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

1 Answer
Jan 31, 2017

Answer:

See explanation.

Explanation:

Let's let the two numbers in question be #a# and #b#.

The arithmetic mean #M_A# of #n# numbers is the familiar idea of a mean—it's the sum of the numbers, divided by how many there are. In this case,

#M_A=(a+b)/2#.

Since #M_A# is halfway between #a# and #b#, #M_A# is just as far from #a# as it is from #b#. In math terms:

#M_A-a" "=" "b-M_A#

The geometric mean #M_G# of #n# numbers is the product of the numbers, taken to the #n^"th"# root. In this case,

#M_G=sqrt(a xx b)#

Under this definition, the proportional change from #a# to #M_G# is the same as the proportional change from #M_G# to #b#. In math terms:

#M_G/a=b/M_G#

The harmonic mean #M_H# of #n# numbers is... the reciprocal of the mean of the reciprocals of the numbers. Wow, that's a mouthful. Basically, you take the reciprocals of the numbers, find their (arithmetic) mean, and take the reciprocal of this mean. In this case,

#M_H = [(1/a+1/b)/2]^"-1"" "=" "2/(1/a+1/b)#

#1/M_H# is halfway between #1/a# and #1/b#. In math terms:

#1/M_H-1/a" "=" "1/b-1/M_H#.

On with the proof!

To prove that #M_A, M_G,# and #M_H# are in geometric sequence, we need to show that there is a common ratio #R# between them.

First, let's try #M_G/M_A#. This is equal to

#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#:

#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 #M_G/M_A = M_H/M_G=R#. This means #M_A xx R = M_G#, and #M_G xx R = M_H#, and so these 3 means are consecutive terms in a geometric sequence, in the order #M_A, M_G, M_H#, with common ratio #R=(2sqrt(ab))/(a+b).#