Question 66130: 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 josmiceli(19441)   (Show Source):
You can put this solution on YOUR website! 1st term = a
2nd term = b
a, b, 26, 40 is the sequence
The first differences are
b-a, 26-b, 40-26
the 2nd differences are 4, so
(26 - b) - (b-a) = 4
14 - (26 - b) = 4
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14 - 26 + b = 4
-12 + b = 4
b = 16
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(26 - b) - (b-a) = 4
(26 - 16) - (16 - a) = 4
10 - 16 + a = 4
-6 + a = 4
a = 10
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So, the sequence is
10,16,26,40
1st differences are
6,10,14
2nd differences are
4,4
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Call the 5th term c
1st difference is
c - 40
2nd difference equals 4
c - 40 - 14 = 4
c - 54 = 4
c = 58
1st 5 terms are
10,16,26,40,58
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