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June 15, 2021 10:41 am
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Staff 
(@armor)
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In the increasing sequence \left \{ a_{n} \right \}  satisfies a_{n+2}=a_{n+1}+a_{n}  when n\geq 1,a_{7}=120.  Find the value of  a_{8}.

 

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June 15, 2021 10:41 am
Topic starter
Staff 
(@armor)
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  Honorable Member
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Let \small a_{1}=x,   \small a_{2}=y,   \small x,y\in \mathbb{N}.  From the given condition, we have \small x<y,  and \small a_{3}=x+y,   \small a_{4}=x+2y,   \small a_{5}=2x+3y,   \small a_{6}=3x+5y,   \small a_{7}=5x+8y,   \small a_{8}=8x+13y.  Since \small a_{7}=120,  then  \small 5x+8y=120.  We have 

\small x=8t

\small y=15-5t

where t is an integer number. Since  \small y> x> 0,  then   \small 15t-5t>8t>0.    Thus \small 0<t<\frac{15}{13}.    Then \small t=1.

Hence x=8,y=10. Therefore \small a_{8}=8*8+13*10=194.

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