Question 1164687: NEED HELP
How do I make a formula for an increasingly decreasing sequence?
eg: 50,49,47,44,40,35,...
and how do I find the next three terms of the sequence?
Found 2 solutions by Theo, Edwin McCravy:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i found what seems to work.
formula would be An = A1 - sum of (i - 1) for i = 1 to n.
formula would work like this:
when n = 1, An = A1 - sum of (i - 1) for i = 1 to n becomes:
A1 = 50 minus sum of (i - 1) for i = 1 to 1 which becomes:
A1 = 50 minus 0 which becomes:
A1 = 50
when n = 2, An = A1 - sum of (i - 1) for i = 1 to n becomes:
A2 = 50 minus sum of (i - 1) for i = 1 to 2 which becomes:
A2 = 50 minus 0 minus 1 which becomes:
A2 = 50 minus 1 which becomes:
A2 = 49
when n = 3, An = A1 - sum of (i - 1) for i = 1 to n becomes:
A3 = 50 minus sum of (i - 1) for i = 1 to 3 which becomes:
A3 = 50 minus 0 minus 1 minus 2 which becomes:
A3 = 50 minus 3 which becomes:
A3 = 47
when n = 4, An = A1 - sum of (i - 1) for i = 1 to n becomes:
A4 = 50 minus sum of (i - 1) for i = 1 to 4 which becomes:
A4 = 50 minus 0 minus 1 minus 2 minus 3 which becomes:
A4 = 50 minus 6 which becomes:
A4 = 44
when n = 5, An = A1 - sum of (i - 1) for i = 1 to n becomes:
A5 = 50 minus sum of (i - 1) for i = 1 to 5 which becomes:
A5 = 50 minus sum of 0 minus 1 minus 2 minus 3 minus 4 which becomes:
A5 = 50 minus 10 which becomes:
A5 = 40
when n = 6, An = A1 - sum of (i - 1) for i = 1 to n becomes:
A6 = 50 minus sum of (i = 1) for i = 1 to 6 which becomes:
A6 = 50 minus 0 minus 1 minus 2 minus 3 minus 4 minus 5 which becomes:
A6 = 50 minus 15 which becomes:
A6 = 50 - 15 which becomes:
A6 = 35
you can see that this formula gets you the number you are looking for, which are:
A1 = 50
A2 = 49
A3 = 47
A4 = 44
A5 = 40
A6 = 35
finding the next 3 terms in the sequence is simply applying the formula.
when n = 7, An = A1 - sum of (i - 1) for i = 1 to n becomes:
A7 = 50 minus sum of (i - 1) for i = 1 to 7 which becomes:
A7 = 50 minus 0 minus 1 minus 2 minus 3 minus 4 minus 5 minus 6 which becomes:
A7 = 50 minus 21 which becomes:
A7 = 29
when n = 8, An = A1 - sum of (i - 1) for i = 1 to n becomes:
A8 = 50 minus sum of (i - 1) for i = 1 to 8 which becomes:
A8 = 50 minus 0 minus 1 minus 2 minus 3 minus 4 minus 5 minus 6 minus 7 which becomes:
A8 = 50 minus 28 which becomes:
A8 = 22
when n = 9, An = A1 - sum of (i - 1) for i = 1 to n becomes:
A9 = 50 minus sum of (i - 1) for i = 1 to 9 which becomes:
A9 = 50 minus 0 minus 1 minus 2 minus 3 minus 4 minus 5 minus 6 minus 7 minus 8 which becomes:
A9 = 50 minus 36 which becomes:
A9 = 14
to confirm the formula gave you the correct answer, you could do the following.
you are always subtracting 1 more from the previous number than you subtracted from the number preceding that.
50 minus 0 = 50 for n = 1
50 minus 1 = 49 for n = 2
49 minus 2 = 47 for n = 3
47 minus 3 = 44 for n = 4
44 minus 4 = 40 for n = 5
40 minus 5 = 35 for n = 6
35 minus 6 = 29 for n = 7
29 minus 7 = 22 for n = 8
22 minus 8 = 14 for n = 9
the formula here appears to be A.n+1 = A.n - (n-1)
you would start this sequence A1.
from there the formula would get you.
A2 = 50 - 1 = 49
A3 = 49 - 2 = 47
A4 = 47 - 3 = 44
A5 = 44 - 4 = 40
A6 = 40 - 5 = 35
A7 = 35 - 6 = 29
A8 = 29 - 7 = 22
A9 = 22 - 8 = 14
either way, you can see that A7 will be 29 andA8 will be 22 and A9 will be 14.
Answer by Edwin McCravy(20056)  (Show Source):
You can put this solution on YOUR website! 50,49,47,44,40,35,...
Make the set of first and second differences
50 49 47 44 40 35
-1 -2 -3 -4 -5
-1 -1 -1 -1
and since the second differences are constant, we
assume a 2nd degree polynomial in n for the general term:
[If the kth differences are constant, the the general term
will be a polynomial in n of degree k.
Then we substitute the first 3 terms for a1, a2, a3 and 1,2,3 for n
or
Solve that system and get
A = -1/2, B = 1/2, and C = 50
So the general term is
You can find the next three terms either by extending the
difference table:
50 49 47 44 40 35 29 22 14
-1 -2 -3 -4 -5 -6 -7 -8
-1 -1 -1 -1 -1 -1 -1
or by substituting 7, 8, and 9 in the general formula:
Edwin
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