We have, A=n∈N∣n2≤n+10,000,B=3k+1∣k∈N and C=2k∣k∈N,
B≡4,7,10,13,16,19,……
C≡2,4,6,8,10,12,14,16,20,……
B−C≡7,13,19,…97,…
Now, n2−n≤100×100
⇒n(n−1)≤100×100
⇒A=1,2,…,100
So, A∩(B−C)=7,13,19,…,97
Let Sn=7+13+19+…+97
Clearly, the terms are in arithmetic progression. Here, a=7 and d=13−7=6 and l=97
⇒l=a+(n−1)d=97 where n is the number of terms.
⇒7+(n−1)6=97
⇒(n−1)6=90
⇒(n−1)=15
⇒n=16
Hence, sum Sn=7+13+19+…+97
Sn=2n(a+l)
Sn=216(7+97)
=832.