Posted by Tk on .
The first term of a quadratic sequence is 8 and it's fourth is 32. If the second difference of the sequence is 2, find Tn.

arithmetic 
Reiny,
the general term of a quadratic sequence is
term(n) = an^2 + bn + c, where a, b, and c are constants
It can be shown using differences that a is (1/2) of the second difference.
Thus a = 1
so term(1) = 8
8 = 1(1^2) + b(1) + c > b+c = 7
term(4) = 32
32 = 1(2^2) + b(2) + c > 2b + c = 28
subtract them: b = 21
then c = 721 = 14
term(n) = n^2 + 21n  14 
correction  arithmetic 
Reiny,
I used the 32 as if it were the 2nd terms, should have been the 4th
Here is the correct version
Thus a = 1
so term(1) = 8
8 = 1(1^2) + b(1) + c > b+c = 7
term(4) = 32
32 = 1(4^2) + b(4) + c > 4b + c = 16
subtract them:
3b = 9
b = 3
then c = 73 = 4
term(n) = n^2 + 3n + 4
check:
t(1) = 1+3+4 = 8
t(4) = 16 + 12 + 4 = 32