arithmetic progression

Consider the following set of numbers:

(i) 2, 5, 8, 11, …
(ii) 2, 4, 8, 16, …
(iii) 1, 5, 3, 7, …
(iv) 3, 12, 43, 50, …
Did you notice that the terms in (i) and (ii) are arranged in a specific order and according to a definite rule, that is,
(i) The terms are in increasing order and each succeeding term is obtained by adding 3 to the preceding term.
(ii) The terms are in increasing order and each succeeding term is obtained by multiplying the preceding term by 2.
The number set given in (iii) does not follow any order or rule and the number set in (iv) are in increasing order but does not follow any rule.

The set of numbers in (i) and (ii) are called sequence. A sequence is a set of numbers specified in a definite order by some assigned rule or law. Each element of the set is called a term. A sequence can be finite or infinite. A finite sequence is that which ends and has the last term. For example 3, 9, 81, 6561 is finite sequence. An infinite sequence is one that has no last term. For example, 30, 24, 18, 12, 6, 0, -6, -12, … We usually use '…' to denote that the sequence continues without bound.

A series is defined as the sum of the terms of a sequence. For example, the series for the sequence given in (i) and (ii) are
(i) 2 + 5 + 8 + 11 + …
(ii) 2 + 4 + 8 + 16 + …

Arithmetic Progression (A.P)

A sequence in which its term continually increase or decrease by the same number is called an arithmetic progression. The fixed number by which they increase or decrease is called the common difference of the arithmetic progression.
For example, 121,131,141,151
You can see the difference between consecutive terms is equal to 10(131-121, 141-131, 151-141), i.e. common difference = 10. So this set of numbers forms an arithmetic progression.
If the first term of an arithmetic progression is 'a' and the common difference is 'd'. The sequence forming an A.P is:
a, (a + d), (a + 2d), …

Finding n^th term of an arithmetic progression
T_n be the n^th term of an arithmetic progression then

Tn = a + (n −1)d

here a is the first term, d is the common difference and n is the number of terms. Also, common difference d can be derived by using the below formula

d = Tn − Tn-1

Let us try solving a few questions

Question 1: Is the sequence 102, 120, 130, 148 forms an AP?
Solution: Difference between the terms: 120 -102 = 18, 130 -120 = 10, 148 - 130 = 18
as the difference between consecutive terms are not equal hence it does not forms an arithmetic progression

Question 2: Write first four terms of the arithmetic progression whose first term is 6 and common difference is 4.
Solution: As T₂ = 6 + (2 - 1)4 = 6 + 4 = 10
Therefore, the first four terms are 6, 10, 14, 18

Question 3: How many two digit numbers are divisible by 4?
Solution: Two digit numbers divisible by 4 are 12, 16, ..., 96
a = 12, d = 4, T_n = 96, n = ?
96 = 12 + (n - 1) 4
4n - 4 + 12 = 96
4n  + 8 = 96
4n = 88 ⇒ n = 22

Sum of n terms of an Arithmetic Progression

The sum S of the first n numbers of an arithmetic progression is given by the formula:

\(S =\frac{n}{2} (a + l)\)

OR

\(S = \frac{n}{2}(2a + (n-1)d)\)

Here S is the sum of first n numbers, d is the difference, a is the first term and l is the last term.

Question: Find the sum of the first 10 terms of the arithmetic progression 2, 5, 8, 11, …
\(S = \frac{10}{2}(2\times 2 + (10 - 1)3) \\ S = 5(4 + 27) \\ S = 155\)
Answer: The sum of the first 10 terms is 155