Given sequence is 1,2,2,3,3,3,4,4,4,4,…
First term =1
Second term =2
Fourth term =3
Seventh term =4
Eleventh term =5…, so on ∴ Let S=1+2+4+7+11…n terms 0=(1+1+2+3+4…n terms )−an−S=1+2+4+7+11…+n terms ⇒an=1+{1+2+3+4…(n−1) terms ⇒an=1+2n(n−1)=2n2−n+2…(i)
If n=14, then an=92
i.e., 92 nd term is 14 .
If n=15, then an=106
i.e., 106 th term is 15 .
Hence, 100 th term is 14