If the arithmetic mean of two numbers, # a and b, a > b > 0,# is five times their geometric mean, then what is #(a+b)/(a-b)# equal to?

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Answer 1 · 2018-03-18T18:56:05

#(a+b)/(a-b) = (5sqrt(6))/12#

From the information we are given:

#1/2(a+b) = 5sqrt(ab)#

Multiply both sides by #2# to get:

#a+b = 10sqrt(ab)#

Square both sides to get:

#(a+b)^2 = a^2+2ab+b^2 = 100ab#

Subtract #4ab# from both ends to get:

#(a-b)^2 = a^2-2ab+b^2 = 96ab#

So:

#((a+b)/(a-b))^2 = (100ab)/(96ab) = 25/24 = (5^2 * 6)/12^2 = ((5sqrt(6))/12)^2#

So since #a > b# we can take the positive square root to find:

#(a+b)/(a-b) = (5sqrt(6))/12#