Q.Which term of the Geometric Progression $3, 6, 12, \ldots$ is equal to $384$?

Set the GP general-term formula equal to $384$ and solve for $n$ using powers of $2$.

Here $a=3$, $r=\dfrac{6}{3}=2$ (checked: $\dfrac{12}{6}=2$ — constant).

Set $T_n=384$: $3\times2^{n-1}=384 \Rightarrow 2^{n-1}=\dfrac{384}{3}=128$.

Since $128=2^7$, equating exponents (both sides now powers of $2$) gives $n-1=7 \Rightarrow n=8$. …

Alternative — repeated doubling from $a=3$: $3\to6\to12\to24\to48\to96\to192\to384$; counting the terms in this list direc …

A common slip is dividing $384$ by $3$ incorrectly or misidentifying $128$ as a power of $2$ — always express the quotient as $2^k$ by repeated halving ($128\to64\to32\to16\to …