How do you prove that arithmetic mean, geometric mean and harmonic mean of two numbers are in geometric sequence?
1 Answer
See explanation.
Explanation:
Let's let the two numbers in question be
The arithmetic mean
#M_A=(a+b)/2# .
Since
#M_A-a" "=" "b-M_A#
The geometric mean
#M_G=sqrt(a xx b)#
Under this definition, the proportional change from
#M_G/a=b/M_G#
The harmonic mean
#M_H = [(1/a+1/b)/2]^"-1"" "=" "2/(1/a+1/b)#
#1/M_H-1/a" "=" "1/b-1/M_H# .
On with the proof!
To prove that
First, let's try
#M_G/M_A=sqrt(a xx b)/((a+b)//2)=(2sqrt(ab))/(a+b)" "-=" "R#
(Let's denote this ratio
Now let's try
#M_H/M_G = (2//(1/a+1/b))/sqrt(a xx b)#
#color(white)(M_H/M_G) = (2)/((1/a+1/b) sqrt(ab))color(blue)(xx sqrt(ab)/sqrt(ab))#
#color(white)(M_H/M_G) = (2sqrt(ab))/((1/a+1/b)ab#
#color(white)(M_H/M_G) = (2sqrt(ab))/((ab)/a+(ab)/b)#
#color(white)(M_H/M_G) = (2sqrt(ab))/(b+a)" "=" "(2sqrt(ab))/(a+b)" "=" "R#
And hey, presto—we're done! We've shown that