Find the number of term in a GP 128,64, 32, ........1/16

The first term, $a = 128$.
The common ratio, $r = \frac{64}{128} = \frac{1}{2}$.

The $n$-th term ($a_n$) of a GP is given by $a_n = a \cdot r^{n-1}$.

Here, $a_n = \frac{1}{16}$.
So, $\frac{1}{16} = 128 \cdot \left(\frac{1}{2}\right)^{n-1}$.

Divide both sides by 128:
$\frac{\frac{1}{16}}{128} = \left(\frac{1}{2}\right)^{n-1}$
$\frac{1}{16 \times 128} = \left(\frac{1}{2}\right)^{n-1}$
$\frac{1}{2048} = \left(\frac{1}{2}\right)^{n-1}$

We know that $2^{10} = 1024$, so $2^{11} = 2048$.
Therefore, $\frac{1}{2048} = \frac{1}{2^{11}} = \left(\frac{1}{2}\right)^{11}$.

So, $\left(\frac{1}{2}\right)^{11} = \left(\frac{1}{2}\right)^{n-1}$.
This implies $11 = n-1$.

$n = 11 + 1$
$n = 12$.

The number of terms in the GP is 12.