For a number n in the series let x = n(n+1)/2 ...(1)
Here, x is the last position of the element n 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.