# Find n where 1536 is the nth term of the series 3,6,12---------------

$(a)\;9\qquad(b)\;3\qquad(c)\;512\qquad(d)\;10$

Explanation : 3,6,12---------is a GP where
$a=3\;,r=\frac{6}{3}=\frac{12}{6}=2$
$T_{n}=ar^(n-1)=1536$
$2^(n-1)=\frac{1536}{a}=\frac{1536}{3}=512=2^9$
$2^(n-1)=2^9$
$n-1=9$
$n=9+1=10.$