Introduction to Arithmetic Series Formula

A sequence is a list of numbers. A series is created by adding terms in the sequence. This lesson assumes that you know something about arithmetic sequences, how to find the common difference and how to find an explicit formula. You may want to review the sequences or finding formulas. There are two ways to denote that you are adding terms in a sequence. One method is by using summation notation and another one method is by using subscript notation similar to how we write explicit forms of sequences.

General Formula for Arithmetic Series:

Arithmetic progression is also called arithmetic sequence, is a sequence that begins with an initial term a, and then each term is found by adding the common difference d.

General Form of arithmetic sequence is,
a, a + d, a + 2d, a + 3d + . . .

The recursive formula is,

an = an-1 + d.

Here, d -> common difference,

n -> number of terms.

The explicit or closed form of an arithmetic sequence is,

an = a1 + (n - 1) d.

Example Using Arithmetic Series Formula:

Ex 1: Find the sum of an = (`(1)/(2)` )n + 1, where nth term is 25.

Sol:  The 25th partial sum of this series is the sum of the first twenty-five terms. The first few terms of the sequence are:

a1 = `(1)/(2)` `xx` 1 + 1 = `(3)/(2)`

a2 = `(1)/(2)` `xx` 2 + 1 = 2

a3 = `(1)/(2)`  `xx` 3 + 1 = `(5)/(2)`

These terms have a common difference d = `(1)/(2)` , so this is indeed an arithmetic sequence. The last term in the partial sum should be,

a35 = a1 + (25 – 1)(d)

=> a35 = `(3)/(2)` + `24 xx (1)/(2)`

=> a35 = `(27)/(2)`

Then, plug the value of n in the formula, the 25th partial sum is:

`(n)/(2)` (a1 + an) =  ` (25)/(2) xx ((3)/(2) + (27)/(2))` = `(25)/(2) xx (30)/(2)` = `(375)/(2)`



Ex 2: Find the nth term of the arithmetic sequence 1, 4, 7…

Sol:    an = a1 +(n – 1) d

=> 1 + (n – 1) (3)

=> 1 + 3n – 3

=> 3n – 2.



Ex 3: The first term of an arithmetic sequence is 5 and the eighth term is 19. Find the 20th term.

Sol:    an = a1 +(n – 1) d

=> a5 = a1 + (8 – 1) d

=> 19 = 5 + 7d

=> 14 = 7d

=> d = 2

Therefore 20th term is,

=> a20 = 5 + (20 – 1) (7)

=> a20 = 5 + (19) (7)

=> a20 = 5 + 133

=> a20 = 138.