Q: What is the sum of numbers 1 to 99?

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100

The sum of all odd numbers 1 through 99 is 9,801.

The sum of all the integers between 1 and 99 inclusive is 4950.

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4950

99

the series of odd numbers from 1 to 99 :1 3 5 7 9.....99 SUM OF THE SERIES: It is a geometric progression with a=1 and l=99 and common difference (d)=2. let 99 be the nth term of the sequence so, 99=1+(n-1)d 99=1+(n-1)2 solving this we get n =50. SUM=(n/2)(a+l) =(50/2)(1+99) =(25)(100) =2500.

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.

answer:2500

The sum of 1+ 3 + 5 + ... + 99 = 50 x (1 + 99) ÷ 2 = 2500.

9801 * The sum of the first two odd numbers (1+3) is 4, or 22 * The sum of the first three odd numbers (1+3+5) is 9, or 32 * The sum of the first four odd numbers (1+3+5+7) is 16, or 42 * ...and so on So the sum of the first 99 odd numbers, using the pattern above, would be 992 or 9801.

The total of all of the numbers from 1 to 99 is 4950.

let the number be n-1, n, n+1 then 3n=99 and so n=33 the numbers are 32, 33, 34

100

99^2 == 9,801

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2500

101

n/2(n + 1) = 49½ x 100 = 4950

100/3-1, 100/3 and 100/3+1 that is, 99, 100 and 101.

Since 1+99=100. And 2+98=100 and 3+97=100 etc to 49+51=100 it's 49 of those plus 50 or 4,950. The sum of the first 99 whole numbers is 4,950

Do you mean:"What three consecutive numbers add up to 99?" If you do, then the numbers are 32, 33, 34.

100

109

The numbers are 97, and 99.

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