User talk:Syaffuan

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 3, 5, 7, 9, 11, 13, … is an arithmetic progression with common difference 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 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 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.

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

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