The sequence $x_1$, $x_2$, $x_3$, . . ., has the property that $x_n = x_{n - 1} + x_{n - 2}$ for all $n \ge 3$. If $x_{11} - x_1 = 99$, then

The sequence x_1, x_2, x_3, . . ., has the property that x_n = x_{n - 1} + x_{n - 2}

for all n \ge 3.

If x_{11} - x_1 = 99, then determine x_6.

\(\begin{array}{|rcll|} \hline x_3 &=& x_2 + x_1 &=& 1\cdot x_2 + 1\cdot x_1 \\ x_4 &=& x_3 + x_2 &=& 2\cdot x_2 + 1\cdot x_1 \\ x_5 &=& x_4 + x_3 &=& 3\cdot x_2 + 2\cdot x_1 \\ \color{red}x_6 &\color{red}=& x_5 + x_4 &=& \color{red}5\cdot x_2 + 3\cdot x_1 \\ x_7 &=& x_6 + x_5 &=& 8\cdot x_2 + 5\cdot x_1 \\ x_8 &=& x_7 + x_6 &=& 13\cdot x_2 + 8\cdot x_1 \\ x_9 &=& x_8 + x_7 &=& 21\cdot x_2 + 13\cdot x_1 \\ x_{10} &=& x_9 + x_8 &=& 34\cdot x_2 + 21\cdot x_1 \\\\ x_{11} &=& x_{10} + x_9 &=& 55\cdot x_2 + 34\cdot x_1 \quad & | \quad x_{11} = 99 + x_1 \\ 99 + x_1 &=& 55\cdot x_2 + 34\cdot x_1 && \quad & | \quad - x_1 \\ 99 &=& 55\cdot x_2 + 33\cdot x_1 && \quad & | \quad : 11 \\ \color{red}9 &\color{red}=& \color{red}5\cdot x_2 + 3\cdot x_1 \\\\ \hline \color{red}9 &\color{red}=& \color{red}5\cdot x_2 + 3\cdot x_1 && \quad & | \quad \color{red}x_6 = 5\cdot x_2 + 3\cdot x_1\\ \mathbf{9} &\mathbf{=}& \mathbf{x_6} \\ \hline \end{array} \)

\($x_n = x_{n - 1} + x_{n - 2}$ for all $n \ge 3$\)

x₁₁  =  99 + x₁

x₉ + x₁₀  =  99 + (x₃ - x₂)

(x₇ + x₈) + (x₈ + x₉)  =  99 + (x₅ - x₄) - (x₃ - x₁)

(x₅ + x₆) + (x₆ + x₇) + (x₆ + x₇) + (x₇ +  x₈)  =  99 + x₅ - x₄ - x₃ + x₁

3x₆ + 3x₇ + x₈  =  99  - x₄ - x₃ + x₁

3x₆  +  3(x₅ + x₆)  + (x₆ + x₇)  = 99 - (x₆ - x₅) - ( x₅ - x₄) + (x₃ - x₂)

7x₆  + 3x₅ + x₇  = 99  -x₆ +x₄ + x₃ - x₂

8x₆ + 3(x₇ - x₆) + x₇  = 99  + x₄ +( x₅ - x₄) - (x₄ - x₃)

5x₆ + 4x₇  =  99 + x₅ - x₄ + x₃

5x₆ + 4(x₅ + x₆)  = 99 + x₅ - (x₆ - x₅) + (x₅ - x₄)

10x₆ + 4x₅  =  99 + 3x₅ -x₄

10x₆ + x₅ = 99 - x₄

10x₆ + x₅ + x₄  = 99

10x₆  + x₆  =  99

11x₆  = 99             divide both sides by 11

x₆  = 9