Find the number of terms of the geometric progression (G.P) whose first term is $\dfrac{3}{4}$, the common ratio is $2$ and the last term is $384$.
Let the number of terms of G.P $= n$
Given: $\;$ First term $= a_1 = \dfrac{3}{4}$
Common ratio $= r = 2$
Last term $= a_n = 384$
Now, $\;$ $a_n = a_1 \times r^{n - 1}$
Substituting the values of $a_1$, $a_n$ and $r$ we get,
$384 = \dfrac{3}{4} \times 2^{n - 1}$
i.e. $\;$ $2^{n - 1} = \dfrac{384 \times 4}{3} = 512 = 2^9$
$\implies$ $n - 1 = 9$ $\implies$ $n = 10$
$\therefore \;$ Number of terms of G.P $= n = 10$