Let us index our own Fibonacci numbers $f_1,f_2,f_3,f_4,f_5,f_6,f_7,f_8,f_9,f_{10} = 2,3,5,8,13,21,34,55,89,144$. Use first three questions $Q_1= \; $"is $n$ smaller than $f_{10}=144$ ?", then $Q_2= \; $"is $n$ smaller than $f_9=89$ ?", then $Q_3= \; $"is $n$ smaller than $f_8=55$ ?". If answer $A_1 =$ Yes, we are done. If not, but answer $A_2 =$ No, we are restricted to the $54$ numbers $89 \to 143$, and we ask $Q_4= \; $"is $n$ smaller than $f_9 + f_7 =123$ ?". If not, but answer $A_2 =$ Yes, we are restricted to the $88$ numbers $1 \to 88$, and we ask $Q_4= \; $"is $n$ smaller than $f_7 =34$ ?". This tree is bound to find $n$, due to the relation $f_{k+2} = f_{k+1} + f_k$; on each branch we are left with exactly the right amount of questions. In fact it's a reverse induction, where the strategies for lesser Fibonacci numbers are just shifted (by other Fibonacci number(s)) according with the answer(s) received.

Nice generalization https://pdfs.semanticscholar.org/52dd/d95486b76fa233a6e1b3f855919e36f09485.pdf