SOLUTION: 20 1/2, 23 1/3, 26 1/6, 29... what would the 7 term be in this sequence and how would you find it?

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Question 775450: 20 1/2, 23 1/3, 26 1/6, 29... what would the 7 term be in this sequence and how would you find it?
Found 2 solutions by Edwin McCravy, MathLover1:
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
20 1/2, 23 1/3, 26 1/6, 29...

Subtract 2nd term minus 1st term

 23 1/3 =  23 2/6 =  22 8/6
-20 1/2 = -20 3/6 = -20 3/6
---------------------------
                      2 5/6

Subtract 3rd term minus 2nd term

 26 1/6 =  26 1/6 =  25 7/6
-23 1/3 = -23 2/6 = -23 2/6
---------------------------
                      2 5/6

Subtract 4th term minus 3rd term

 29        28 6/6  
-26 1/6 = -26 1/6
-----------------
            2 5/6
 
So the common difference is 2 5/6

So we add 2 5/6 each time to get the next term:

 29
+ 2 5/6
-------
 31 5/6 = 5th term

 31 5/6  
+ 2 5/6
-------
 33 10/6 = 34 4/6 = 34 2/3 = 6th term

 34 2/3 =  34 4/6
+ 2 5/6   + 2 5/6
-----------------
           36 9/6 = 36 3/2 = 37 1/2 = 7th term.

Edwin

Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!
given:
20 1/2, 23 1/3, 26 1/6, 29.
or you can write them this way:
41/2, 70/3, 157/6, 29, ..........
To find any term of an arithmetic sequence:
where is the first term of the sequence, is the common difference, is the number of the term to find.











here is more terms of your sequence:
41/2, 70/3, 157/6, 29, 191/6, 104/3, 75/2, 121/3, 259/6, 46, 293/6, 155/3, 109/2, 172/3, ...

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