Arithmetic Sequences and Series

1. Arithmetic Sequence

   Examples

   Find the general term for the following sequences both recursively and explicitly:
   
   

   1. 2,6,10,14,18,22, ...
      

   2.  -5,-3,-1,1,3,...
      

   3. 1,4,7,10,13,16,...
      

   4. -1,10,21,32,43,54,...
      

   5. 3,0,-3,-6,-9,-12,...
      

   Solution
   
   All of these have one thing in common.  To get to the next term we add a fixed number. 
    

   1. Add four to obtain the next term. Thus
      
              a_n+1 = a_n + 4,     a₁ = 2   
      
      To find an explicit expression, we use the following reasoning.  To get the first term, we start with 2 and add no 4's.  To get to the second term we start with 2 and add one 4.  To get to the third term, we start with 2 and add two 4's.  To get the the fourth term we start with 2 and add three 4's.  To get to the n^th term, we start with 2 and add n - 1 four.  Hence
      
              a_n = 2 + 4(n - 1)
      
      

   2.  Add two to obtain the next term. Thus
      
              a_n+1 = a_n + 2,     a₁ = -5   
      
      To find an explicit expression, we use the same reasoning as in part "A".  To get the first term, we start with -5 and add no 2's.  To get to the second term we start with -5 and add one 2.  To get to the third term, we start with -5 and add two 2's.  To get the the fourth term we start with -5 and add three 2's.  To get to the n^th term, we start with -5 and add n - 1 twos.  Hence
      
              a_n = -5 + 2(n - 1)
      
      

   3. As in the preceding exercises,
      
              a_n+1 = a_n + 3,     a₁ = 1
      
      and
      
              a_n = 1 + 3(n - 1)
      
      

   4. We have
      
              a_n+1 = a_n + 11,     a₁ = -1
      
      and
      
              a_n = -1 + 11(n - 1)
      
      

   5. Finally,
      
              a_n+1 = a_n - 3,     a₁ = 3
      
      and
      
              a_n = 3 -  3(n - 1)
      
                

   Definition   A sequence with general term           an+1 = an + d is called an arithmetic sequence.

   
   This definition defines an arithmetic sequence recursively.  The next theorem shows how to find an explicit form for an arithmetic sequence.
   
   
   

   Theorem An arithmetic sequence with            an+1 = an + d   has explicit form             an = a1 + (n - 1)d

    

   Proof:  (by induction)
   
   For n = 1, we have 
   
           a₁ = a₁ + (1 - 1)d  (true)
   
   Assume that the theorem is true for n = k - 1, hence
   
           a_k-1 = a₁ + (k - 1 - 1)d = a₁ + (k - 2)d
   
   Then 
   
           a_k = a_k-1 + d = a₁ + (k - 2)d + d
   
           = a₁ + kd - 2d + d = a₁ + kd - d = a₁ + (k - 1)d

   Hence by mathematical induction, the theorem is true.
   
   

   Example:  
   
   Suppose that a₁ = 4 and d = 2 then the sequence is 
   
           4,6,8,...,(4 + (n - 1)d),...

    

   Example

   Suppose that the 13th term of an arithmetic sequence is 46 and the fourth term is 100.  Find the expression for the general term.
   

   Solution

   We have 
   
           46 = a₁ + d(13 - 1) = a₁ + 12d
   
   and
   
           100 = a₁ + d(4 - 1) = a₁ + 3d
   
   Subtracting the two equations gives
   
           -54 = 9d
   
   or        
   
           d = -6
   
   Putting this back into the first equation gives
   
           46 = a₁ + 12(-6)
   
   or   
   
           a₁ = 118

   We can conclude that 
   
           a_n = 118 - 6(n - 1)

   
   

2. The Arithmetic Series

   The following theorem provides us with an easy way to calculate the arithmetic series.

   
   

   Theorem If           an =  a1 + (n - 1)d is an arithmetic sequence then the sum of the sequence is          Sn =  S i=1n  an =  n/2 (a1 + an)

    

   Proof:

           S_n = S _i=1ⁿ  a_n = a₁ + a₂ + ... + a_n-1 + a_n
   
           = a₁ + (a₁ + d)  + ... +(a₁ + (n - 2)d) + (a₁ + (n - 1)d)
   
           The sum can also be written by working backwards from the last term, that is to get to the previous term, subtract d.
   
           S_n = S _i=1ⁿ  a_n = a_n + a_n-1 + ... + a₂ + a₁
   
           = (a₁ + (n - 1)d) + (a₁ + (n - 2)d) + ... + (a₁ + d)+ a₁  
   
   Since these are the same we can add them together to get 2S.
   
           S_n = a₁                    +    (a₁ + d)        + ... + (a₁ + (n - 2)d) + (a₁ + (n - 1)d)
           S_n = (a₁ + (n - 1)d) + (a₁ + (n - 2)d) + ... + (a₁ + d)           +  a₁
   

   ---

           2S_n = [a₁ + (a₁ + (n - 1)d)] + [a₁ + (a₁ + (n - 1)d)] + ... 
                   +[a₁ + (a₁ + (n - 1)d)] + [a₁ + (a₁ + (n - 1)d)] 
   
           =  [a₁ + a_n] + [a₁ + a_n] +... + [a₁ + a_n] + [a₁ + a_n] 
   
           = n[a₁ + a_n]
   
           Hence 
   
           S_n =  n/2 [a₁ + a_n]

    

   Example:  
   
   Find 
   
           3 + 7 + 11 + 15  + ... + 35

   Solution:
   
   We have 
   
           a₁ = 3, a_n = 35, d = 4
   
   To find n we note that 
   
           35 = 3 +  (n - 1)4
   
   so that 
   
           32 = (n - 1)4
   
   Dividing gives
   
           8 = n - 1
   
   Hence
   
           n = 9
   
   Now we are ready to use the formula
   
           S_n = 9/2 (3 + 35) = 171

    

   Exercise:  
   
   Suppose that the sum of the first 18 terms of an arithmetic sequence is -45 and 
   
           d = -9 
   
   find the first term.

    

   Application  

   Suppose that you play black jack at Harrah's on June 1 and lose $1,000.  Tomorrow you bet and lose $15 less.  Each day you lose $15 less that your previous loss.  What will your total losses be for the 30 days of June?
   
   Solution
   
   This is an arithmetic series with
   
           a₁ = 1000 
   
   and
   
           d = -15
   
   We can calculate
   
           a₃₀ = 1000 - 15(30 - 1) = 565
   
   Now we use the formula
   
           S₃₀ = 30/2 (1000 + 565) = 23,475
   
   You will lose a total of $23,475 during June.

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