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# brain teaser!

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If the 5th and the 8th entries of a Fibonacci Sequence are 19 and 79, respectively, then what is the first entry of this sequence?

May 14, 2019

#1
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A general Fibonacci sequence is defined as: $$a_n = a_{n-1} + a_{n-2}$$

We can use what was given, that $$a_5 = 15$$ and $$a_8 = 79$$ to solve for $$a_1$$

$$a_6 + a_7 = a_8 \rightarrow a_6 + a_7 = 79$$

$$a_5 + a_6 = a_7 \rightarrow 19 + a_6 = a_7$$

Substituting for $$a_7$$ in the top equation , we can see $$a_6 + (a_6 + 19) = 79$$

$$2a_6 = 60 \rightarrow a_6 = 30$$

This part gets a lil repetitive

$$a_{4} + {19} = {30} \rightarrow a_{4} = {11}$$

$$a_{3} + {11} = {19} \rightarrow a_{3} = {8}$$

$$a_{2} + {8} = {11} \rightarrow a_{2} = {3}$$

$$a_{1} + {3} = {8} \rightarrow a_{1} = {5}$$

Therefore, the first entry of this sequence is 5.

May 14, 2019
#2
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If the 5th and the 8th entries of a Fibonacci Sequence are 19 and 79, respectively, then what is the first entry of this sequence?

List of Fibonacci numbers with $$a_0 = 0$$ and $$a_1 = 1$$:

$$\small{ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \ldots& F_{-2} & F_{-1} & F_0 & F_1 & F_2 & F_3 & F_4 & F_5 & F_6 & F_7 & F_8 &F_9 &F_{10} &F_{11} &F_{12} &F_{13} &F_{14} &F_{15}&\ldots \\ \hline \ldots& -1 & 1 & 0 & 1 & 1 & 2 & 3 & 5 & 8 & 13 & 21 & 34 & 55 & 89 & 144 & 233 & 377 & 610&\ldots \\ \hline \end{array} }$$

In general:

$$\begin{array}{|rcll|} \hline a_n &=& F_na_1+F_{n-1}a_0 \quad | \quad \text{with } a_0=0 \text{ and } a_1 = 1,\quad \text{yield}\quad a_n =F_n \\ \hline \end{array}$$

$$\mathbf{a_1=\ ?}$$

$$\begin{array}{|lrcll|} \hline n=5: & F_5a_1+F_4a_0 &=& a_5 \quad | \quad a_5 = 19 \\ n=8: & F_8a_1+F_7a_0 &=& a_8 \quad | \quad a_8 = 79 \\ \hline & \mathbf{F_5a_1+F_4a_0} &=& \mathbf{ 19 } \qquad (1) \\ & a_0 &=& \dfrac{19-F_5a_1}{F_4} \quad | \quad F_4=3,\ F_5=5 \\ & a_0 &=& \dfrac{19-5a_1}{3} \\\\ & \mathbf{F_8a_1+F_7a_0} &=& \mathbf{79} \qquad (2) \\ & F_8a_1+F_7\cdot \left( \dfrac{19-5a_1}{3} \right) &=& 79 \quad | \quad F_7=13,\ F_8=21 \\ & 21a_1+13\cdot \left( \dfrac{19-5a_1}{3} \right) &=& 79 \quad | \quad \cdot 3 \\ & 63a_1+13\cdot ( 19-5a_1 ) &=& 237 \\ & 63a_1+247 - 65a_1 &=& 237 \\ & -2a_1 &=& -10 \\ & 2a_1 &=& 10 \\ & \mathbf{ a_1 } &=& \mathbf{ 5 } \\ \hline \end{array}$$

The first entry of this sequence is 5

May 14, 2019
edited by heureka  May 14, 2019
#3
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Thanks Anthrax and heureka!!

May 14, 2019