Question

Between two positive real nos. a and b there are 2 G.M's G_1 and G_2 and a single A.M A.The G.M between G_1 and G_2 is M.Can you prove that M^3=ab(A^3)?

Answer 1

Counterexample

Explanation:

If I understand the problem right, this result is not true.

Example: Let `a = 2 and b = 128`.
Then `A = (2 + 128)/2 = 65` [a single arithmetic mean]
The common ratio for the geometric means is r = 4, and
`G_1 = 2*4 = 8`
`G_2 = 8*4 = 32`
Of course `b = 32*4 = 128 = b`.

Now the Geometric Ratio between `G_1 and G_2` is 2,
so that `M = 2*8 = 16`, and of course `2*16 = G_2`.

`M^3 = 16^3 = 4096`

But the right side is much larger, containing `65^3` times two integers.

Did I misunderstand what you asked?