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(D) We need to find the sum of all $n$ such that $100  1$.Since $91 = 7 	imes 13$,$H.C.F. (91, n) > 1$ means $n$ must

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$3121$

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(D) We need to find the sum of all $n$ such that $100  1$.Since $91 = 7 	imes 13$,$H.C.F. (91, n) > 1$ means $n$ must be divisible by $7$ or $13$.Let $S_A$ be the sum of numbers between $100$ and $200$ divisible by $7$.The numbers are $105, 112, \dots, 196$.This is an arithmetic progression with $a = 105$,$l = 196$,and $d = 7$.Number of terms $k = \frac{196 - 105}{7} + 1 = 14$.$S_A = \frac{14}{2} (105 + 196) = 7 	imes 301 = 2107$.Let $S_B$ be the sum of numbers between $100$ and $200$ divisible by $13$.The numbers are $104, 117, \dots, 195$.This is an arithmetic progression with $a = 104$,$l = 195$,and $d = 13$.Number of terms $m = \frac{195 - 104}{13} + 1 = 8$.$S_B = \frac{8}{2} (104 + 195) = 4 	imes 299 = 1196$.Let $S_C$ be the sum of numbers between $100$ and $200$ divisible by both $7$ and $13$ (i.e.,divisible by $91$).The only number is $182$.$S_C = 182$.By the Principle of Inclusion-Exclusion,the required sum is $S_A + S_B - S_C = 2107 + 1196 - 182 = 3121$.

The sum of all natural numbers $n$ such that $100 < n < 200$ and $H.C.F. (91, n) > 1$ is

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