SOLUTION: The third and fourth terms of a sequence are 26 and 40. If the second differences are a constant 4, what are the first five terms of the sequence?

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Question 331346: The third and fourth terms of a sequence are 26 and 40. If the second differences are a constant 4, what are the first five terms of the sequence?
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
The third and fourth terms of a sequence are 26 and 40. If the second differences are a constant 4, what are the first five terms of the sequence?

     1st          2nd
an  diffs        diffs    
---------------------------------------------------------
a1 |       |
   | a2-a1 |
a2 |       | (26-a2)-(a2-a1) = 26-a2-a2+a1 = 26-2a2+a1 = 4, or a1-2a2=-22
   | 26-a2 |
26 |       | 14-(26-a2) = 14-26+a2 = -12+a2 = 4, or a2=16
   |  14   |
40 |       | a5-40-14 = a5-54 = 4, or a5 = 58
   | a5-40 |
a5 |       |


So we have the system of equations:



Susbstitute 16 for  in the first equation:









So , , , , 
1
Checking:

   1st 2nd 
   dif dif
10
    6
16      4
    10 
26      4
    14
40      4
    18 
58

Edwin
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