Problem:
The increasing sequence of positive integers a1,a2,a3,… has the property that an+2=an+an+1 for all n≥1. If a7=120, then a8 is
Answer Choices:
A. 128
B. 168
C. 193
D. 194
E. 210
Solution:
If a1=a and a2=b then
(a3,a4,a5,a6,a7,a8)=(a+b,a+2b,2a+3b,3a+5b,5a+8b,8a+13b)
Therefore 5a+8b=a7=120. Since 5a=8(15−b) and 8 is relatively prime to 5,a must be a multiple of 8. Similarly, b must be a multiple of 5. Let a=8j and b=5k to obtain 40j+40k=120, which has two solutions in positive integers, (j,k)=(1,2) and (2,1). Since the sequence is increasing, (j,k)=(1,2). Thus a=8⋅1=8 and b=5⋅2=10, so a8=8a+13b=194.