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 === Olympiad Problem ===   === Olympiad Problem === 
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−  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.
 
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−  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:
 
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−  <math>\ a_n = a_1 + (n  1)d,</math>
 
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−  and in general
 
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−  <math>\ a_n = a_m + (n  m)d.</math>
 
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−  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.
 
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−  The behavior of the arithmetic progression depends on the common difference d. If the common difference is:
 
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−  Positive, the members (terms) will grow towards positive infinity.
 
−  Negative, the members (terms) will grow towards negative infinity.
 
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−  Contents
 
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−  1 Sum
 
−  1.1 Derivation
 
−  2 Product
 
−  3 Standard deviation
 
−  4 See also
 
−  5 References
 
−  6 External links
 
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−  Sum
 
−  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>
 
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−  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.
 
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−  The sum of the members of a finite arithmetic progression is called an arithmetic series. For example, consider the sum:
 
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−  <math> 2 + 5 + 8 + 11 + 14</math>
 
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−  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:
 
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−  <math> \frac{n(a_1 + a_n)}{2}</math>
 
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−  In the case above, this gives the equation:
 
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−  <math> 2 + 5 + 8 + 11 + 14 = \frac{5(2 + 14)}{2} = \frac{5 \times 16}{2} = 40.</math>
 
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−  This formula works for any real numbers <math>a_1</math> and <math> a_n.</math> For example:
 
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−  <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>
 
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−  Derivation
 
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−  To derive the above formula, begin by expressing the arithmetic series in two different ways:
 
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−  <math> S_n=a_1+(a_1+d)+(a_1+2d)+\cdots+(a_1+(n2)d)+(a_1+(n1)d)</math>
 
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−  <math> S_n=(a_n(n1)d)+(a_n(n2)d)+\cdots+(a_n2d)+(a_nd)+a_n.</math>
 
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−  Adding both sides of the two equations, all terms involving d cancel:
 
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−  <math> \ 2S_n=n(a_1 + a_n).</math>
 
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−  Dividing both sides by 2 produces a common form of the equation:
 
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−  <math>S_n=\frac{n}{2}( a_1 + a_n).</math>
 
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−  An alternate form results from reinserting the substitution: <math>a_n = a_1 + (n1)d:</math>
 
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−  <math> S_n=\frac{n}{2}[ 2a_1 + (n1)d].</math>
 
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−  Furthermore the mean value of the series can be calculated via: <math>S_n / n:</math>
 
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−  <math> \overline{n} =\frac{a_1 + a_n}{2}.</math>
 
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−  In 499 AD Aryabhata, a prominent mathematicianastronomer 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
 
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−  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
 
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−  <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}+n1)=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>
 
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−  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.)
 
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−  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
 
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−  <math>m \times (m+1) \times (m+2) \times \cdots \times (n2) \times (n1) \times n \,\!</math>
 
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−  for positive integers m and n is given by
 
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−  <math>\frac{n!}{(m1)!}.</math>
 
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−  Taking the example from above, the product of the terms of the arithmetic progression given by an = 3 + (n1)(5) up to the 50th term is
 
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−  <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>
 
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−  Standard deviation
 
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−  The standard deviation of any arithmetic progression can be calculated via:
 
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−  <math>\sigma = d\sqrt{\frac{(n1)(n+1)}{12}}</math>
 
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−  where <math>n</math> is the number of terms in the progression, and d is the common difference between terms
 
   
 == See also ==   == See also == 