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Official PYQ · 2023

GATE CS 2023 Q3 — General Aptitude

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A series of natural numbers F1,F2,F3,F4,F5,F6,F7,…F_1, F_2, F_3, F_4, F_5, F_6, F_7, \ldots obeys Fn+1=Fn+Fn−1F_{n+1}=F_n+F_{n-1} for all integers n≥2n \geq 2.

If F6=37,F_6=37, and F7=60,F_7=60, then what is F1?F_1 ?

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Answer: A

The given recurrence relation is:

Fn+1=Fn+Fn−1for n≥2F_{n+1} = F_n + F_{n-1} \quad \text{for } n \geq 2

We can rewrite this relation to find the previous term in the sequence:

Fn−1=Fn+1−FnF_{n-1} = F_{n+1} - F_n

We are given F7=60F_7 = 60 and F6=37F_6 = 37. Let us work backwards to find F1F_1:

  • For n=6n = 6:
    F5=F7−F6=60−37=23F_5 = F_7 - F_6 = 60 - 37 = 23
  • For n=5n = 5:
    F4=F6−F5=37−23=14F_4 = F_6 - F_5 = 37 - 23 = 14
  • For n=4n = 4:
    F3=F5−F4=23−14=9F_3 = F_5 - F_4 = 23 - 14 = 9
  • For n=3n = 3:
    F2=F4−F3=14−9=5F_2 = F_4 - F_3 = 14 - 9 = 5
  • For n=2n = 2:
    F1=F3−F2=9−5=4F_1 = F_3 - F_2 = 9 - 5 = 4

Thus, the value of F1F_1 is 4.

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