About Airthmetic 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

This section is about Finite arithmetic series. For Infinite arithmetic series, see Infinite arithmetic series.

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:

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

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

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

Derivation

To derive the above formula, begin by 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: $a_n = a_1 + (n-1)d$ :

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

Furthurmore 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).

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 a_n = 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:

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

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

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