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