Problem Statement:
In a geometric sequence of positive numbers, each term after the second is the sum of the two preceding terms. Find the common ratio.
My Approach:
[MATH] t_{1} = a\\ t_{2} = ar\\ t_{3} = a + ar\\ t_{4} = a + 2ar\\ t_{5} = 2a + 3ar\\ t_{6} = 3a + 5ar\\ t_{7} = 5a + 8ar\\ [/MATH]Then I arbitrarily equated the ratio of 4 different terms since it is geometric.
[MATH] (t_{6})/(t_{5})=(t_{4})/(t_{3})\\ (3a+5ar)/(2a+3ar) = (a+2ar)/(a+ar)\\ [/MATH]Let a = a and b = ar
[MATH] 8a^2+8ab+5b^2=2a^2+7ab+6b^2\\ 6a^2+ab-b^2\\ (3a-b)(2a+b)\\ b = 3a\\ ar = 3a\\ r = 3\\ OR\\ b = -2a\\ ar = -2a\\ r = -2\\ [/MATH]
However, none of these answers are correct since according to the source, the answer is [MATH]\frac{1+\sqrt{5}}{2}[/MATH]
It'd be really helpful for someone to let me know where I went wrong, or if my approach is outright incorrect. Thanks.
In a geometric sequence of positive numbers, each term after the second is the sum of the two preceding terms. Find the common ratio.
My Approach:
[MATH] t_{1} = a\\ t_{2} = ar\\ t_{3} = a + ar\\ t_{4} = a + 2ar\\ t_{5} = 2a + 3ar\\ t_{6} = 3a + 5ar\\ t_{7} = 5a + 8ar\\ [/MATH]Then I arbitrarily equated the ratio of 4 different terms since it is geometric.
[MATH] (t_{6})/(t_{5})=(t_{4})/(t_{3})\\ (3a+5ar)/(2a+3ar) = (a+2ar)/(a+ar)\\ [/MATH]Let a = a and b = ar
[MATH] 8a^2+8ab+5b^2=2a^2+7ab+6b^2\\ 6a^2+ab-b^2\\ (3a-b)(2a+b)\\ b = 3a\\ ar = 3a\\ r = 3\\ OR\\ b = -2a\\ ar = -2a\\ r = -2\\ [/MATH]
However, none of these answers are correct since according to the source, the answer is [MATH]\frac{1+\sqrt{5}}{2}[/MATH]
It'd be really helpful for someone to let me know where I went wrong, or if my approach is outright incorrect. Thanks.