Question

If S is the sum, P the product and R the sum of reciprocals of n terms of a G.P. then prove that
$\rm P^{2} = \Big(\frac{S}{R}\Big)^{n}$

Answer 1

let a sequence in geometric progression is ; a , ar , ar² , ar³ , ar⁴ .......... $ar^{n-1}$

S is the sum of n terms of GP.

so, S = a(rⁿ - 1)/(r - 1)

P is the product of n terms of GP.

so, P = a.(ar).(ar²).(ar³). ... $(ar^{n-1})$

P = $a^n\left(r^{1+2+3+...(n-1)}\right)$

P = $a^n\left(r^{\frac{n(n-1)}{2}}\right)$

R is the sum of reciprocal of n terms in GP.

e.g., 1/a, 1/ar, 1/ar² , 1/ar³ ....
first term = 1/a and common ratio = 1/r

now, R = 1/a(1 - 1/rⁿ)/(1 - 1/r)

R = (rⁿ - 1)/{a(r - 1)r^(n-1)}

now, LHS = P² = a²ⁿ r^{n(n-1)}

RHS = (S/R)ⁿ =[ {a(rⁿ - 1)/(r - 1)}/{(rⁿ - 1)/a(r - 1)r^(n-1)} ]ⁿ

= [a²r^(n-1) ]ⁿ

= a²ⁿ r^{n(n-1)}

hence, LHS = RHS