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December 18, 2014

December 18, 2014

Posted by **Bhupendra** on Friday, May 11, 2012 at 6:47am.

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 k-th 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 n-th term of this

progression.

- Math -
**tchrwill**, Friday, May 11, 2012 at 10:48amThe 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

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