Given sequence is $1,2,2,3,3,3,4,4,4,4, \ldots$
First term $=1$
Second term $=2$
Fourth term $=3$
Seventh term $=4$
Eleventh term $=5 \ldots$, so on
$\therefore $ Let $S=1+2+4+7+11 \ldots n$ terms
$\frac{-S=1+2+4+7+11 \ldots+n \text { terms }}{0=(1+1+2+3+4 \ldots n \text { terms })-a_{n}}$
$\Rightarrow \,a_{n}=1+\{1+2+3+4 \ldots(n-1)$ terms
$\Rightarrow a_{n}=1+\frac{n(n-1)}{2}=\frac{n^{2}-n+2}{2}\,\,\,\,\,\,\dots(i)$
If $n=14$, then $a_{n}=92$
i.e., 92 nd term is 14 .
If $n=15$, then $a_{n}=106$
i.e., 106 th term is 15 .
Hence, 100 th term is 14