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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
avatar+115 
+3

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
avatar+23170 
+2

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

 

laugh

 May 14, 2019
edited by heureka  May 14, 2019
 #3
avatar+200 
+2

Thanks Anthrax and heureka!!

 May 14, 2019

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