Posted by Bhupendra on .
A sequence is called an arithmetic progression of the rst order if the
dierences of the successive terms are constant. It is called an arith
metic progression of the second order if the dierences of the successive
terms form an arithmetic progression of the rst order. In general, for
k 2, a sequence is called an arithmetic progression of the kth order
if the dierences of the successive terms form an arithmetic progression
of the (k 1)th order.
The numbers
4; 6; 13; 27; 50; 84
are the rst six terms of an arithmetic progression of some order. What
is its least possible order? Find a formula for the nth term of this
progression.

Math 
tchrwill,
The numbers 4; 6; 13; 27; 50; 84 do not form an arithmetic progression as the differences between succsessive terms are not constant.
If you take the successive differences of the terms given,
n.......1....2....3....4....5....6...
N.......4....6...13...27...50...84...
1st Diff..2....7...14....23...34...
2nd Diff....5....7....9.....11
3d Diff.......2....2.....2
you find that the 3rd differences are constant meaning that the sequence is a finite difference sequence of the 3rd order, the defining expression being of the form N = an^3 + bn^2 + cn + d.
Using the given data,
a(1)^3 + b(1)^2 + c(1) + 1 = 4 or a + b + c + d = 4
a(2)^3 + b(2)^2 + c(2) + 2 = 6 or 8a + 4b + 2c + 2 = 6
a(3)^3 + b(3)^2 + c(3) + 3 = 13 or 27a + 9b + 3c + 3 = 13
a(4)3 + b(4)^2 + c(4) + 4 = 27 or 64a + 16b + 4c + 4 = 27
Solve for a, b, c and d and substitute back into N = an^3 + bn^2 +cn + d