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The $5000$th term of the sequence $1,2,2, 3,3,3,4,4,4,4, \cdots$ is

1. $98$
2. $99$
3. $100$
4. $101$

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For a number n in the series let x = n(n+1)/2 ...(1)

Here, x is the last position of the element in the sequence.

A/q n(n+1)/2 = 5,000

=>  n(n+1) = 10,000​ ; ​​​​​​which simply means n~100

Put n=100 in (1), 100(101)/2 = 5050 i.e. position of last 100 is 5050 and position of first 100 must be 5050 - (100 - 1) = 4951.

Therefore, form position 4951 to 5050 there are 100 100s, hence the correct answer is option C.

edited

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