Calculate the Sum S_n of the Terms of the Arithmetic Sequence, the Series Sum, Given the First Term of the Series -1/2, the Common Difference -1/2 and the Last Term -20.5

Have a look at this example of an arithmetic sequence, consisting of six terms, similar to the one entered on this page. Starting with the first term:
- ¹/₂, - 1, - ³/₂, - 2, - ⁵/₂, - 3.

An arithmetic sequence is an ordered set of numbers (in ascending or descending order) in which the difference between any two consecutive terms is the same. This difference is called the common difference of the arithmetic sequence. The ordered numbers are called the terms of the sequence.

The common difference of the sequence is calculated by subtracting any two consecutive terms, a_k - a_k-1:
- 3 - (- ⁵/₂) = - ⁵/₂ - (- 2) = - 2 - (- ³/₂) = - ³/₂ - (- 1) = - 1 - (- ¹/₂) = - ¹/₂

When the terms of the ordered set of numbers are written as a comma separated string, we call this an arithmetic sequence.

When the same terms are written as a summation, then we call this an arithmetic series.

If we call a = the first term of an arithmetic sequence, d = the common difference, n = the number of terms in the sequence, then the terms of the arithmetic sequence will be:

The first term: a
In our progression this term is: - ¹/₂

The second term:
a₂ = a + d

The third term:
a₃ = a₂ + d = a + d + d = a + 2 × d

The fourth term:
a₄ = a₃ + d = a + 2 × d + d = a + 3 × d

...

The k^th term:
a_k = a_k-1 + d = a + (k - 2) × d + d = a + (k - 1) × d

...

The n^th term, the last one:
a_n = a_n-1 + d = a + (n - 2) × d + d = a + (n - 1) × d
In our progression this term is: - 20.5

We write the arithmetic series in two ways.

First we write it from the term a to the term a_n, then we write it "inverted", from the term a_n to a:

S_n = a + a₂ + a₃ + ... + a_(n-2) + a_(n-1) + a_n

S_n = a_n + a_(n-1) + a_(n-2) + ... + a₃ + a₂ + a

Then we add up the two identical series S_n, which were written differently:

S_n + S_n = (a + a_n) + (a₂ + a_(n-1)) + (a₃ + a_(n-2)) + ... + (a_(n-2) + a₃) + (a_(n-1) + a₂) + (a_n + a) =>

2 × S_n = (a + a + (n - 1) × d) + (a + d + a + (n - 2) × d) + (a + 2 × d + a + (n - 3) × d) + ... + (a + (n - 3) × d + a + 2 × d) + (a + (n - 2) × d + a + d) + (a + (n - 1) × d + a) =>

2 × S_n = (2 × a + (n - 1) × d) + (2 × a + (n - 1) × d) + (2 × a + (n - 1) × d) + ... + (2 × a + (n - 1) × d) + (2 × a + (n - 1) × d) + (2 × a + (n - 1) × d) =>

2 × S_n = n × (2 × a + (n - 1) × d) =>

S_n = ⁿ/₂ × (2 × a + (n - 1) × d)

S_n = ⁿ/₂ × (2 × a + (n - 1) × d)

n = the number of terms in the arithmetic series

a = the first term of the arithmetic series

d = the common difference of the arithmetic series

a_n = the last term of the arithmetic series

Furthermore, knowing that a_n = a + (n - 1) × d

We can also write the sum S_n in the form:

S_n = ⁿ/₂ × (a + a_n)

Use one of the two formulas:

Formula no. 1:
S_n = ⁿ/₂ × (a + a_n)

Formula no. 2:
S_n = ⁿ/₂ × (2 × a + (n - 1) × d)