1.

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.

Answer»

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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