Arithmetic Progression

A succession of Numbers which are arranged in a definite order according to a certain given rule is called as sequence. A sequence is a finite sequence or infinite according to number of terms in it is finite or infinite respectively. Sequences following certain pattern are called progressions.
 
By Arithmetic Progression, we Mean that it is a sequence in which each term except first term differs from its preceding term by a constant.
 
Constant difference is called as common difference of Arithmetic Progressions. To Define Arithmetic Progression, we say that it is the series of numbers which has first term as “a”, common difference “ d” and we define n th term by an. Arithmetic Progression Definition also tells us that n th term of an A.P. with first term 'a' and common difference 'd' is given as follows:
an = a + (n – 1) * d,
Arithmetic Progression Problems solving is always convenient to make different choices based on following :
i)              3 numbers in AP as (a – d), a, (a + d),
ii)            4 numbers in AP as (a – 3d), (a – d), (a + d), (a + 3d),
iii)           5 numbers in AP as (a – 2d), (a – d), a , (a + d), (a + 2d),
iv)           6 numbers in AP as ((a – 5d), (a – 3d), (a – d), (a + d), (a + 3d), (a + 5d).
 
 
Arithmetic Progression Proof of two theorems are as follows:

Theorem 1:
Show that the n th term of an AP with first term 'a' and common difference 'd' is given by:
an = a + (n – 1)* d
Proof : Let us consider an AP with first term 'a' and common difference 'd'. Then, AP is given by:
a, a + d, a + 2d, a + 3d, …….
In this AP we have:
First term, a₁ = a = a + 0 * d = a + (1 – 1) * d,
Second term, a₂ = (a + d) = a + (2 - 1) * d,
Third term, a₃ = (a + 2d) = a + (3 – 1) * d,
Thus we say that,
Nth term a_n = a + (n – 1) d,
So we say that a_n = a + (n – 1) d.

Theorem 2:
To show that n th term from end of an AP, with first term 'a', common difference 'd' and last term 'l' is given by [l – (n - 1) d],
Proof: If 'a' is first term, 'd' is common difference and 'l' is last term of given AP
Then the AP is given by: a, a + d, a + 2d, . . . . . , (l – 2d), (l – d), l.
So we say that last term will be = l = l – (1 - 1) d,
Second last term = (l – d) = l – (2 - 1) d,
Third last term = (l – 2d) = l – (3 – 1),
So nth last term will be = l – (n – 1)d.