Mathematics - Sequence & Series Question with Solution | TestHub

$$\begin{array}{rl}& 1-\frac{2}{3}-\frac{2}{3^2}\dots \cdot \frac{2}{3^{n-1}}<\frac{1}{100}\\ & \Rightarrow 1-\frac{2}{3}\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\dots \frac{1}{3^{n-1}}\right]<\frac{1}{100}\\ & \Rightarrow \frac{1-2\left[\frac{1}{3}\left(\frac{1}{3^n}-1\right)\right]}{\frac{1}{3}-1}<\frac{1}{100}\\ & \Rightarrow 1-2\left[\frac{3^n-1}{2.3^n}\right]<\frac{1}{100}\\ & \Rightarrow 1-\left[\frac{3^n-1}{3^n}\right]<\frac{1}{100}\\ & \Rightarrow 1-1+\frac{1}{3^n}<\frac{1}{100}\end{array}$$ $$\Rightarrow 100<3^n$$ Thus, least value of $n$ is 5