3, \(\frac{3}{2}\), \(\frac{3}{4}\)

Here r = \(\frac{1}{2}\)< 1

sum of G.P = \(\frac{a(1-r^n)}{1-r}\)

⇒ \(\frac{255}{128} = \frac{3(1-(\frac{1}{2})^n)}{1- (\frac{1}{2})}\) = \(\frac{3(1- \frac{1}{2^n})}{\frac{1}{2}} = 6({1- \frac{1}{2^n}})\)

⇒ \(\frac{255}{256}\) = \(1- \frac{1}{2^n}\)

⇒ \(\frac{1}{2^n} = 1 - \frac{255}{256}\) = \(\frac{1}{256} = \frac{1}{2^8}\)

⇒ n = 8