The average is 10. You will find that the average when adding together the first n odd numbers is always equal to n.
152
3, 5, 7, 9 you knowwww there are an infinite amount of odd numbers in between 1 and 10 such as: 1.1, 1.3, 1.5, 1.7, 1.9, 2.1, 2.3, 2.5, 2.7, 2.9, and so on
No. For example, 5+5=10.
101
The sum of the first 10 odd whole numbers is 100.
100
The average is 10. You will find that the average when adding together the first n odd numbers is always equal to n.
for (int i = 2; i < 10; i ++) printf("%d\n", i); You did say even and odd numbers between 1 and 10. That's allnumbers between 1 and 10.
The sum of all odd numbers, up to the odd number (2n-1) is n^2. So the sum of the first 1000 or 10^3 odd positive numbers is (10^3)^2=10^6 Now divide by 10^3 or 1000 since we have 1000 numbers so we have 10^6/10^3=1000 The mean of the first 1000 positive odd numbers is 1000 If you ask the mean of the odd numbers between, 1 and 1000, that is another problem, since there are 500 of them, the answer would be 500^2/500=500 Think of the mean of the odd numbers between 1 and 10, there are 5 of them and there sum is 25 so the mean is 25/5 or 5 **** in general, the arithmetic mean of the first n odd numbers is n^2/n or n. So that is why the number is 1000.
152
The last digit in the product of the first 20 odd natural numbers can be determined by looking at the pattern of the units digit in the multiplication of consecutive odd numbers. The units digit of the product of consecutive odd numbers alternates between 1 and 5. Since there are 10 odd numbers between 1 and 19, and 20 is also an odd number, the last digit in the product of the first 20 odd natural numbers is 5.
Numbers that are prime can't be multiples of 10. Multiples of 10 can't be odd.
10
3, 5, 7, 9 you knowwww there are an infinite amount of odd numbers in between 1 and 10 such as: 1.1, 1.3, 1.5, 1.7, 1.9, 2.1, 2.3, 2.5, 2.7, 2.9, and so on
The odd numbers between 1 and 10 are 1, 3, 5, 7, and 9.
#include<stdio.h> int main () { int odd=1; int count=0; while (count++<10) { printf (%d\n", odd); odd+=2; } return 0; }