# How do you prove that arithmetic mean, geometric mean and harmonic mean of two numbers are in geometric sequence?

##### 1 Answer

#### 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 #R#.*)

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