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Show that the product of n geometric means between a and b is equal to the nth power of the single GM between a and b.

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To prove: Product of n geometric means between a and b is equal to the nth power of the single GM between a and b.

Formula used:

(i) Geometric mean between a and b = \(\sqrt{ab}\)

(ii) sum of n terms of A.P. = \(\frac{(n)(n+1)}{2}\)

Let the n geometric means between and b be G1, G2, G3, … Gn 

Hence a, G1, G2, G, … Gn, b are in GP 

⇒ G1 = ar, G2 = ar2 and so on … 

Now, we have n+2 term

⇒ b = arn+2-1 

⇒ b = arn+1

⇒ r = \((\frac{b}{a})^{\frac{1}{n+1}}\).............(i)

The product of n geometric means is G1× G2× G3× … Gn 

= ar × ar2 × ar3 × … arn 

= an × r(1+2+3… + n)

= an \(\times\) r(n)(\({\frac{n+1}{2}}\)) [sum of n terms of A.P. = \(\frac{(n)(n+1)}{2}\)]

Substituting the value of r from eqn. (i)

Single geometric mean between a and b = \(\sqrt{ab}\)

nth power of single geometric mean between a and b = (\(\sqrt{ab}\))n

Hence Proved

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