Question

The least positive integers $n$ such that $1-\frac{2}{3}-\frac{2}{3^2}-..-\frac{2}{3^{n-1}}<\frac{1}{100}$ is

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

The correct option is A $4$

We have,

$1-\frac{2}{3}-\frac{2}{3^2}-.....-\frac{2}{3^{n-1}}<\frac{1}{100}$

Now,

$1-2(\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{n-1}})<\frac{1}{100}$

It is G.P.

Then,

First term $a=\frac{1}{3}$

Common ratio $r=\frac{\frac{1}{3^2}}{\frac{1}{3}}=\frac{1}{3}$

So,

Then, using formula

If$r<1$

$S_n=\frac{a(1-r^n)}{(1-r)}$

So,

$S_n=\frac{\frac{1}{3}(1-{\left(\frac{1}{3}\right)}^{n-1})}{1-\frac{1}{3}}$

$=\frac{\frac{1}{3}(1-\frac{1}{3^{n-1}})}{\frac{2}{3}}$

$=\frac{1}{2}(1-\frac{1}{3^{n-1}})$

So,

$1-2\times \frac{1}{2}(1-\frac{1}{3^{n-1}})<\frac{1}{100}$

$\Rightarrow 1-(1-\frac{1}{3^{n-1}})<\frac{1}{100}$

$\Rightarrow 1-1+\frac{1}{3^{n-1}}<\frac{1}{100}$

$\Rightarrow \frac{1}{3^{n-1}}<\frac{1}{100}$

$\Rightarrow 3^{n-1}>100$

$\Rightarrow \frac{3^n}{3}>100$

$\Rightarrow 3^n>100\times 3$

$\Rightarrow 3^n>300$

$\Rightarrow 3^n<3^4$

On comparing that,

$n=4$

Hence, this is the answer.