Arithmetic series

An arithmetic series is a sum of consecutive terms in an arithmetic sequence. For instance,

$2 + 6 + 10 + 14 + 18$

is an arithmetic series whose value is 50.

To find the sum of an arithmetic sequence, we can write it out as so (S is the sum, a is the first term, n is the number of terms, and d is the common difference): $\begin{align*} S &= a + (a+d) + (a+2d) + ... + (a+(n-1)d) \\ S &= (a+(n-1)d) + (a+(n-2)d)+ ... + (a+d) + a \end{align*}$

Now, adding vertically and shifted over one, we get

$2S = (2a+(n-1)d)+(2a+(n-1)d)+(2a+(n-1)d)+...+(2a+(n-1)d)$

This equals $2S = n(2a+(n-1)d)$ , so the sum is $\frac{n}{2} (2a+(n-1)d)$ .

Problems

Introductory Problems

Intermediate Problems

2003 AIME I, Problem 2

Olympiad Problem

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:

and in general

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.

Contents

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

Sum This section is about Finite arithmetic series. For Infinite arithmetic series, see Infinite arithmetic series. $2 + 5 + 8 + 11 + 14 = 40$ $14 + 11 + 8 + 5 + 2 = 40$ $16 + 16 + 16 + 16 + 16 = 80$

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:

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:

In the case above, this gives the equation:

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

Derivation

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:

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

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

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

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). An n member arithmetical progression. 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

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