1 votes 1 votes The $5000$th term of the sequence $1,2,2,3,3,3,4,4,4,4,\cdots$ is $98$ $99$ $100$ $101$ Quantitative Aptitude isi2016-dcg quantitative-aptitude sequence-series + – gatecse asked Sep 18, 2019 • recategorized Nov 19, 2019 by Lakshman Bhaiya gatecse 363 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes Answer: C Explanation: Small observation: Every number is repeated by the same number of times. Eg. $1$ repeats $1$ time, $2$ repeat $2$ times, $\cdots$ Now, this question reduces to nothing but just a sum of n terms of natural numbers, which is given by: $\frac{n(n-1)}{2} < 5000 \le \frac{n(n+1)}{2}$ On substituting the options, we will get that 100 satisfies this equation. $\therefore \mathbf C $ is the right option. `JEET answered Sep 21, 2019 • edited Nov 10, 2019 by `JEET `JEET comment Share Follow See all 4 Comments See all 4 4 Comments reply techbd123 commented Oct 27, 2019 reply Follow Share I think the correct inequality will be $\frac{n(n-1)}{2}<5000\le \frac{n(n+1)}{2}$ [representing the position of $n$ in the sequence] where $n$ is the required repeating number. 1 votes 1 votes techbd123 commented Oct 28, 2019 reply Follow Share @`JEET $\frac{n(n+1)}{2}\ge5000\Rightarrow n\ge100 \Rightarrow n=100,101,102,...$ The inequality above doesn't produce the unique answer. But this inequality $\frac{n(n-1)}{2}<5000\le \frac{n(n+1)}{2}$ produce only $n=100$. 1 votes 1 votes `JEET commented Oct 28, 2019 reply Follow Share I think for objective technique that is correct! 1 votes 1 votes `JEET commented Oct 28, 2019 reply Follow Share I have taken the minimum value, it can satisfy even more than $2$ values as well. Don't you think its correct keeping in view of objective problem solving? But I still change to make it mathematically correct. 1 votes 1 votes Please log in or register to add a comment.