arithmetic progressionArithmeticProgression2003-05-26 15:20:402004-09-24 23:49:51Definition\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}\PMlinkescapeword{columns}
Arithmetic progression of length $n$, initial term $a_1$ and common difference $d$ is the sequence $a_1, a_1+d,a_1+2d,\dotsc,a_1+(n-1)d$.
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The sum of terms of an arithmetic progression can be computed using Gauss's trick:
\parbox{\linewidth}{\begin{align*}
S&=\makebox[7em]{$(a_1+0)$}+\makebox[7em]{$(a_1+d)$}+\dotsb+\makebox[7em]{$(a_1+(n-2)d)$}
+\makebox[7em]{$(a_1+(n-1)d)$}\\
+\underline{S\vphantom{\makebox[7em]{$(a_1+(n-1)d)$}}}&\underline{{}=\makebox[7em]{$(a_1+(n-1)d)$}+
\makebox[7em]{$(a_1+(n-2)d)$}+\dotsb+\makebox[7em]{$(a_1+d)$}
+\makebox[7em]{$(a_1+0)$}}\\
2S&=\makebox[7em]{$(2a_1+(n-1)d)$}+\makebox[7em]{$(2a_1+(n-1)d)$}+\dotsb+\makebox[7em]{$(2a_1+(n-1)d)$}+
\makebox[7em]{$(2a_1+(n-1)d)$}.
\end{align*}}
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
\begin{equation*}
S=\frac{(2a_1+(n-1)d)n}{2}.
\end{equation*}