Consider the sequence `1,2,2,3,3,3,"……",` where n occurs n times that occuts as 2011th terms is

Options

A. 61
B. 62
C. 63
D. 64

Answer

C

Explanation

The last 4 occurs as `1+2+3+4+=10th` term. The last n occurs as `((n(n+1))/(2))^(th)` term, the last 62 occurs as
`((63xx63)/(2))^(th)=1953rd` term and the last 63 occurs as
`((64xx64)/(2))^(th)=2016th` term.
`:. 63` occurs from 1954th term to 2016th term.
Hence, `(2011)^(th)` term os 63.