The correct options are
A a(100)<100
C a(200)<100
a(n)=1+12+13+14+⋯+1n , where n∈Z+
Let us consider a series b(n)
b(n)=1+12+12+12+⋯+nterms
Using comparison test, b(n)>a(n) always
⇒ Putting n=200,b(200)>a(200)
⇒1+1992>a(200)⇒a(200)<100.5≃100
Now, let c(n)=1+1+1+⋯+nterms
⇒c(n)>a(n)
Putting n=100,c(100)>a(100)
⇒a(100)<100
Hence, all options are incorrect