How many terms of the G.P. $3,3/2,3/4$ are needed to give the sum $3069/512?$

From the question, it is given that, 

Sum of the terms $S_n=3069/512$ 

First term $a=3$ 

Common ratio $r=(3/2)/3$

$=(3/2)\times (1/3)$

$=1/2$

We know that, 

$S_n=a\left(1-r^n\right)/(1-r)$ 

$(3069/512)=3\left[1-(1/2{)}^n\right]/(1-1/2)$

$(3069/512)=(2\times 3)\left[1-(1/2{)}^n\right]$

$1-(1/2{)}^n=3069/(512\times 6)$

$1-(1/2{)}^n=1023/1024$

$(1/2{)}^n=1-(1023/1024)$

$(1/2{)}^n=(1024-1023)/1024$

$(1/2{)}^n=1/1024$

$(1/2{)}^n=(1/2{)}^{10}$

By comparing both LHS and RHS, $n=10$

Therefore, there are 10 terms are in the G.P.