arithmetic progression

Arithmetic progression of length $n$ , initial term $a_{1}$ and common difference $d$ is the sequence $a_{1},a_{1}+d,a_{1}+2d,\ldots,a_{1}+(n-1)d$ .

The sum of terms of an arithmetic progression can be computed using Gauss’s trick:

$\displaystyle S$

$\displaystyle=\makebox[70.0pt]{$(a_{1}+0)$}+\makebox[70.0pt]{$(a_{1}+d)$}+% \cdots+\makebox[70.0pt]{$(a_{1}+(n-2)d)$}+\makebox[70.0pt]{$(a_{1}+(n-1)d)$}$

$\displaystyle+\underline{S\vphantom{\makebox[70.0pt]{$(a_{1}+(n-1)d)$}}}$

$\displaystyle\underline{{}=\makebox[70.0pt]{$(a_{1}+(n-1)d)$}+\makebox[70.0pt]% {$(a_{1}+(n-2)d)$}+\cdots+\makebox[70.0pt]{$(a_{1}+d)$}+\makebox[70.0pt]{$(a_{% 1}+0)$}}$

$\displaystyle 2S$

$\displaystyle=\makebox[70.0pt]{$(2a_{1}+(n-1)d)$}+\makebox[70.0pt]{$(2a_{1}+(n% -1)d)$}+\cdots+\makebox[70.0pt]{$(2a_{1}+(n-1)d)$}+\makebox[70.0pt]{$(2a_{1}+(% n-1)d)$}.$

We just add the sum with itself written backwards, and the sum of each of the columns equals to $(2a_{1}+(n-1)d)$ . The sum is then

$S=\frac{(2a_{1}+(n-1)d)n}{2}.$

Title	arithmetic progression
Canonical name	ArithmeticProgression
Date of creation	2013-03-22 13:39:00
Last modified on	2013-03-22 13:39:00
Owner	bbukh (348)
Last modified by	bbukh (348)
Numerical id	10
Author	bbukh (348)
Entry type	Definition
Classification	msc 00A05
Classification	msc 11B25
Related topic	MulidimensionalArithmeticProgression
Related topic	SumOfKthPowersOfTheFirstNPositiveIntegers