The solution to the given problem can be obtained by sum formula of arithmetic progression.
In arithmetic progression difference of two consecutive terms is constant. The multiples of any whole number(in sequence) form an arithmetic progression.
The first multiple of 3 is 3 and the 100th multiple is 300.
3, 6, 9, 12,... 300. There are 100 terms.
The sum 3 + 6 + 9 + 12 + ... + 300 can be obtained by applying by sum formula for arithmetic progression.
Sum = (N/2)(First term + Last term)
where N is number of terms which in this case is 100.
First term = 3; Last term = 300.
Sum = (100/2)(3 + 300) = 50 x 303 = 15150.
Wiki User
∙ 2013-08-07 08:06:27Anonymous
the sum of the first 100 multiples of 3 is 15150
5 + 10+15+20=50 ans
this isn't a place to do your math homework.
Sum of first 25 multiples of 44+8+12....100taking 4 common4(1+2+3+4....+25) = 4*325 =1300
There are 166 multiples in 500 (500/3=166r2)and 33 multiples in100(100/3=33 r1) if you subtract the later from the first.... 166-33=133
3+6+9...+45taking 3 common3(1+2+...+15) = 360
The sum of the first 10 multiples of 3 is 165.
975
30
what are the first 3 multiples? 7 14 21 so add them together!
3 and 2
Created in a spreadsheet, the answer is: 375750
5 + 10+15+20=50 ans
3 + 6 + 9 + 12 = 30
this isn't a place to do your math homework.
4 + 8 + 12 = 24
10+20+30 = 60
It is: 8+16+24 = 48