To write it out: -10 + -9 + ... + 9 + 10 = 0. It's pretty easy to just add it out or use a calculator. However, there is a simpler, more logical solution. We must realize that additive inverses add to 0. Meaning -10 + 10 = 0 - 9 + 9 = 0 And so on. Note that all numbers except the 0 will be reduced to 0. All of these 0s equal 0.
-2, -1, 0, 1
There are one thousand. Starting at 2000 and going to 2999
Do this in reverse. The sum of -1 and -1 is -1+-1=-2 The difference of -6 and -6 is -6-(-6)=-6+6=0 0 increased by -2 is 0+-2=-2 The sum of 10 and -2 is 10+-2=8 Translation: 8
All integers more than 0, although 0 is sometimes included in the set.
Let the integers be 'n' & 'n+2' Hence Their squares are n^(2) & (n+2)^(2) Their sum is n^(2) + (n+2)^(2) = 74 Expanding n^(2) + n^(2) + 4n + 4 = 74 2n^2 + 4n -70 = 0 n^2 + 2n - 35 = 0 Factor (n - 7)(n + 5) Hence n = '5' & '7' Odd and consecutive.
When the sum of all the positive integers in the sum is exactly matched (in magnitude) by the sum of all the negative integers.
what is the sum of the first 10 positive integers? To me, if you include 0 as the first integer, then the tenth integer is 9 and the sum is 45. If you don't include 0, the tenth integer is 10, and the sum is 55.
Two integers which sum to zero (e.g. 3 and -3) are additive inverses of each other. All pairs of additive inverses sum to 0 and all pairs of integers which sum to 0 are additive inverses.
The sum of the first 10 positive integers 1,2,3,4,5,6,7,8,9 and 10 is 55. The sum of the first 10 negative integers -1,-2,-3,-4,-5,-6,-7,-8,-9 and -10 is -55. The sum of the first 10 positive plus the sum of the first 10 negative integers is 0
The sum of the first 201 positive integers is 20100 if you include 0 (i.e. from 0 to 200). If you sum the integers from 1 to 201 instead, the sum is 20301.
The flowchart to read 10 positive integers K>10 Start A N K=1 Sum = 0 Sum = Sum + K2 B Is Y Print K > 100? sum K=k+1 End B A
The following will sum all integers from 1-100. Adjust according to your needs.int sum = 0;for (int i = 1; i
It is 0.
10
you can use the condition statement like for(i=0:i<=10;i++)
Take any negative integers, say -5 and -10, their sum is -15 which is smaller than both of them. We could have used 0 as well, so I should have said any non-positive integers. To see that is does not work with positive integers, take 5 and 10 whose sum is 15 which is BIGGER than either one.
-1, 0, 1