Question

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

Answer 1

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).`