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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.

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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