9514 1404 393
Answer:
t(n) = -n^2 -n +102
Step-by-step explanation:
We can use the first 3 terms to find a, b, c.
a(1)^2 +b(1) +c = 100
a(2)^2 +b(2) +c = 96
a(3)^2 +b(3) +c = 90
Subtract the first equation from the other two:
3a +b = -4 . . . . . . [eq4]
8a +2b = -10 . . . . [eq5]
[eq5] can be reduced to
4a +b = -5 . . . . . . [eq6]
Subtracting [eq4] from [eq6] gives ...
(4a +b) -(3a +b) = (-5) -(-4)
a = -1 . . . . . . . . simplify
Filling that into [eq4], we get
3(-1) +b = -4
b = -1 . . . . . . . add 3
Putting the values for 'a' and 'b' into the first equation gives ...
-1 +-1 +c = 100
c = 102
Then the n-th term t(n) of the sequence is ...
t(n) = -n^2 -n +102
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Additional comment
As you might guess, there is a generic solution for quadratic sequences. In terms of the first term (a1), first first difference (d1), and second difference (d2), the coefficients can be written as ...
This sequence has first differences of -4, -6, -8, -10, and second differences of -2. Putting the sequence values into the above formulas, we find ...
