Let a1,a2,a3,a4,a5,a6 be the first six terms of the G.P.
According to question, sum of first three terms = 16
⇒ a1+a2+a3 = 16 ....(i)
Sum of next three terms = 128
⇒ a4+a5+a6 = 128 ....(ii)
Let S6 be the sum of first six terms
i.e., S6 = 16 +128 = 144 [from (i) and (ii)]
Let a be the first term and r be the common ratio, then
S6 =

a(1−r6)

1−r

⇒ 144 =

a(1−r6)

1−r

... (iii)
and S3 =

a(1−r3)

1−r

⇒ 16 =

a(1−r3)

1−r

... (iv)
∴

S6

S3

=

−r6

1−r3

=

144

16

[from (iii) and (iv)]
⇒

(1−r3)(1+r3)

(1−r3)

= 9
⇒ 1 + r3 = 9 ⇒ r3 = 8 ⇒ r = 2
∴ S3 = a

(1−r3)

1−r

⇒ 16 =

a(1−8)

1−2

=

7a

1

⇒ a =

16

7

∴ Sn = a

(1−rn)

1−r

=

16

7

(1−2n)

(1−2)

=

16

7

(2n−1).
Hence a =

16

7

, r = 2 and Sn =

16

7

(2n−1)