Let the G.P. with first term a and common ratio r be a , ar , ar2 , ... , arn−1
Then, Sum (S) =

a(1−rn)

1−r

Product (P) = a . ar ... arn−1
= a . a .... a (n times) (r1.r2...r(n−1)) = anr1+2+...+(n−1)
= anr

n(n−1)

2

Since 1 + 2 + ... + n - 1 =

n(n−1)

2

Sum of reciprocals (R)
=

1

a

+

1

ar

+ ... +

1

arn−1

=

1

a

[

1−( 1 r )n

1− 1 r

] =

r(rn−1)

a(r−1)rn

=

rn−1

arn−1(r−1)

Now, P2Rn = [anr

n(n−1)

2

]2[

rn−1

arn−1(r−1)

]n
= [a2nrn(n−1)][

(rn−1)n

anrn(n−1)(r−1)n

]
= an

(rn−1)n

(r−1)n

= [

a(rn−1)

r−1

]n = (

a(1−rn)

1−r

)n = Sn
Hence P2Rn = Sn