SOLUTION: In a sequence of positive integers, each term is larger than the previous term. Also, after the first two terms, each term is the sum of the previous two terms. The eighth term o

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Question 730737: In a sequence of positive integers, each term is larger than the previous term. Also, after the
first two terms, each term is the sum of the previous two terms.
The eighth term of the sequence is 390. What is the ninth term?
Answer by fcabanski(1391) About Me  (Show Source):
You can put this solution on YOUR website!
Call the first term x and the second term y.


1: x


2: y


3: x+y


4: x+y + y = x+2y (added the previous two terms).


5: 2x+3y


6: 3x+5y


7: 5x+8y


8: 8x + 13y = 390
9: 13x+21y


The problem states that the second term, y, is greater than the first term, x. They are both positive and both integers.


Try a few: y = x+1 gives term 8 as:


8x + 13(x+1) = 390


8x + 13x + 13 = 390


21x = 377


x = 377/21 which is not an integer. So y cannot be x+1.


Try y = x+2:


8x + 13(x+2) = 390


8x +13x +26 = 390


21x = 364


364/21 = not an integer.


Now there's a pattern. In each case (y+n), x= (390-13n)/21. It must be an integer.


n = 3, 4, 5, 6, 7, 8 don't work. But n=9 works.


8x + 13(x+9)= 390


8x+13x + 117 = 390


21x = 273


x = 273/21 = (390-13*9)/21 = 13.


x = 13 and y = 13+9 = 22


The 9th term is 13x + 21y = 13(13)+21(22) = 169 + 462 = 631

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