How to Identify an Arithmetic Progression (Step-by-Step Examples)

NCERT–Class-10^th –Exercise-5.1– Arithmetic Progressions, Arithmetic Progression Formula Sheet for Quick Revision, Understanding the Arithmetic Sequence Formula (a, d, n, Tn), Common Difference in AP: How to Find It (with Examples, Arithmetic Progression Class 10,

(A) What is a pattern or progression?
(a)Meanings of pattern and progression are same in mathematics.
(b) Pattern or progression is basically an arrangement of more than two numbers such that each and every element is connected to each other by a predefined rule.
(c)Example:     
2, 4, 6, 8,,,,,,,,,  etc.
This is an example of a pattern in which every next element is 2 more than its previous element.
(i) 4 = 2 + 2
(ii) 6 = 4 + 2
(iii) 8 = 6 +2 and so on.


(B) Types of progression/ how many types are there of progression or pattern?/ Describe different types of pattern or progression./ Write down different types of patterns.
There are several types of pattern or progression in the field of mathematics but three patterns are important:
(1) Arithmetic Progressions (A.P.)
(2) Geometric Progression (G.P.)
(3) Harmonic Progression (H.P.)

Out of three important patterns, only Arithmetic Progressions have been dealt in class 10 which is going to be discussed below.

(C) What are consecutive terms?
(a) It may be realized something different. But the concept of consecutive terms going to be discussed below will be very useful in relation to the arithmetic progression.
(b) Consecutive terms are basically terms being neighbor to each other.
(c) Example:  2, 4, 6, 8,,,,,,,,,,,,,  etc
(i) In the above example, 2 is the neighbor term of 4. Hence 2 and4 are consecutive terms.
(ii) 4 is the neighbor term of 2 and 6. Hence (2 and 4) and (4 and 6) are consecutive terms.
(iii) In the same way, 8 will form pair of consecutive terms with 6 and 10 separately.



(D) What is an A.P. or Arithmetic Progression?/ Define AP (arithmetic progression)/ writhe the definition of an AP (arithmetic progression).
(1) Arithmetic progression is nothing but it is a group of numbers in which each next term is obtained by adding a fixed number to the just previous term.
(2) This fixed number is called common difference.
(3) Common difference can be positive, negative or zero.



(E) Mathematical Derivation for the General Form of an A.P./ Expression of n^th term./Formula of n^th term./ General form of n^th term of an AP (arithmetic progression)

Let us consider the 1^st term and common difference of an A.P. as a and d respectively.
Therefore, from the definition of an AP (arithmetic progression)
(i) 1^st term
= a
= a + 0 (1^st term + common difference)
= a + 0×d
= a + (1–1)×d

(ii) 2^nd term
= 1^st term + common difference
= a + d
= a + 1×d
= a + (2–1)×d

(iii) 3^rd term
= 2^nd term + common difference
= (a+d) + d
= a + d + d
= a + 2d
= a + (3–1)×d

(iv) 4^th term
= 3^rd term + common difference
= (a+2d) + d
= a + 2d + d
= a + 3d
= a+(4–1)×d

Therefore, we can say that
Term of specific position = 1^st term + (position –1)× common difference

In this way, ift_n is then^th term.
t_n  
= n^th term
= 1^st term + (position –1)× common difference
= a + (n–1)× d

Hence the general form of an A.P. is a, (a+d), (a+2d), (a+3d),  (a+4d) ,,,,,,,,,,,,,,,, {a+(n–1)× d} where
t_n = n^th term = a + (n–1)× d

N.B.(Note Book)
(a) The 1^st term is present in every term.
(b) The common difference in a given term is multiplied by a number which is equal to the 1 less than the position of that specific term.


(F) Sum of n terms of an AP (arithmetic progression)
Let us consider an A.P. a, (a+d), (a+2d), (a+3d),  (a+4d) ,,,,,,,,,, in which
The 1^st term = a = t₁
Common difference = d
n^th term = t_n = a+(n–1)×d

Also, let sum of n terms of the mentioned A.P. = S_n
Therefore,
S_n = a+(a+d) + ———+{a+(n–2)×d} + {a+(n–1)× d} ———————— equation (1)

Now we are going to write terms in the reversed order. Then,
S_n = {a+(n–1)× d} +{a+(n–2)×d} + ———+ (a+d)+a ———————— equation (2)

On adding equation (1) and (2),
S_n + S_n = [a+{a+(n–1)×d}]+[(a+d)+{a+(n–2)×d}]+————-+[{a+(n–2)×d}+(a+d)]+[{a+(n–1)×d}+a]
or,  2×S_n = {2a+(n–1)×d} + {2a+(n–2)×d+d} +————-+{2a+(n–2)×d+d} + {2a+(n–1)×d}
or,  2×S_n = {2a+(n–1)×d}+{2a+(n–2+1)×d}+———————+{2a+(n–2+1)×d}+{2a+(n–1)×d}
or,  2×S_n = {2a+(n–1)×d}+{2a+(n–1)×d}+————————-+{2a+(n–1)×d}+{2a+(n–1)×d}
or,  2×S_n = {2a+(n–1)×d}+———- up to n terms
or,  2×S_n = {2a+(n–1)×d}×n
or,  2×S_n = n×{2a+(n–1)×d}
or,   S_n = (n/2)×{2a+(n–1)×d}


This formula can also be reduced in the below manner:
S_n = (n/2)×{2a+(n–1)×d}
or,   S_n = (n/2)×{a+a+(n–1)×d}
or,   S_n = (n/2)×{a+t_n} [Because, t_n= a+(n–1)×d]
or,   S_n = (n/2)×{1st term + nth term}
or,   S_n = [n×(1st term + nth term)]/2