SOLUTION: i need the last number in the following sequences 3, -3, 12, 4, 20, 7, 16, 8, 27, 9 2,7, 26, 101, 400

let the 6th term = x

2nd term - 1st term = (-3)- (3) =  -6
3rd term - 2nd term = (12)-(-3) =  15
4th term - 3rd term =  (4)-(12) =  -8
5th term - 4th term = (20)- (4) =  16  
6th term - 5th term =  (x)-(16) = -10 

                         x - 16 = -10
                              x = 6                       
6th term = 6
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The odd numbered terms are obviously going 7,8,9,10,11,12,13,...,
adding 1 each time

The even numbered terms could be going 16,27,38,49,60,71,...,
adding 11 each time.

Or, the even numbered terms could be going 42,33,24,15,06,(-1)7...,
Subtracting 1 from the base and adding 1 to the exponent each time.  Which
would be 16,27,16,1,0,-1,... 

Or, the even numbered terms could be going 24,33,42,51,60,7-1...,
Adding 1 to the base and subtracting 1 from the exponent each time.  Which
would be 16,27,16,5,1,1/7,...

There is no way to tell.  Since either of the last two ways would make
the 5th term be 16, I'd go with 16.

But there is no way to be sure it's not the first way, or some other way.  It's
just whichever way the person who created it happened to have created it.

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  2×4-1 =    8-1 =    7
  7×4-2 =   28-2 =   26
 26×4-3 =  104-3 =  101
101×4-4 =  404-4 =  400
400×4-5 = 1600-5 = 1595

Answer: 1595

Edwin