Q: Sum of the first 100 positive even integers?

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(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.

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

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

use this formula: S=(N/2)(F+L) S= the sum of the first 100 positive even integers N= the number of terms F= first term L=last term Since we are using the first 100 positive integers we will replace "N" with 100. so it will look like this : S=(100/2)(F+L) Next we must substitute the "F" with the first positive term which is the number 2. So now it looks like this: S=(100/2)(2+L) We use the number 2 because our positive integers are 2,4,6,8,10,12...... you get the picture. Now we must substitute the "L" with the last positive term which is 200. As you probably guessed it will look like this: S=(100/2)(2+200) We use the number 200 because it is the last positive even integer. If we used 202 then that would have meant we used the 101th positive even integer and we don't want to do that. == == We should solve the numbers inside the parenthesis first. S=(100/2)(2+200) 100 divided by 2 = 50 So: S=50(2+200) 2+200=202 That means: S=50(202) 50 x 202 = 10,100 Finally: S=10,100 There you go! Wasn't that easy? Note: This formula only works if you are looking for the sum of the first 100 positive or negative, even or odd integers. It sucks cause this formula is the Bomb right? Oh well

Should be 50! Every odd integer is 1 less than the corresponding even integer and there are 50 of each in 100...

There are 25 of them.

They are 13.

The positive integers less than 100 are a finite set. The positive integers greater than 100 are an infinite set.

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.

MB55

The sum of all the digits of all the positive integers that are less than 100 is 4,950.

10,100

Of the 729 numbers that satisfy the requirement of positive integers, 104 are divisible by 7.

there are 999 - 100 + 1 = 900 positive triple digit positive integers, between 100 and 999.(e.g. there are 102 - 100 + 1 = 3 triple digit integers between 100 and 102,namely 100, 101 and 102.)multiply that by 2 to take in consideration of the negative integers,you have 1800 triple digit integers.

It is 50.5

The squares -- nine of them.

There are 80 such integers.

900 numbers from 100 to 999

All of them from 45 to 99 ... 55 integers.

100

Between 100 and 999 there are 448.

32

128 of them.

2550

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