Question

The sequence x_1, x_2, x_3, . . ., has the property that x_n = x_{n - 1} + x_{n - 2} for all n>=3. If x_{11} - x_1 = 99, then determine x_6.

Answer 1

x3 = x1+x2
x4 = x2+x3 = x2 + x1 + x2 = x1 + 2x2
x5 = x3 + x4 = 2x1+3x2
x6 = x4+x5 = 3x1+5x2
xn = F(n-3)x1 + F(n-2)x2

x11 = F8*x1 + F7*x2 = 34x1 + 21x2
x11-x1 = 33x1+21x2

Since 33 divides 99, we can try x1=3, x2=0, and we have

3,0,3,3,6,9,15,24,39,63,102

x6 = 15

Answer 2

that is wrong

Answer 3

Its 9.

Answer 4

Let $a = x_1$ and $b = x_2$. Then $x_3 = x_2 + x_1 = a + b$, and
\[x_4 = x_3 + x_2 = (a + b) + b = a + 2b.\]
We can express all terms up to $x_{11}$ similarly, in terms of $a$ and $b$:
\begin{align*}
x_5 &= 2a + 3b, \\
x_6 &= 3a + 5b, \\
x_7 &= 5a + 8b, \\
x_8 &= 8a + 13b, \\
x_9 &= 13a + 21b, \\
x_{10} &= 21a + 34b, \\
x_{11} &= 34a + 55b.
\end{align*}
Then $x_{11} - x_1 = (34a + 55b) - a = 33a + 55b$, so we must have $33a + 55b = 99$.
We can't find $a$ or $b$ right away, but we don't need to! We want $x_6$, which equals $3a+5b$.
Since $33a + 55b = 99$, dividing by 11 gives $3a + 5b = 9$, so $x_6 = \boxed{9}.$