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Q: What is the sum of the first 100 positive multiples of 4?

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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 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.

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

Related questions

The sum of the first 100 positive even numbers is 10,100.

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

10100.

The sum of the first 10 multiples of 3 is 165.

The sum of the first 100 odd numbers (1 through 199) is 10000 (ten thou)

We have to assume that you're talking about whole numbers. The sum is 5,050 .

Their sum is 1200.

The sum is 375,750.

69342

They are 13.

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.

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