Arithmetic progression

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 arithmetic progression of real numbers beginning with ${\displaystyle a_1}$ and ending with ${\displaystyle a_n}$. For example,

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

Derivation

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

${\displaystyle S_n=a+a_2+a_3+\cdots +a_{(n-1)}+a_n}$

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

Rewriting the terms in reverse order:

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

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

${\displaystyle S_n=\frac{n}{2}[2a+(n-1)d].}$

This formula can be simplified as:

${\displaystyle \begin{array}{cc}{S}_n& =\frac{n}{2}[a+a+(n-1)d].\\ & =\frac{n}{2}(a+a_n).\\ & =\frac{n}{2}(\text{initial term}+\text{last term}).\end{array}}$

Furthermore, the mean value of the series can be calculated via: ${\displaystyle S_n/n}$:

${\displaystyle \overline{a}=\frac{a_1+a_n}{2}.}$

The formula is essentially the same as the formula for the mean of a discrete uniform distribution, interpreting the arithmetic progression as a set of equally probable outcomes.

Product

The product of the members of a finite arithmetic progression with an initial element a₁, common differences d, and n elements in total is determined in a closed expression

${\displaystyle \begin{array}{cc}{a}_1{a}_2{a}_3\cdots a_n& =a_1(a_1+d)(a_1+2d)\cdots (a_1+(n-1)d)\\ & =\prod _{k=0}^{n-1}(a_1+kd)=d^n\frac{\mathrm{\Gamma }\left(\frac{a_1}{d}+n\right)}{\mathrm{\Gamma }\left(\frac{a_1}{d}\right)}\end{array}}$

where ${\displaystyle \mathrm{\Gamma }}$ denotes the Gamma function. The formula is not valid when ${\displaystyle a_1/d}$ is negative or zero.

This is a generalization of the facts that the product of the progression ${\displaystyle 1\times 2\times \cdots \times n}$ is given by the factorial ${\displaystyle n!}$ and that the product

${\displaystyle m\times (m+1)\times (m+2)\times \cdots \times (n-2)\times (n-1)\times n}$

for positive integers ${\displaystyle m}$ and ${\displaystyle n}$ is given by

${\displaystyle \frac{n!}{(m-1)!}.}$

Derivation

${\displaystyle \begin{array}{cc}{a}_1{a}_2{a}_3\cdots a_n& =\prod _{k=0}^{n-1}(a_1+kd)\\ & =\prod _{k=0}^{n-1}d\left(\frac{a_1}{d}+k\right)\\ & =d\left(\frac{a_1}{d}\right)d\left(\frac{a_1}{d}+1\right)d\left(\frac{a_1}{d}+2\right)\cdots d\left(\frac{a_1}{d}+(n-1)\right)\\ & =d^n\prod _{k=0}^{n-1}\left(\frac{a_1}{d}+k\right)=d^n{\left(\frac{a_1}{d}\right)}^{\overline{n}}\end{array}}$

where ${\displaystyle x^{\overline{n}}}$ denotes the rising factorial.

By the recurrence formula ${\displaystyle \mathrm{\Gamma }(z+1)=z\mathrm{\Gamma }(z)}$, valid for a complex number ${\displaystyle z>0}$,

${\displaystyle \mathrm{\Gamma }(z+2)=(z+1)\mathrm{\Gamma }(z+1)=(z+1)z\mathrm{\Gamma }(z)}$,

${\displaystyle \mathrm{\Gamma }(z+3)=(z+2)\mathrm{\Gamma }(z+2)=(z+2)(z+1)z\mathrm{\Gamma }(z)}$,

so that

${\displaystyle \frac{\mathrm{\Gamma }(z+m)}{\mathrm{\Gamma }(z)}=\prod _{k=0}^{m-1}(z+k)}$

for ${\displaystyle m}$ a positive integer and ${\displaystyle z}$ a positive complex number.

Thus, if ${\displaystyle a_1/d>0}$,

${\displaystyle \prod _{k=0}^{n-1}\left(\frac{a_1}{d}+k\right)=\frac{\mathrm{\Gamma }\left(\frac{a_1}{d}+n\right)}{\mathrm{\Gamma }\left(\frac{a_1}{d}\right)},}$

and, finally,

${\displaystyle a_1{a}_2{a}_3\cdots a_n=d^n\prod _{k=0}^{n-1}\left(\frac{a_1}{d}+k\right)=d^n\frac{\mathrm{\Gamma }\left(\frac{a_1}{d}+n\right)}{\mathrm{\Gamma }\left(\frac{a_1}{d}\right)}}$

Examples

Example 1

Taking the example ${\displaystyle 3,8,13,18,23,28,\dots }$, the product of the terms of the arithmetic progression given by ${\displaystyle a_n=3+5(n-1)}$ up to the 50th term is

${\displaystyle P_{50}=5^{50}\cdot \frac{\mathrm{\Gamma }\left(3/5+50\right)}{\mathrm{\Gamma }\left(3/5\right)}\approx 3.78438\times {10}^{98}.}$

Example 2

The product of the first 10 odd numbers ${\displaystyle (1,3,5,7,9,11,13,15,17,19)}$ is given by

${\displaystyle 1\cdot 3\cdot 5\cdots 19=\prod _{k=0}^9(1+2k)=2^{10}\cdot \frac{\mathrm{\Gamma }\left(\frac{1}{2}+10\right)}{\mathrm{\Gamma }\left(\frac{1}{2}\right)}}$ = 654,729,075

Standard deviation

The standard deviation of any arithmetic progression is

${\displaystyle \sigma =|d|\sqrt{\frac{(n-1)(n+1)}{12}}}$

where ${\displaystyle n}$ is the number of terms in the progression and ${\displaystyle d}$ is the common difference between terms. The formula is essentially the same as the formula for the standard deviation of a discrete uniform distribution, interpreting the arithmetic progression as a set of equally probable outcomes.

Intersections

The intersection of any two doubly infinite arithmetic progressions is either empty or another arithmetic progression, which can be found using the Chinese remainder theorem. If each pair of progressions in a family of doubly infinite arithmetic progressions have a non-empty intersection, then there exists a number common to all of them; that is, infinite arithmetic progressions form a Helly family. However, the intersection of infinitely many infinite arithmetic progressions might be a single number rather than itself being an infinite progression.

Amount of arithmetic subsets of length k of the set {1,...,n}

Let ${\displaystyle a(n,k)}$ denote the number of arithmetic subsets of length ${\displaystyle k}$ one can make from the set ${\displaystyle \{1,\cdots ,n\}}$ and let ${\displaystyle \varphi (\eta ,\kappa )}$ be defined as:

${\displaystyle \varphi (\eta ,\kappa )=\{\begin{array}{ll}0& \text{if }\kappa \mid \eta \\ \left(\left[\eta (\text{mod }\kappa )\right]-2\right)\left(\kappa -\left[\eta (\text{mod }\kappa )\right]\right)& \text{if }\kappa \nmid \eta \end{array}}$

Then:

${\displaystyle \begin{array}{cc}a(n,k)& =\frac{1}{2(k-1)}\left(n^2-(k-1)n+(k-2)+\varphi (n+1,k-1)\right)\\ & =\frac{1}{2(k-1)}\left((n-1)(n-(k-2))+\varphi (n+1,k-1)\right)\end{array}}$

As an example, if $(n,k)=(7,3)$, one expects $a(7,3)=9$ arithmetic subsets and, counting directly, one sees that there are 9; these are $\{1,2,3\},\{2,3,4\},\{3,4,5\},\{4,5,6\},\{5,6,7\},\{1,3,5\},\{3,5,7\},\{2,4,6\},\{1,4,7\}.$