+101 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.
+129852 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.....!!!!
+23252 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.
+2863 Set up equatiojn
with var a and b
(a) * (a + b) * (a + 2b) = 64
Expand
a^3 + a^2*b + a^2*2b + ab^2 = 64
graph it:
https://www.desmos.com/calculator/fkcrxzgdfy
lowest possible value for b is... 4...
4, 4, 4, is an arithmetic sequence with a common difference of 0.
ok bye bye
