Finding Ratio of Geometric Sequence

[dxoo]

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.

 

[yoscar04]

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.

Try to solve for r: t₃=ar²=a+ar.

 

[JayJay]

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.

Your method was OK but you made an error multiplying (3a + 5ar)(a + ar).

 

[HallsofIvy]

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

Good idea! So $\displaystyle \frac{3a+ 5b}{2a+ 3b}= \frac{a+ 2b}{a+ b}$
$\displaystyle (3a+ 5b)(a+ b)= (a+ 2b)(2a+ 3b)$.
$\displaystyle 3a^2+ 8ab+ 5b^2= 2a^2+ 7ab+ 6b^2$
$\displaystyle a^2+ ab- b^2= 0$.

Hmm, that is NOT what you have below! I can see how you might have got "$\displaystyle 5a^2$, with a sign error, but I can't imagine how you got "$\displaystyle 8a^2$".

Using the quadratic formula, $\displaystyle a= \frac{-b\pm\sqrt{b^2+ 4b^2}}{2}=\frac{-b\pm\sqrt{5b^2}}{2}= \frac{-1\pm\sqrt{5}}{2}b$

[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.

 

[dxoo]

Thank you all for your responses! I ended up using a different, easier method to find [MATH]r[/MATH] which started off from yoscar04's response:

[MATH] t_{3} = ar^2 = a+ ar\\ t_{4} = ar^3 = a + 2ar [/MATH]
Divide second equation by first:

[MATH] \frac{ar^3}{ar^2} = r = \frac{a+2ar}{a+ar}\\ r = \frac{1+2r}{1+r}\\ r^2-r-1 =0\\ r = \frac{1\pm\sqrt5}{2} [/MATH]
I skipped a few steps in the above calculation, but hopefully you can work out the solution.
However, in the problem statement, all terms are positive and hence, we discover:

[MATH] r = \frac{1+\sqrt5}{2} [/MATH]
Thank you all for your help!

 

[yoscar04]

Thank you all for your responses! I ended up using a different, easier method to find [MATH]r[/MATH] which started off from yoscar04's response:

[MATH] t_{3} = ar^2 = a+ ar\\ t_{4} = ar^3 = a + 2ar [/MATH]
Divide second equation by first:

[MATH] \frac{ar^3}{ar^2} = r = \frac{a+2ar}{a+ar}\\ r = \frac{1+2r}{1+r}\\ r^2-r-1 =0\\ r = \frac{1\pm\sqrt5}{2} [/MATH]
I skipped a few steps in the above calculation, but hopefully you can work out the solution.
However, in the problem statement, all terms are positive and hence, we discover:

[MATH] r = \frac{1+\sqrt5}{2} [/MATH]
Thank you all for your help!

Actually the shortest way is to solve for r using the equation for t₃: r²=1+r.

 

[Deleted member 4993]

To beat a dead horse:

You have:

t₂=ar

t₃=a+ar

r = t₃/t₂ = (1+r)/r

r² - r - 1 = 0

$\displaystyle r_{1,2} \ = \ \frac{-(-1) \pm \sqrt{1^2-4*(1)*(-1)}}{2} \ = \ \frac{1 \pm \sqrt{5}}{2}$

This the famous "golden ratio" https://www.mathsisfun.com/numbers/golden-ratio.html

 

[Steven G]

Why would you complicate things by using t₆/t₅=t₄/t₃?

How about t₂/t₁ which clearly equals r and t₃/t₂ = (a + ar)/ (ar) = 1/r + 1

So r = 1/r + 1. This is easily solved and well-known to be (1-sqrt(5))/2 -- the 'golden ratio'.

 

[mmm4444bot]

Why would you complicate things by using t₆/t₅=t₄/t₃? …

The op states that choice was arbitrary.

?

 

[Steven G]

The op states that choice was arbitrary.

?

Yes, by why choose more complicated ones? I never had to cross multiply with my choice. I am just pointing out using t1, t2 and t3 will always be the easier three to pick.

 

[mmm4444bot]

… why choose more complicated ones? …

The choice was arbitrary, Jomo. They did not choose to do it hard; they tried an idea that came to mind, and, after getting feedback, they went in a different direction.

Are you choosing to make this complicated?

$\;$