Sunday, 1 September 2013

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


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 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(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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