Answer:
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.
Step-by-step explanation:
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