How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?
Given: G.P. 3, 32, 33, … and Sum = 120
Here a = 3 and r =
= 3
The Sum of terms in G.P. is given by: ![]()
∴
= 120
⇒
= 120
⇒
= 120(2)
⇒
= 240
⇒
=
= 80
⇒ 3n = 80 +1 = 81
⇒ 3n = 81
Applying ln on both sides, we get
![]()
⇒ ![]()
⇒ ![]()
⇒ n = 4