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answer is "3"
first find the sum of multiples of 3
that is 3+6+9+...........+198=Sn1=n(a+l)/2
we know that last term is 198 so "l"is 198 and a is 3 we have to find n value
we know that Tn =a+(n-1)d
198=3(n-1)3
n=66
there fore Sn1=66(3+198)/2 =6633
second find the sum of multiples of 5 similarly
we get Sn2=3900
then find the sum of multiples of 15 similarly because 15 is divisible by 3 and 5
so we get Sn3=1365
now add Sn1 and Sn2 then substract Sn3 from it
we get Sn=6633+3900-1365=9168
now we have to find sum to 199 with a=1 , d=1
we get S199=19900
now substract Sn from S199
there fore the final answer is 19900-9168=10732
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