To find the sum of the first 100 positive multiples of 4, we can use the formula for the sum of an arithmetic series: Sn = n/2 * (a1 + an), where Sn is the sum, n is the number of terms, a1 is the first term, and an is the last term. In this case, a1 = 4, an = 4*100 = 400, and n = 100. Plugging these values into the formula, we get: Sn = 100/2 * (4 + 400) = 50 * 404 = 20,200. Therefore, the sum of the first 100 positive multiples of 4 is 20,200.
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The sum of the first 100 positive even numbers is 10,100.
Sum of the first 15 positive integers is 15*(15+1)/2 = 120 Sum of the first 15 multiples of 8 is 8*120 = 960
The sum of the first 100 odd numbers (1 through 199) is 10000 (ten thou)
The sum of the first 25 numbers is 25*26/2 = 325 So the sum of the first 25 multiples of 12 is 325*12 = 3900
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.