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