressions namely Arithmetic Progression,
Geometric Progression and Harmonic Progression.

3. Arithmetic Progression (A.P.)
An arithmetic progression is a sequence in which the difference between
any two consecutive terms is always constant.
The first term t1 is usually denoted by ‘a’ and the common difference
is denoted by ‘d’. The value of d may be positive, negative or zero.
For an A.P.
t 2= t 1 + d = a + d
t 3 = t 2 + d = a + d + d = a + 2d
t 4 = t 3 + d = a + 2d + d = a +3d
In general, t n – t n –1 = constant = d
and so on…….

4. General Term or nth Term of an A.P.
Derivation of the formula for the general term:
Let ‘a’ be the first term and ‘d’ be the common difference of an A.P.
Here,
…(1)
t1=a
t2–t1=d
…(2)
t3–t2=d
…(3)
::::::::::::::::::::::::::::::::::::::::::::::
t n –1 – t n – 2
t n – t n–1= d
…(n –1)
… (n)
Adding the above equations, we get,
t 1 + ( t 2 – t 1 ) + ( t 3 – t 2) + …+ ( t n – t n – 1) = a + d + d +….+ d (n – 1)
times
Therefore, t n = a + (n – 1) d.
Thus, the general term, i.e. nth term of an A.P. with the first term ‘a’ and the
common difference ‘d’ is given as
t n = a + (n – 1) d.

5. The sum of the first n terms of an A.P.
Derivation of the formula
Consider an A.P
a, a + d, a + 2d,…, a (n – 1)d,
where ‘a’ is the first term and ‘d’ is the common difference.
The sum of the first n terms of an A.P. is
Sn  a  a  d  ....  a  (n  2)d  a  (n  1)d ...(1)
On reversing the terms and rewriting the expression, we get,
Sn  a  (n  1)d  a  (n  2)d  ...  a  d   a
...(2)
Adding equations (1) and (2),
2Sn  2a  (n  1)d  2a  (n  1)d  ....  2a  (n  1)d  (n times)
2Sn   2a  (n  1) d 
Sn 
n
2a  (n  1) d
2
The sum of the first n terms of an A.P. whose first term is ‘a’ and the
common difference ‘d’ is
Sn 
Also,
n
 2a  (n  1) d 
2

6. n
a  a  (n  1)d
2
n
Sn   t1  t n ...where a  t1 and a  (n  1)d  t n
2
If we take t n  last term  , then
n
Sn  a  
2
Sn 
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