Can You Solve This Tricky Geometric Sequence Problem?

[midododo11]

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​

 

[tiny-tim]

welcome to pf!

hi midododo11! welcome to pf!

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

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

 

[midododo11]

hi midododo11! welcome to pf!

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

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

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

Spoiler

[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]

 

[tiny-tim]

don't use a = first, use x = second

 

[midododo11]

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

 

[Caracrist]

Spoiler

1 2 4

 

[tiny-tim]

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?

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!​

 

[midododo11]

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

maybe i was put off by the colour?

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

 

[ILoveScience]

I got 1,4, and 2...?

 

[midododo11]

I got 1,4, and 2...?

yeah but it's required to write the steps that's the difficult part

 

[sourlemon]

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