Sum of the terms of an Arithmetic Progression

One arithmetic progression (PA) is a sequence numerical in which each term is the sum of the previous one by a constant, called the ratio. They exist mathematical expressions to determine the term of a PA and to calculate the sum of its no first terms.

The formula used to calculate the sum of terms of a finite PA or the sum of the no first terms of a PA is as follows:

s_no = at₁ + the_no)
2

*n is the number of BP terms; The₁ is the first term, and the_no is the last.

Origin of the sum of the terms of the PA

It is said that the German mathematician Carl Friederich Gauss, at approximately 10 years of age, was punished with his class at school. The teacher told the students to add up all the numbers that appear in the sequence from 1 to 100.

Gauss was not only the first to finish in a very short period of time, he was also the only one to get the result right (5050). Besides, he didn't show any calculations. What he did was repair the following property:

The sum of two terms equidistant from the extremes of a finite PA is equal to the sum of the extremes.

There was no knowledge about PAN at the time, but Gauss viewed the list of numbers and realized that adding the first to the last would result in 101; adding the second to the penultimate, the result would also be 101 and so on. As the sum of all pairs of terms equidistant of the extremes came to 101, Gauss only had to multiply that number by half the available terms to find the 5050 result.

Note that from the number 1 to the number 100, there are exactly 100 numbers. Gauss realized that if he added them two by two, he would get 50 results equal to 101. Therefore, this multiplication was done by half of the total terms.

Demonstration of the sum of terms of a PA

This feat gave rise to the expression used to calculate the sum of no first terms of a PA. The tactic used to arrive at this expression is as follows:

given one PAN any, we'll add the first n terms of it. Mathematically, we will have:

s_no = the₁ + the₂ + the₃ + … + the_{n – 2} + the_{n - 1} + the_no

Just below this sum of terms, we will write another one, with the same terms as the previous one, but in a decreasing sense. Note that the sum of terms in the first is equal to the sum of terms in the second. Therefore, both were equated with S_no.

s_no = the₁ + the₂ + the₃ + … + the_{n – 2} + the_{n - 1} + the_no

s_no = the_no + the_{n - 1} + the_{n – 2} + … + the₃ + the₂ + the₁

Note that these two expressions were obtained from a single PAN and that the equidistant terms are aligned vertically. Therefore, we can add the expressions to obtain:

s_no = the₁ + the₂ + the₃ + … + the_{n – 2} + the_{n - 1} + the_no

+ s_no = the_no + the_{n - 1} + the_{n – 2} + … + the₃ + the₂ + the₁

2S_no = (the₁ + the_no) + (a₂ + the_{n - 1}) + … + (a_{n - 1} + the₂) + (a_no + the₁)

Remember that the sum of terms equidistant from the extremes is equal to the sum of the extremes. Therefore, each parenthesis can be replaced by the sum of the extremes, as we will do next:

2S_no = (the₁ + the_no) + (a₁ + the_no) +... + (the₁ + the_no) + (a₁ + the_no)

Gauss' idea was to add the equidistant terms of a sequence. So he got half the amount of terms from PAN in results 101. We made it so that each term of the initial BP was added to its equidistant value, preserving its number of terms. Thus, as the PA had n terms, we can change the sum, in the expression above, by a multiplication and solve the equation to find:

2S_no = (the₁ + the_no) + (a₁ + the_no) +... + (the₁ + the_no) + (a₁ + the_no)

2S_no = n (a₁ + the_no)

s_no = at₁ + the_no)
2

This is exactly the formula used to add the no first terms of a PA.

Example

Given P.A (1, 2, 3, 4), determine the sum of its first 100 terms.

Solution:

We will need to find the term a₁₀₀. For this, we will use the general term formula of a PA:

The_no = the₁ + (n – 1)r

The₁₀₀ = 1 + (100 – 1)1

The₁₀₀ = 1 + 99

The₁₀₀ = 100

Now the formula for summing the first n terms:

s_no = at₁ + the_no)
2

s₁₀₀ = 100(1 + 100)
2

s₁₀₀ = 100(101)
2

s₁₀₀ = 10100
2

s₁₀₀ = 5050

By Luiz Paulo Moreira
Graduated in Mathematics

Source: Brazil School - https://brasilescola.uol.com.br/matematica/soma-dos-termos-uma-progressao-aritmetica.htm