Question

Which term of the sequence $3,6,12$, ............ is $1536$?

Answer 1

It is given that the first term of G.P is $a_1=3$ and the second term is $a_2=6$, therefore,

$r=\frac{a_2}{a_1}=\frac{6}{3}=2$

We know that the formula for the $n$th term of an G.P is $T_n=a{r}^{n-1}$, where $a$ is the first term, $r$ is the common ratio.

It is given that the first term is $a=3$, the $n$th term is $T_n=1536$ and the common ratio is $r=2$, thus,

$T_n=a{r}^{n-1}\Rightarrow 1536=3\times (2{)}^{n-1}\Rightarrow \frac{1536}{3}=2^{n-1}\Rightarrow 2^{n-1}=512\Rightarrow 2^{n-1}=2^9\Rightarrow n-1=9\Rightarrow n=9+1=10$

Hence, the $10$th term is$1536$.