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, … 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

Template:Other uses-section The sum of the members of a finite arithmetic progression is called an arithmetic series.

Expressing the arithmetic series in two different ways:

$S_n=a_1+(a_1+d)+(a_1+2d)+\cdots+(a_1+(n-2)d)+(a_1+(n-1)d)$
$S_n=(a_n-(n-1)d)+(a_n-(n-2)d)+\cdots+(a_n-2d)+(a_n-d)+a_n.$

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: $a_n = a_1 + (n-1)d$:

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

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).<ref>Aryabhatiya Template:Lang-mr, Mohan Apte, Pune, India, Rajhans Publications, 2009, p.95, ISBN 978-81-7434-480-9</ref>

So, for example, the sum of the terms of the arithmetic progression given by an = 3 + (n-1)(5) up to the 50th term is

$S_{50} = \frac{50}{2}[2(3) + (49)(5)] = 6,275.$

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

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

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

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

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

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