Prove that the product of n geometric means between two quantities is equal to the nth power of a geometric mean of those two quantities.

Let us suppose a and b are two numbers.


Let us say G is a number that is the Geometric mean of a and b


Therefore a, G and b must be in Geometric Progression or GP.


This means, common ratio = G/a = b/G


Or, G2 = ab


Or, G = ?(ab)... (1)


Now, let us say G1 , G2 , G3 , .......Gn are n geometric means between a and b.


Which means that


a , G1 , G2 , G3 ...... Gn , b form a G.P.


Note that the above GP has n+2 terms and the first term is a and the last term is b, which is also the (n+2)th term


Hence, b = arn+2–1


where a is the first term.


So,


b = arn+1


r = (b/a)1/n+1 ....(2)


Now the product of GP becomes


Product = G1G2G3......Gn


= (ar)(ar2)(ar3).....(arn)


= an r(1+2+3+4+.............+n)


= an rn(1+n)/2


Putting the value of r from equation 2 , we get


= an (b/a)n(1+n)/2(n+1)


= (ab)n/2


= (?ab)n


Now, putting the value from equation 1, we get,


Product = Gn


Or, G1G2G3......Gn = Gn


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