Using the formula for the sum of an arithmetic sequence;
Sum = 1/2*n [n is the total number of terms] *(2a [a is the value of the first term] + (n-1) * d)[d is the common difference between each term]
So the formula looks like this: Sum = 1/2*n (2a+(n-1)*d)
n = 50 because there's 50 odd numbers in a hundred and 50 even numbers (50+50=100)
a = 1
d = 2
So inputting the values gives:
Sum= 1/2*50 (2*1+(50-1)* 2)
Sum= 25*(2+49*2)
Sum= 25*(2+98)
Sum= 25*(100)
Sum= 2500
Answer= 2,500
The sum of all the odd numbers from 1 through 100 is 10,000
The sum of the first 100 odd numbers (1 through 199) is 10000 (ten thou)
100
The sum of all the odd numbers between 1 and 12000 is 36000000.
The sum (done on a spreadsheet) of 1 to 35 as odd numbers is 1225.
The sum of all the odd numbers from 1 through 100 is 10,000
The sum of 5 odd numbers must be odd and 100 is even. So there cannot be such a set.
The sum of the first 100 odd numbers (1 through 199) is 10000 (ten thou)
100
100
99
50
over 50
200
The sum of the even numbers is (26 + 28 + ... + 100); The sum of the odd numbers is (25 + 27 + ... + 99) Their difference is: (26 + 28 + ... + 100) - (25 + 27 + ... + 99) = (26 - 25) + (28 - 27) + ... + (100 - 99) = 1 + 1 + ... + 1 There are (100 - 26) ÷ 2 + 1 = 38 terms above which are all 1; their sum is 38 x 1 = 38. So the difference of the sum of all even numbers and all odd numbers 25-100 is 38.
The sum is 2500.
101