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Arithmetic progression

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Title: Arithmetic progression  
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Subject: Geometric progression, Infinite arithmetic series, Series (mathematics), Arithmetic combinatorics, Siamese method
Collection: Arithmetic Series, Articles Containing Proofs, Sequences and Series
Publisher: World Heritage Encyclopedia

Arithmetic progression

In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. For instance, the sequence 5, 7, 9, 11, 13, 15 … is an arithmetic progression with common difference of 2.

If the initial term of an arithmetic progression is a_1 and the common difference of successive members is d, then the nth term of the sequence (a_n) is given by:

\ a_n = a_1 + (n - 1)d,

and in general

\ a_n = a_m + (n - m)d.

A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. The sum of a finite arithmetic progression is called an arithmetic series.

The behavior of the arithmetic progression depends on the common difference d. If the common difference is:

  • Positive, the members (terms) will grow towards positive infinity.
  • Negative, the members (terms) will grow towards negative infinity.


  • Sum 1
    • Derivation 1.1
  • Product 2
  • Standard deviation 3
  • See also 4
  • References 5
  • External links 6


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

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:

\frac{n(a_1 + a_n)}{2}

In the case above, this gives the equation:

2 + 5 + 8 + 11 + 14 = \frac{5(2 + 14)}{2} = \frac{5 \times 16}{2} = 40.

This formula works for any real numbers a_1 and a_n. For example:

\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}.


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


Adding both sides of the two equations, all terms involving d cancel:

\ 2S_n=n(a_1 + a_n).

Dividing both sides by 2 produces a common form of the equation:

S_n=\frac{n}{2}( a_1 + a_n).

An alternate form results from re-inserting the substitution: a_n = a_1 + (n-1)d:

S_n=\frac{n}{2}[ 2a_1 + (n-1)d].

Furthermore the mean value of the series can be calculated via: S_n / n:

\overline{n} =\frac{a_1 + a_n}{2}.

In 499 AD Aryabhata, a prominent mathematician-astronomer from the classical age of Indian mathematics and Indian astronomy, gave this method in the Aryabhatiya (section 2.18).


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

a_1a_2\cdots a_n = d \frac{a_1}{d} d (\frac{a_1}{d}+1)d (\frac{a_1}{d}+2)\cdots d (\frac{a_1}{d}+n-1)=d^n {\left(\frac{a_1}{d}\right)}^{\overline{n}} = d^n \frac{\Gamma \left(a_1/d + n\right) }{\Gamma \left( a_1 / d \right) },

where x^{\overline{n}} denotes the rising factorial and \Gamma denotes the Gamma function. (Note however that the formula is not valid when a_1/d is a negative integer or zero.)

This is a generalization from the fact that the product of the progression 1 \times 2 \times \cdots \times n is given by the factorial n! and that the product

m \times (m+1) \times (m+2) \times \cdots \times (n-2) \times (n-1) \times n \,\!

for positive integers m and n is given by


Taking the example from above, the product of the terms of the arithmetic progression given by an = 3 + (n-1)(5) up to the 50th term is

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

Standard deviation

The standard deviation of any arithmetic progression can be calculated via:

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

where n is the number of terms in the progression, and d is the common difference between terms

See also


  • Sigler, Laurence E. (trans.) (2002). Fibonacci's Liber Abaci. Springer-Verlag. pp. 259–260.  

External links

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