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 February 23rd, 2018, 01:13 AM #1 Member   Joined: May 2015 From: Australia Posts: 77 Thanks: 7 Geometric sequences https://www.dropbox.com/s/s944t19skpf47d7/1.jpeg?dl=0 Hello everyone, I've attached the question in the link above. I've done part a,b,c but I'm struggling to solve part d. For part d, this is how I tried to solve it: I rearranged the equation from part c: 1/2(a5*b5)=(b6)(a6) (1/2)(a5)(b5)=(b6)(a6) b5/b6 = [2(a6)]/(a5) Then, I rearranged the equation given in part d: (a5)/(b5) = (a6)/(b6) a6 = [(a5)(b6)]/(b5) Then, I substituted this into the equation I rearranged from part c: b5/b6 = [2(a5)(b6)/(b5)]/(a5) b5/b6 = 2(b6)/(b5) If b5/b6 = r, then: r = 2/r r^2 = 2 r = square root 2 I'm not sure whether that's the correct way of solving it though. Last edited by skipjack; February 23rd, 2018 at 10:30 AM.
 February 23rd, 2018, 02:17 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,923 Thanks: 1123 Math Focus: Elementary mathematics and beyond Viewing the JPEG requires setting up an account. Please upload the image here.
February 23rd, 2018, 02:22 AM   #3
Member

Joined: May 2015
From: Australia

Posts: 77
Thanks: 7

I just attached the image here. I'm not sure if it will be clear.

Edit: the image isn't clear, sorry. I'm not sure how to make it clear. But just letting you know that you don't need an account to view the image in the link. Even though it says you need to make an account, you just need to close that box to view the image below it.
Attached Images
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Last edited by pianist; February 23rd, 2018 at 02:25 AM.

 February 23rd, 2018, 04:24 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Actually, from the first post, I got a message box saying that if I wanted to add an image I had to open an account but when I closed that box, I could see, and enlarge, the image. We have a series of papers with height $\displaystyle a_n$ and width $\displaystyle b_n$ such that we can make each paper by cutting the previous paper in half horizontally then swapping "height" and "width". That is the height of each paper is the width of the previous paper, $\displaystyle a_n= b_{n-1}$, and the width of each paper is half the height of the previous paper, $\displaystyle b_n= \frac{1}{2}a_{n-1}$. We have the additional requirement that all papers have the same "shape": $\displaystyle \frac{a_n}{b_n}= \frac{a_{n-1}}{b_{n-1}}$ and the problem is to find the ratio of the width of the 5th paper to the width of the 6th paper, $\displaystyle \frac{b_5}{b_6}$. Yes, I get your answer, $\displaystyle \sqrt{2}$. From the last requirement, we have $\displaystyle \frac{a_6}{b_6}= \frac{a_5}{b_5}$. From that, $\displaystyle \frac{b_5}{b_6}= \frac{a_5}{a_6}$. Replacing $\displaystyle a_5$ with $\displaystyle b_6$ and $\displaystyle a_6$ with $\displaystyle \frac{1}{2}b_5$, we have $\displaystyle \frac{\frac{1}{2}b_5}{b_6}= \frac{b_6}{b_5}$. Multiplying both sides by $\displaystyle 2b_5$ and dividing by $\displaystyle b_6$ we get $\displaystyle \frac{b_5^2}{b_6^2}= 2$ so that $\displaystyle \frac{b_5}{b_6}= \sqrt{2}$. Thanks from pianist

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