Prove that the pr

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 ...... G_n , 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 = ar^n+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......G_n

= (ar)(ar²)(ar³).....(arⁿ)

= aⁿ r^{(1+2+3+4+.............+n)}

= aⁿ r^n(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_n = Gⁿ