Can You Solve This Tricky Geometric Sequence Problem?

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Discussion Overview

The discussion revolves around a geometric sequence problem where participants are tasked with finding the sequence given that the sum of the first three terms is 7 and the sum of their cubes is 73. The conversation includes attempts to derive the sequence, share insights, and clarify steps involved in the solution process.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents the problem and asks for help in finding the geometric sequence.
  • Another suggests simplifying calculations by defining the second term of the sequence as "x".
  • A participant expresses frustration at getting stuck in the calculations despite knowing the answer through trial and error.
  • Some participants discuss the implications of using "x" as the second term, noting that it leads to expressions for the first and third terms as x/r and xr, respectively.
  • One participant admits to initially misreading "geometric" as "arithmetic" and speculates on the value of the common ratio r, suggesting it must be slightly more than 73/7.
  • A participant claims to have found the sequence 1, 4, and 2, but acknowledges the challenge of documenting the steps taken to arrive at that answer.
  • Another participant humorously suggests that writing a program to brute force the solution could count as a valid approach.

Areas of Agreement / Disagreement

Participants express differing views on the approach to solving the problem, with no consensus on the best method or the correctness of the proposed sequences. The discussion remains unresolved regarding the exact sequence and the steps to derive it.

Contextual Notes

There are limitations in the discussion, including potential misinterpretations of the problem type (geometric vs. arithmetic) and the lack of systematic methods presented for finding the solution. Some participants express uncertainty about the calculations and the reasoning behind their proposed solutions.

midododo11
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In a geometric sequence, the sum of the first three terms is 7
and the sum of the cubes of the first three terms is 73
find the sequence and how did you get it​
 
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welcome to pf!

hi midododo11! welcome to pf! :smile:

hmm :rolleyes: … let's try to simplify the calculation by making it as symmetric as possible, sooo …

hint: call the second term of the sequence "x" :wink:​
 


tiny-tim said:
hi midododo11! welcome to pf! :smile:

hmm :rolleyes: … let's try to simplify the calculation by making it as symmetric as possible, sooo …

hint: call the second term of the sequence "x" :wink:​

Hi tiny-tim!
I tried a lot and I always get stuck at the last step, I know the answer but I got it by trial of numbers
Take a look
[URL]http://latex.codecogs.com/gif.latex?\large&space;\\a=first&space;\term&space;\\r=&space;common\ratio&space;\\a+ar+ar^2=7&space;\&space;Equation&space;(1)&space;\\a^3+a^3r^3+a^3r^6=73&space;\&space;Equation&space;(2)&space;\\by\&space;dividing\&space;1&space;/&space;2&space;\\\frac{a(1+r+r^2)}{a^3(1+r^3+r^6)}=\frac{7}{73}&space;\\\frac{1+r+r^2}{a^2(1+r^3+r^6)}=\frac{7}{73}
​
[/URL]
 
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don't use a = first, use x = second
 
Sorry but I don't see the difference if used x as a second term the first term will be x/r and the third xr
 
1 2 4
 
midododo11 said:
Sorry but I don't see the difference if used x as a second term the first term will be x/r and the third xr

i've no idea why, but I've been misreading "geometric" as "arithmetic" …

maybe i was put off by the colour? :redface:

i don't think there's any systematic way of finding the answer, except possibly to say that r4 must obviously be a bit more than 73/7 … so r must be 2!​
 
tiny-tim said:
i've no idea why, but I've been misreading "geometric" as "arithmetic" …

maybe i was put off by the colour? :redface:

i don't think there's any systematic way of finding the answer, except possibly to say that r4 must obviously be a bit more than 73/7 … so r must be 2!​

Ok, thanks
 
I got 1,4, and 2...?
 
  • #10
ILoveScience said:
I got 1,4, and 2...?
yeah but it's required to write the steps that's the difficult part
 
  • #11
midododo11 said:
yeah but it's required to write the steps that's the difficult part

does writing the program to calculate it by brute force count? :P
 

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