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)?

1 Answer
Jan 8, 2018

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?