Let a, b, c be a three-term arithmetic series where all the terms are positive, such that abc = 64. Find the smallest possible value of b.
That's impossible, guest.....an arithmetic series will have the same difference between terms......
FWIW....I put this into WolframAlpha......it didn't find any integer solutions for this where all the integers were positive
It came back with
- 8 , -2, 4
Which has a constant difference of 6
Maybe someone else knows how to find the positive solutions.....!!!!
Try these: [ 'cubrt' means 'cube root' and 'sqrt' means 'square root' ]
a = 4·cubrt(2) - sqrt(8)·cubrt(2)
b = 4·cubrt(2)
c = 4·cubrt(2) + sqrt(8)·cubrt(2)
I don't know if this is the smallest value of b, but it works.
Set up equatiojn
with var a and b
(a) * (a + b) * (a + 2b) = 64
a^3 + a^2*b + a^2*2b + ab^2 = 64
lowest possible value for b is... 4...
4, 4, 4, is an arithmetic sequence with a common difference of 0.
ok bye bye