# arithmetic series

An *arithmetic series ^{}* is the series, ${\sum}_{i=1}^{n}{a}_{i}$, in which each real term has the form ${a}_{i}={a}_{i-1}+d$ for $i=2,\mathrm{\dots},n$ where $d$ is constant. The sum of the sequence is given by the following
$\frac{1}{2}}n[2{a}_{1}+d(n-1)].$
In order to find the formula above firstly we express the terms of the sequence, ${a}_{2},\mathrm{\dots},{a}_{n}$ in terms of ${a}_{1}$ and the constant $d$. In this case we get ${a}_{2}={a}_{1}+d,{a}_{3}={a}_{2}+2d,\mathrm{\dots},{a}_{n}={a}_{1}+(n-1)d$. Now we express the sum of the sequence by developing the series forward and we have:

$${S}_{n}=\sum _{i=1}^{n}{a}_{i}={a}_{1}+{a}_{1}+d+\mathrm{\cdots}+{a}_{1}+(n-2)d+{a}_{1}+(n-1)d$$ |

Reversely, we develop the series backwards and we get

$${S}_{n}={a}_{n}-d+{a}_{n}-2d+\mathrm{\cdots}+{a}_{n}-(n-1)d$$ |

It is easily seen that by adding the two expressions we get

$2{S}_{n}=n({a}_{1}+{a}_{n})$ | (1) | ||

${S}_{n}={\displaystyle \frac{1}{2}}n({a}_{1}+{a}_{n})$ | (2) |

Hence, by substituting ${a}_{n}={a}_{1}+(n-1)d$ we get the first formula.

Title | arithmetic series |
---|---|

Canonical name | ArithmeticSeries |

Date of creation | 2013-03-22 16:17:58 |

Last modified on | 2013-03-22 16:17:58 |

Owner | georgiosl (7242) |

Last modified by | georgiosl (7242) |

Numerical id | 10 |

Author | georgiosl (7242) |

Entry type | Definition |

Classification | msc 40A05 |