To solve this problem, we need to determine the 6th term [tex]\( x_6 \)[/tex] of a sequence defined such that:
[tex]\[ x_n = x_{n-1} + x_{n-2} \quad \text{for all } n \ge 3 \][/tex]
We are given that:
[tex]\[ x_{11} - x_1 = 99 \][/tex]
To find [tex]\( x_6 \)[/tex], let’s define the sequence starting with [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex] as the first two terms:
1. [tex]\( x_1 \)[/tex]
2. [tex]\( x_2 \)[/tex]
3. [tex]\( x_3 = x_1 + x_2 \)[/tex]
4. [tex]\( x_4 = x_2 + x_3 = x_2 + (x_1 + x_2) = x_1 + 2x_2 \)[/tex]
5. [tex]\( x_5 = x_3 + x_4 = (x_1 + x_2) + (x_1 + 2x_2) = 2x_1 + 3x_2 \)[/tex]
6. [tex]\( x_6 = x_4 + x_5 = (x_1 + 2x_2) + (2x_1 + 3x_2) = 3x_1 + 5x_2 \)[/tex]
Now let’s continue this sequence until [tex]\( x_{11} \)[/tex]:
7. [tex]\( x_7 = x_5 + x_6 = (2x_1 + 3x_2) + (3x_1 + 5x_2) = 5x_1 + 8x_2 \)[/tex]
8. [tex]\( x_8 = x_6 + x_7 = (3x_1 + 5x_2) + (5x_1 + 8x_2) = 8x_1 + 13x_2 \)[/tex]
9. [tex]\( x_9 = x_7 + x_8 = (5x_1 + 8x_2) + (8x_1 + 13x_2) = 13x_1 + 21x_2 \)[/tex]
10. [tex]\( x_{10} = x_8 + x_9 = (8x_1 + 13x_2) + (13x_1 + 21x_2) = 21x_1 + 34x_2 \)[/tex]
11. [tex]\( x_{11} = x_9 + x_{10} = (13x_1 + 21x_2) + (21x_1 + 34x_2) = 34x_1 + 55x_2 \)[/tex]
Given that:
[tex]\[ x_{11} - x_1 = 99 \][/tex]
Therefore:
[tex]\[ 34x_1 + 55x_2 - x_1 = 99 \][/tex]
[tex]\[ 33x_1 + 55x_2 = 99 \][/tex]
Solving for [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex], let's represent this equation in a more standardized algebraic form:
[tex]\[ 33x_1 + 55x_2 = 99 \][/tex]
This can be simplified by dividing the entire equation by 11:
[tex]\[ 3x_1 + 5x_2 = 9 \][/tex]
We can solve this equation for the specific values of [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex]:
[tex]\[ x_6 = 3x_1 + 5x_2 \][/tex]
From the equation [tex]\( 3x_1 + 5x_2 = 9 \)[/tex], it is clear that:
[tex]\[ x_6 = 9 \][/tex]
Thus, the value of [tex]\( x_6 \)[/tex] is:
[tex]\[ x_6 = 9 \][/tex]
