An arithmetic progression or arithmetic sequence 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
a1
d
n
an
an=a1+(n-1)d.
According to an anecdote of uncertain reliability,[1] in primary school Carl Friedrich Gauss reinvented the formula
\tfrac{n(n+1)}{2}
n
n=100
| 2 | + | 5 | + | 8 | + | 11 | + | 14 | = | 40 | |
| 14 | + | 11 | + | 8 | + | 5 | + | 2 | = | 40 | |
| 16 | + | 16 | + | 16 | + | 16 | + | 16 | = | 80 | |
Computation of the sum 2 + 5 + 8 + 11 + 14. When the sequence is reversed and added to itself term by term, the resulting sequence has a single repeated value in it, equal to the sum of the first and last numbers (2 + 14 = 16). Thus 16 × 5 = 80 is twice the sum.
The sum of the members of a finite arithmetic progression is called an arithmetic series. For example, consider the sum:
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:
| n(a1+an) | |
| 2 |
In the case above, this gives the equation:
2+5+8+11+14=
| 5(2+14) | |
| 2 |
=
| 5 x 16 | |
| 2 |
=40.
This formula works for any arithmetic progression of real numbers beginning with
a1
an
| \left(- | 3 |
| 2 |
\right)+\left(-
| 1 | |
| 2 |
\right)+
| 1 | |
| 2 |
=
| ||||||||
| 2 |
=-
| 3 | |
| 2 |
.
To derive the above formula, begin by expressing the arithmetic series in two different ways:
Sn=a+a2+a3+...+a(n-1)+an
Sn=a+(a+d)+(a+2d)+...+(a+(n-2)d)+(a+(n-1)d).
Sn=(a+(n-1)d)+(a+(n-2)d)+...+(a+2d)+(a+d)+a.
Adding the corresponding terms of both sides of the two equations and halving both sides:
| S | ||||
|
[2a+(n-1)d].
\begin{align} Sn&=
| n | |
| 2 |
[a+a+(n-1)d].\\ &=
| n | |
| 2 |
| (a+a | ||||
|
(initialterm+lastterm). \end{align}
Sn/n
\overline{a}=
| a1+an | |
| 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.
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
a1a2a3 … an=a1(a1+d)(a1+2d)...(a1+(n-1)d)=
| n-1 | |
| \prod | |
| k=0 |
(a1+kd)=dn
| ||||||
|
where
\Gamma
a1/d
This is a generalization of the facts that the product of the progression
1 x 2 x … x n
n!
m x (m+1) x (m+2) x … x (n-2) x (n-1) x n
for positive integers
m
n
| n! | |
| (m-1)! |
.
\begin{align} a1a2a3 … an
| n-1 | |
| &=\prod | |
| k=0 |
(a1+kd)\\ &=
| n-1 | ||
| \prod | d\left( | |
| k=0 |
| a1 | |
| d |
+k\right)=d\left(
| a1 | |
| d |
\right)d\left(
| a1 | |
| d |
+1\right)d\left(
| a1 | |
| d |
+2\right) … d\left(
| a1 | |
| d |
+(n-1)\right)\\ &=
| n-1 | ||
| d | \left( | |
| k=0 |
| a1 | |
| d |
+k\right)=dn{\left(
| a1 | |
| d |
\right)}\overline{n
x\overline{n
By the recurrence formula
\Gamma(z+1)=z\Gamma(z)
z>0
\Gamma(z+2)=(z+1)\Gamma(z+1)=(z+1)z\Gamma(z)
\Gamma(z+3)=(z+2)\Gamma(z+2)=(z+2)(z+1)z\Gamma(z)
so that
| \Gamma(z+m) | |
| \Gamma(z) |
=
| m-1 | |
| \prod | |
| k=0 |
(z+k)
for
m
z
Thus, if
a1/d>0
| n-1 | ||
| \prod | \left( | |
| k=0 |
| a1 | |
| d |
+k\right)=
| ||||||
|
and, finally,
a1a2a3 … an=
| n-1 | ||
| d | \left( | |
| k=0 |
| a1 | |
| d |
+k\right)=dn
| ||||||
|
3,8,13,18,23,28,\ldots
an=3+5(n-1)
P50=550 ⋅
| \Gamma\left(3/5+50\right) | |
| \Gamma\left(3/5\right) |
≈ 3.78438 x 1098.
(1,3,5,7,9,11,13,15,17,19)
1 ⋅ 3 ⋅ 5 … 19
| 9 | |
| =\prod | |
| k=0 |
(1+2k)=210 ⋅
| ||||||
|
The standard deviation of any arithmetic progression is
\sigma=|d|\sqrt{
| (n-1)(n+1) | |
| 12 |
where
n
d
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.[10] However, the intersection of infinitely many infinite arithmetic progressions might be a single number rather than itself being an infinite progression.
Let
a(n,k)
k
\{1, … ,n\}
\phi(η,\kappa)
\phi(η,\kappa)=\begin{cases} 0&if\kappa\midη\\ \left(\left[η (mod\kappa)\right]-2\right)\left(\kappa-\left[η (mod\kappa)\right]\right)&if\kappa\not\midη\\ \end{cases}
Then:
\begin{align}a(n,k)&=
| 1 | |
| 2(k-1) |
\left(n2-(k-1)n+(k-2)+\phi(n+1,k-1)\right)\\&=
| 1 | |
| 2(k-1) |
\left((n-1)(n-(k-2))+\phi(n+1,k-1)\right)\end{align}
As an example, if one expects arithmetic subsets and, counting directly, one sees that there are 9; these are