Arithmetic Series

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If the terms of a sequence are added, the result is known as a series.
An arithmetic series isnthe indicated sum of the terms of an arithmetic sequence.

The sequence:  1,2,3,4,5,6,…
Gives the series: 1+2+3+4+5+6+…

The sum of the terms of a series is referred to as Sn, the sum of n terms of a series.

For an arithmetic series,  Sn= a1 + a2 + a3 + … + an
 

                   Sn= a1 + (a1 + d) + (a1 + 2d) + … + an   (1)

Rewriting the expanded form in the reverse order with decreasing d,
                 Sn= an + (an - d) + (an - 2d) + … +a1    (2)

Adding (1) and (2) term by term,
      2Sn= (a1 + an) + (a1 + an) + (a1 + an) + … (a1 + an)

Note that the right side is the sum of n identical terms of (a1 + an) Hence,
                      2Sn= n(a1 + an)  
                     | Sn= n/2 (a1 + an) |

This is the formula in finding the sum of an arithmetic sequence given the first term and the nth term. If the nth term is not given the expand the formula using the formula for the nth term, an= a1 + (n-1)d.
                     | Sn= n/2 [2a1 +  (n-1)d] |

Examples:

1.)     The sum of the first 100 natural numbers.
    The sum of the first natural number is 1+2+3+…+100
    This is an arithmetic series.    a1= 1, an= 100, n= 100

Since the last term is given, use the first formula.
        Sn= n/2 (a1 + an)

Substituting the given,    S100= 100   (1+100)
                                                       2    
                    = 50(101) = 5050

2.)     Find the sum of the first 14 terms of the arithmetic series 2+5+8+11+14+17+
The last term is not given, use the second formula,
               a1= 2, n= 14, d= 3

             Sn= n/2 [2a1 +  (n-1)d]

Subtituting the given,
           S14= 14/2 [2(2) + (14 - 1) 3]
           S14= 7[4+13(3)]
           S14= 7(43)
           S14= 301

The sum of the first 14 terms of the series is 301.

3.)     How many terms should be added in the sequence 3,6,9,12,… to sum up to 165?
The last term is not given, use the second formula,
                   a1= 3, d= 3, Sn= 165
       
               Sn= n/2 [2a1 +  (n-1)d]

               165= n/2 [2(3) + (n-1)3]
               165= n/2 [6 + 3n - 3]
          330 = 3n + 3n²
3n² + 2n - 330= 0
n² + n - 110= 0
(n-10)(n-11)= 0
n-10 =0
n- 10 = 0         n+11= 0
n= 10               n= -11

Reject n= -11 since sequwnce is defined on positive integers only. The number of terms is 10.

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