Question
Wed November 09, 2011 By:

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.

Expert Reply
Thu November 10, 2011
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 geomteric 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 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......G= Gn
 
Thank you
 
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