Question

The sum of the geometric progression a,ar,ar^2,ar^n-1 is S the product of these n term s is P. Find R,the sum of the reciprocals of these n term s and show
that (S/R)^n=P^2

Answer 1

Let T=1+r+...+r^(n-1)
Then
S = aT

Let u = 1+2+...+(n-1)= n(n-1)/2
Then P = a^n * r^u

R = 1/a + 1/ar + ... + 1/ar^(n-1)
Placing all these over a common denominator of ar^(n-1)
R = (r^(n-1)+r^(n-2)+...+r+1) / ar^(n-1) = T/ar^(n-1)

S/R = aT / [T/ar^(n-1)] = a^2 r^(n-1)
(S/R)^n = a^(2n) r^(n(n-1)) = P^2
since
P^2 = (a^n r^u)^2 = a^(2n) r^(n(n-1))