Question

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

Answer 1

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