The number of terms of the G.P.3,32,34,......needed to obtain a sum of 3069512 is :

Here a = 3, r = 1/2.Let the number of terms needed be n, then             3069512=Sn=31−(1/2)n1−1/2⇒      10231024=1−12n⇒       12n=11024⇒n=10