$\text{ If }{\mathrm{S}}_{\mathrm{n}}=1+\frac{1}{3}+\frac{1}{3^2}+..+\frac{1}{3^{\mathrm{n}-1}},\mathrm{n}\in \mathrm{N},\text{ then the least value of ' }\mathrm{n}\text{ ' such that }3-2{\mathrm{S}}_{\mathrm{n}}<\frac{1}{100}\text{ is}$

${\mathrm{S}}_{\mathrm{n}}=1+\frac{1}{3}+\frac{1}{3^2}+..+\frac{1}{3^{\mathrm{n}-1}}$

$\Rightarrow S_n=\frac{1\left(1-{\displaystyle \frac{1}{3^n}}\right)}{1-{\displaystyle \frac{1}{3}}}$ $\left(sum of n terms in G.P= \frac{a\left(1-r^n\right)}{1-r}\right)$

$3-2{\mathrm{S}}_{\mathrm{n}}<\frac{1}{100}$

$\Rightarrow$$3-\frac{2\left(1-\frac{1}{3^n}\right)}{1-1/3}<\frac{1}{100}\Rightarrow \frac{1}{3^{n-1}}<\frac{1}{100}$

$\Rightarrow 3^{\text{n-1}}>100\Rightarrow \text{n}-1\ge 5\Rightarrow \text{n}\ge 6$