Question 1164975
The sum without removal is (1/2)(150*151)=11325
multiples of 3 are 50, 
3+6+9+...+150
that sum Is (n/2)(a1+an)=25*(3+150)=3825
multiples of 7 are 21;
7+14+21+28+...+147
that sum is (21/2)(7+147)=1617
the sum of both is 5442
double counted are the product of both or multiples of 21
21+42+63+84+105+126+147=(n/2)(168)=588
so the amount to be subtracted is 5442-588=4854

the answer is 11325-4854=6471