Search any question & find its solution
Question:
Answered & Verified by Expert
The least positive integer $\mathrm{n}$ such that $1-\frac{2}{3}-\frac{2}{3^2}-\ldots .-\frac{2}{3^{n-1}} < \frac{1}{100}$, is:
Options:
Solution:
1398 Upvotes
Verified Answer
The correct answer is:
5
5
$$
\begin{aligned}
&1-\frac{2}{3}-\frac{2}{3^2} \ldots \cdot \frac{2}{3^{n-1}} < \frac{1}{100} \\
&\Rightarrow 1-\frac{2}{3}\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\ldots \frac{1}{3^{n-1}}\right] < \frac{1}{100} \\
&\Rightarrow \frac{1-2\left[\frac{1}{3}\left(\frac{1}{3^n}-1\right)\right]}{\frac{1}{3}-1} < \frac{1}{100} \\
&\Rightarrow 1-2\left[\frac{3^n-1}{2.3^n}\right] < \frac{1}{100} \\
&\Rightarrow 1-\left[\frac{3^n-1}{3^n}\right] < \frac{1}{100} \\
&\Rightarrow 1-1+\frac{1}{3^n} < \frac{1}{100}
\end{aligned}
$$
$$
\Rightarrow 100 < 3^n
$$
Thus, least value of $n$ is 5
\begin{aligned}
&1-\frac{2}{3}-\frac{2}{3^2} \ldots \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}+\ldots \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{aligned}
$$
$$
\Rightarrow 100 < 3^n
$$
Thus, least value of $n$ is 5
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.