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First we find the sum of the integers from 2 to 100 including 2 and 100, then we divide by 99 since there are 99 numbers. There are 100 numbers between 1 and 100 and we are excluding only the number 1. If we want to exclude the number 2, and count only the numbers 3,4,5....100, this can be done with the same procedure and a slight modification. So the sum of the numbers 1 to 100 can be found by writing the numbers 1, 2,3,...100 Now write them backwards, starting at 100,99,98....1 Each column has a sum of 101 and there are 100 columns. So the total is 100x101, but we wrote the list twice so we must divide by 2 The sum is 100x101/2=50x101=5050 Now remember we want only 2 to 100 so the sum we seek is 5049. Since there are 99 number, the mean is 5049/99=51 In general the sum of the first n positive integers is n(n+1)/2 This can be proved the way we did or by induction.

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Q: What is the mean of the integers from 2 to 100?
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How many are the first 100 positive integers are divisible by 7?

(7*100*101)/2 = 35,350 jpacs * * * * * What? How can there be 35,350 integers in the first 100 integers? There are 14 of them.


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The sum of two consecutive odd integers is 100 Find the integers?

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What is the pattern for the sum integers 1 to 100?

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What is the sum of the even integers from 2 to 100 inclusive?

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What is the sum of the 2 integers directly above 100?

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How many odd integers under 1000 are not divisible by 5?

you can say that 2/10 of every 10 integers is divisible by 5, so multiplying 2/10 by 100, giving you 200/1000 total integers are divisible by 5. half of all integers are odd, so divide 200/1000 by 2 is 100/1000, so you can correctly state that 100 odd integers under 1000 are divisible by 5.


How many even integers are between 100 and 1000?

If you mean "between" 100 and 1000 to mean excluding 100 and 1000 then 449. If you want to include 100 and 1000 then 451.


How many triple digit numbers are there?

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n(n+1)/2 5050


What is the sum of the first 100 positive integers?

To find the sum of the first 100 integers, you first add 1 plus 100 (the first and last numbers of the set) and get 101. Do the same with the next two integers, 2 and 99 and you'll get 101. Since you are adding two integers at a time and there are 100 integers between 1 and 100, you'll get 101, fifty times. Therefore, a shortcut would be to simply multiply 101 times 50 and get 5,050.