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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 *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**.

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