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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.

 Mar 24, 2020
 #1
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0

The AP is; 2, 4, 8   and b = 4

 Mar 24, 2020
 #2
avatar+111330 
0

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.....!!!!

 

cool cool cool

 Mar 24, 2020
 #3
avatar+21017 
+1

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.

 Mar 24, 2020
 #5
avatar+111330 
0

I'll  buy  that, geno    !!!!

 

Did  you find this mathematically  or by "trial and error???"

 

cool cool cool

CPhill  Mar 24, 2020
 #4
avatar+2854 
0

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

 Mar 24, 2020

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