An arithmetic progression or arithmetic sequence (AP) is a sequence of numbers such that the difference from any succeeding term to its preceding term remains constant throughout the sequence. The constant difference is called common difference of that arithmetic progression. For instance, the sequence 5, 7, 9, 11, 13, 15, . . . is an arithmetic progression with a common difference of 2.

If the initial term of an arithmetic progression is ${\displaystyle a_{1}}$ and the common difference of successive members is ${\displaystyle d}$, then the ${\displaystyle n}$-th term of the sequence (${\displaystyle a_{n}}$) is given by:

${\displaystyle a_{n}=a_{1}+(n-1)d}$

A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. The sum of a finite arithmetic progression is called an arithmetic series.

History

According to an anecdote of uncertain reliability,^[1] young Carl Friedrich Gauss, who was in primary school, reinvented the formula ${\displaystyle {\tfrac {n(n+1)}{2}}}$ for summing the integers from 1 through ${\displaystyle n}$, for the case ${\displaystyle n=100}$, by grouping the numbers from both ends of the sequence into pairs summing to 101 and multiplying by the number of pairs. However, regardless of the truth of this story, Gauss was not the first to discover this formula, and some find it likely that its origin goes back to the Pythagoreans in the 5th century BC.^[2] Similar rules were known in antiquity to Archimedes, Hypsicles and Diophantus;^[3] in China to Zhang Qiujian; in India to Aryabhata, Brahmagupta and Bhaskara II;^[4] and in medieval Europe to Alcuin,^[5] Dicuil,^[6] Fibonacci,^[7] Sacrobosco^[8] and to anonymous commentators of Talmud known as Tosafists.^[9]

Sum

The sum of the members of a finite arithmetic progression is called an arithmetic series. For example, consider the sum:

${\displaystyle 2+5+8+11+14=40}$

This sum can be found quickly by taking the number n of terms being added (here 5), multiplying by the sum of the first and last number in the progression (here 2 + 14 = 16), and dividing by 2:

${\displaystyle {\frac {n(a_{1}+a_{n})}{2}}}$

In the case above, this gives the equation:

${\displaystyle 2+5+8+11+14={\frac {5(2+14)}{2}}={\frac {5\times 16}{2}}=40.}$

This formula works for any real numbers ${\displaystyle a_{1}}$ and ${\displaystyle a_{n}}$. For example: this

${\displaystyle \left(-{\frac {3}{2}}\right)+\left(-{\frac {1}{2}}\right)+{\frac {1}{2}}={\frac {3\left(-{\frac {3}{2}}+{\frac {1}{2}}\right)}{2}}=-{\frac {3}{2}}.}$

Derivation

To derive the above formula, begin by expressing the arithmetic series in two different ways:

${\displaystyle S_{n}=a+a_{2}+a_{3}+\dots +a_{(n-1)}+a_{n}}$

${\displaystyle S_{n}=a+(a+d)+(a+2d)+\dots +(a+(n-2)d)+(a+(n-1)d).}$

Rewriting the terms in reverse order:

${\displaystyle S_{n}=(a+(n-1)d)+(a+(n-2)d)+\dots +(a+2d)+(a+d)+a.}$

Adding the corresponding terms of both sides of the two equations and halving both sides:

${\displaystyle S_{n}={\frac {n}{2}}.}$

This formula can be simplified as:

${\displaystyle {\begin{aligned}S_{n}&={\frac {n}{2}}.\\&={\frac {n}{2}}(a+a_{n}).\\&={\frac {n}{2}}({\text{initial term}}+{\text{last term}}).\end{aligned}}}$

Furthermore, the mean value of the series can be calculated via: ${\displaystyle S_n/n}$Zdroj:https://en.wikipedia.org?pojem=Arithmetic_progression
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