(a): We have, $1001$ So, $n$ can be multiple of 7 or 13 or both $\{\because 91=7 \times 13\}$ Let $S_P=$ Sum of all natural numbers which are divisible by 7 and lie between 100 and 200 $S_Q=$ Sum of all natural numbers which are divisible by 13 and lie between 100 and 200\n$S_R=$ Sum of all natural numbers which are divisible by both 7 and 13 and lie between 100 and 200\nSo, $S_P=105+112+\cdots .+196$\n$=\frac{14}{2}[105+196]=7(301)=2107$\n$S_Q=104+\cdots .+195=\frac{8}{2}[104+195]=4(299)=1196$\n$S_R=182$\n∴ Required Sum $=S_P+S_Q-S_R=2107+1196-182=3121$