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Difference between revisions of "Arithmetic series"

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(→‎Problems)
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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 5, 7, 9, 11, 13, 15 … is an arithmetic progression with common difference of 2.
+
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 a_1 and the common difference of successive members is d, then the nth term of the sequence (a_n) is given by:
+
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:
  
−
     \ a_n = a_1 + (n - 1)d,
+
     <math>\ a_n = a_1 + (n - 1)d,</math>
  
 
and in general
 
and in general
  
−
     \ a_n = a_m + (n - m)d.
+
     <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.
 
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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Sum
 
Sum
 
This section is about Finite arithmetic series. For Infinite arithmetic series, see Infinite arithmetic series.
 
This section is about Finite arithmetic series. For Infinite arithmetic series, see Infinite arithmetic series.
−
2 + 5 + 8 + 11 + 14 = 40
+
<math>2 + 5 + 8 + 11 + 14 = 40</math>
−
14 + 11 + 8 + 5 + 2 = 40
+
<math>14 + 11 + 8 + 5 + 2 = 40</math>
−
16 + 16 + 16 + 16 + 16 = 80
+
<math>16 + 16 + 16 + 16 + 16 = 80</math>
  
−
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.
+
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:
 
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
+
  <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 2 + 14 = 16), and dividing by 2:
+
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:
  
−
    \frac{n(a_1 + a_n)}{2}
+
  <math> \frac{n(a_1 + a_n)}{2}</math>
  
 
In the case above, this gives the equation:
 
In the case above, this gives the equation:
  
−
    2 + 5 + 8 + 11 + 14 = \frac{5(2 + 14)}{2} = \frac{5 \times 16}{2} = 40.
+
  <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 a_1 and a_n. For example:
+
This formula works for any real numbers <math>a_1</math> and <math> a_n.</math> 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}.
+
  <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>
  
 
Derivation
 
Derivation
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To derive the above formula, begin by expressing the arithmetic series in two different ways:
 
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)
+
  <math> S_n=a_1+(a_1+d)+(a_1+2d)+\cdots+(a_1+(n-2)d)+(a_1+(n-1)d)</math>
  
−
    S_n=(a_n-(n-1)d)+(a_n-(n-2)d)+\cdots+(a_n-2d)+(a_n-d)+a_n.
+
  <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:
 
Adding both sides of the two equations, all terms involving d cancel:
  
−
    \ 2S_n=n(a_1 + a_n).
+
  <math> \ 2S_n=n(a_1 + a_n).</math>
  
 
Dividing both sides by 2 produces a common form of the equation:
 
Dividing both sides by 2 produces a common form of the equation:
  
−
     S_n=\frac{n}{2}( a_1 + a_n).
+
     <math>S_n=\frac{n}{2}( a_1 + a_n).</math>
  
−
An alternate form results from re-inserting the substitution: a_n = a_1 + (n-1)d:
+
An alternate form results from re-inserting the substitution: <math>a_n = a_1 + (n-1)d:</math>
  
−
    S_n=\frac{n}{2}[ 2a_1 + (n-1)d].
+
  <math> S_n=\frac{n}{2}[ 2a_1 + (n-1)d].</math>
  
−
Furthermore the mean value of the series can be calculated via: S_n / n:
+
Furthermore the mean value of the series can be calculated via: <math>S_n / n:</math>
  
−
    \overline{n} =\frac{a_1 + a_n}{2}.
+
  <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).
 
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).
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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
 
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 \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> 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 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.)
+
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 1 \times 2 \times \cdots \times n is given by the factorial n! and that the product
+
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
  
−
     m \times (m+1) \times (m+2) \times \cdots \times (n-2) \times (n-1) \times n \,\!
+
     <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
 
for positive integers m and n is given by
  
−
     \frac{n!}{(m-1)!}.
+
     <math>\frac{n!}{(m-1)!}.</math>
  
 
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
 
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}.  
+
  <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
 
Standard deviation
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The standard deviation of any arithmetic progression can be calculated via:
 
The standard deviation of any arithmetic progression can be calculated via:
  
−
     \sigma = |d|\sqrt{\frac{(n-1)(n+1)}{12}}
+
     <math>\sigma = |d|\sqrt{\frac{(n-1)(n+1)}{12}}</math>
  
−
where n is the number of terms in the progression, and d is the common difference between terms
+
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 20:51, 3 June 2014

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

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: 

   $\ a_n = a_1 + (n - 1)d,$

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.

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: 

  $2 + 5 + 8 + 11 + 14$

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 the equation: 

  $2 + 5 + 8 + 11 + 14 = \frac{5(2 + 14)}{2} = \frac{5 \times 16}{2} = 40.$

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: $a_n = a_1 + (n-1)d:$ 

  $S_n=\frac{n}{2}[ 2a_1 + (n-1)d].$

Furthermore 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). 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 

  $a_1a_2\cdots a_n = d \frac{a_1}{d} d (\frac{a_1}{d}+1)d (\frac{a_1}{d}+2)\cdots d (\frac{a_1}{d}+n-1)=d^n {\left(\frac{a_1}{d}\right)}^{\overline{n}} = d^n \frac{\Gamma \left(a_1/d + n\right) }{\Gamma \left( a_1 / d \right) },$

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: 

   $\sigma = |d|\sqrt{\frac{(n-1)(n+1)}{12}}$

where $n$ is the number of terms in the progression, and d is the common difference between terms

See also

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