Well, the multiples of 8 go "8, 16, 24...etc." The first four multiples of 8 over 20 are 24, 32, 40, and 48. i) 24 + 32 = 56 ii) 56 + 40 = 96 iii) 96 + 48 = 144 Therefore the sum is 144.
It is 155 greater.
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
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
5 + 10+15+20=50 ans
The sum of two decimal numbers greater than 0.5 will always be greater than 1
72
The sum of the first 10 multiples of 3 is 165.
It is 155 greater.
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
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
Their sum is 1200.
2025.
This is easiest to answer by summing all the numbers 1-10000 and subtracting the sum of the multiples of 7 (7, 14, 21, ..., 9996). The sum of a series is: S = (first + last) x number_of_terms / 2 For for 1-10000, the sum is: S1 = (1 + 10000) x 10000 / 2 = 10001 x 5000 = 50005000 For the multiples of 7 the sum is: S2 = (7 + 9996) x 1428 / 2 = 10003 x 714 = 7142142 So the sum of all integers not greater than 10000 that are not divisible by 7 is: S = S1 - S2 = 50005000 - 7142142 = 42,862,858
75
90
975
5 + 10+15+20=50 ans