An arithmetic series is the sum of all the sequence where { a_{k} } where k = 1, 2, … and in which every term is evaluated from the previous one by addition or subtraction of the constant of d. Thus, for k > 1,

a_{k} = a_{k-1} + d = a_{k-2} + 2d = … = a_{1} + d (k – 1).

The sum of the arithmetic series of the first n terms is given by –

S = ∑ a_{k} from k = 1 to n

= ∑ [a_{k} + (k – 1) d] from k = 1 to n

= n a_{1} + d ∑ (k – 1) from k = 1 to n

= n a_{1} + d ∑ (k – 1) from k = 2 to n

= n a_{1} + d ∑ k from k = 1 to n

Using the sum identity –

∑ k from k = 1 to n = 1 / 2 n (n + 1)

Then it gives,

S_{n} = n a_{1} + 1 / 2 d n (n – 1) = 1 /2 n [2a_{1} + d (n – 1)]

Now note that,

a_{1} + a_{n} = a_{1} + [a_{1} + d (n – 1)] = 2 a_{1} + d(n – 1), so,

S_{n} = 1 /2 n (a_{1} + a_{n})

Or n times of the arithmetic means the first and the last terms. This is just the strategy used for the school children for solving the problem of summing the integers from around 1 to 100 whilst their peers toiled away by doing it in long addition. Guass only wrote one number and got the correct answer – let us see how:

1 / 2 (100) (1 + 100) = 50 * 101 = 5050

When the answers were checked, Guass was proved right with his answer. Thereby, the arithmetic series was introduced by him.

## Arithmetic series

The series like 3 + 7 + 11 + 15 + … + 99 or 10 + 20 +30 + … + 100 has a constant difference between all the terms. The first term is defined by a1, the common difference between terms is defined as d, and the number of terms in the series is defined as n. The sum of the arithmetic series is solved by just multiplying the number of terms times the average of both the first and the last term.

Let us see the formula first for the arithmetic series –

Formula => Sum = n (a_{1} + a_{n} / 2) OR n / 2 [2a_{1} + (n – 1) d]

Now we will solve some of the examples of arithmetic series –

**Example – 1: 3 +7 +11 +15 + … + 99 which has a _{1} = 3 and d = 4. Find out n for the arithmetic sequence using the explicit formula.**

Solution – we will solve the 3 + (n – 1) * 4 = 99 for getting n = 25

Sum = 25 * (3 + 99 / 2) = 1275

Or sum = 25 / 2 [ 2 * 3 + (25 – 1) * 4 = 1275