The correct option is C 3121
Numbers between 100 and 200 which are divisible by 7:
105,112,⋯,196.
Let their sum be S1.
∴S1=142(105+196)=2107
Numbers between 100 and 200 which are divisible by 13:
104,117,⋯,195.
Let their sum be S2.
∴S2=82(104+195)=1196
Number between 100 and 200 which is divisible by 7 and 13 both =182
Hence, required sum=2107+1196−182=3121