Given: The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128.

Let a, ar, ar², ar³, ar⁴, ar⁵ be the terms of the G.P

Here a + ar +ar² = 16 (∵ given that sum of first 3 terms is 16) -------–1

Also, ar³, ar⁴, ar⁵ = 128 (∵ given that sum of next 3 terms is 128)--------------–2

Divide eq –2 and eq –1

We get,

⇒

⇒ r³ = 8

⇒ r = ∛8

⇒ r = 2

Here a + ar + ar² = 16

⇒ a × (1 + r + r²) = 16

⇒ a × (1 + 2 + (2)²) = 16

⇒ a × (1 + 2 + 4) = 16

⇒ a × (7) = 16

⇒ a =

The Sum of terms in G.P. is given by:

∴ =

∴ First term of the G.P is , common ratio of the G.P is 2 and Sum of n terms of the G.P is