Solving a+ar+ar^2=7 and a^3+a^3r^3+a^3r^6=73

[santa]

find a and r

$$a+ar+ar^2=7$$

$$a^3+a^3r^3+a^3r^6=73$$

 

[John Creighto]

find a and r

$$a+ar+ar^2=7$$

$$a^3+a^3r^3+a^3r^6=73$$

http://www.imf.au.dk/kurser/algebra/E05/GBintro.pdf

 

[santa]

OK but the solution where

 

[dodo]

The first trial-and-error attempt I made turned out to be right: a=1, r=2,
1 + 2 + 4 = 7,
1 + 8 + 64 = 73.

 

[Diffy]

Not sure if there is more than one solution, but I found that a = 1 and r =2 works.

 

[qspeechc]

a(1 + r + r^2) = 7

Then you can solve for 'a' or 'r', and substitute into the other equation. It's not pretty, but it will work!

I remember seeing a very similar problem in high school, must see if I can find it again.

In fact, you could try by summing geometric equations.

a + ar + ar^2 = a(1 - r^3)/(1 - r) = 7

a^3 + a^3.r^3 + a^3.r^6 = a^3(1 - r^9)/(1 - r^3) = 73

Divide the one sum by the other and see where it leads you.