Difference between revisions of "Arithmetic series"

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=== Olympiad Problem ===
=== 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 <math>5, 7, 9, 11, 13, 15 …</math> is an arithmetic progression with common difference of 2.
If the initial term of an arithmetic progression is<math>a_1</math> and the common difference of successive members is d, then the nth term of the sequence <math>(a_n)</math> is given by:
    <math>\ a_n = a_1 + (n - 1)d,</math>
and in general
    <math>\ a_n = a_m + (n - m)d.</math>
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.
    1 Sum
        1.1 Derivation
    2 Product
    3 Standard deviation
    4 See also
    5 References
    6 External links
This section is about Finite arithmetic series. For Infinite arithmetic series, see Infinite arithmetic series.
<math>2 + 5 + 8 + 11 + 14 = 40</math>
<math>14 + 11 + 8 + 5 + 2 = 40</math>
<math>16 + 16 + 16 + 16 + 16 = 80</math>
Computation of the sum<math> 2 + 5 + 8 + 11 + 14.</math> 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 <math>(2 + 14 = 16). </math>Thus <math>16 × 5 = 80</math> is twice the sum.
The sum of the members of a finite arithmetic progression is called an arithmetic series. For example, consider the sum:
  <math> 2 + 5 + 8 + 11 + 14</math>
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 <math>2 + 14 = 16</math>), and dividing by 2:
  <math> \frac{n(a_1 + a_n)}{2}</math>
In the case above, this gives the equation:
  <math> 2 + 5 + 8 + 11 + 14 = \frac{5(2 + 14)}{2} = \frac{5 \times 16}{2} = 40.</math>
This formula works for any real numbers <math>a_1</math> and <math> a_n.</math> For example:
  <math> \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}.</math>
To derive the above formula, begin by expressing the arithmetic series in two different ways:
  <math> S_n=a_1+(a_1+d)+(a_1+2d)+\cdots+(a_1+(n-2)d)+(a_1+(n-1)d)</math>
  <math> S_n=(a_n-(n-1)d)+(a_n-(n-2)d)+\cdots+(a_n-2d)+(a_n-d)+a_n.</math>
Adding both sides of the two equations, all terms involving d cancel:
  <math> \ 2S_n=n(a_1 + a_n).</math>
Dividing both sides by 2 produces a common form of the equation:
    <math>S_n=\frac{n}{2}( a_1 + a_n).</math>
An alternate form results from re-inserting the substitution: <math>a_n = a_1 + (n-1)d:</math>
  <math> S_n=\frac{n}{2}[ 2a_1 + (n-1)d].</math>
Furthermore the mean value of the series can be calculated via: <math>S_n / n:</math>
  <math>  \overline{n} =\frac{a_1 + a_n}{2}.</math>
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.
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
  <math> a_1a_2\cdots a_n = d \frac{a_1}{d} d (\frac{a_1}{d}+1)d (\frac{a_1}{d}+2)\cdots d (\frac{a_1}{d}+n-1)=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) },</math>
where <math>x^{\overline{n}}</math> 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 <math>1 \times 2 \times \cdots \times n</math> is given by the factorial<math> n!</math> and that the product
    <math>m \times (m+1) \times (m+2) \times \cdots \times (n-2) \times (n-1) \times n \,\!</math>
for positive integers m and n is given by
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
  <math> P_{50} = 5^{50} \cdot \frac{\Gamma \left(3/5 + 50\right) }{\Gamma \left( 3 / 5 \right) } \approx 3.78438 \times 10^{98}. </math>
Standard deviation
The standard deviation of any arithmetic progression can be calculated via:
    <math>\sigma = |d|\sqrt{\frac{(n-1)(n+1)}{12}}</math>
where <math>n</math> is the number of terms in the progression, and d is the common difference between terms
== See also ==
== See also ==

Revision as of 16:41, 19 September 2015

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 $\frac{n}{2} (2a+(n-1)d)$.


Introductory Problems

Intermediate Problems

Olympiad Problem

See also

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