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Q: Find the mean of n natural numbers?

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The sum of the first n natural numbers is n*(n+1)/2 There are n numbers so their mean = (n+1)/2

1 Sum of first n natural numbers = n(n+1)2[Formula.]2 Arthmetic mean of first n natural numbers = Sum of the numbers n[Formula.]3 = n(n+1)2n = n+124 So, the Arthmetic mean of first n natural numbers = n+12

Mean = Total of the numbers / n So total of the numbers = n*mean

The average of n numbers = (sum of n numbers) / (count of n numbers).

The general equation to find the sum of the numbers 1 to n is: (n*(n+1))/2So, for n=10, you have:(10*(10+1))/2(10*11)/2110/255

You can only answer this question if you know how many numbers there were before. (If you know how many were there after, then subtracting two gives you the number before). So suppose there were n numbers with mean x. And you add two more numbers, a and b. That means the sum of the n numbers was n*x. Sum of the n+2 numbers is n*x + a + b So the new mean is (n*x + a + b)/(n + 2)

You find the mean of negative numbers the same way you find the mean of any numbers. You add them up and then you divide by the number of numbers. An example, using -3, -4, -6, and -9. The sum is -22. N is 4. The mean is -5.5.

N

Let n be an arbitrary natural number. The next larger natural number is n + 1. The sum of these natural numbers is n + (n + 1) n + (n + 1) = 525 2n + 1 = 525 2n = 524 n = 262 n + 1 = 263. The two natural numbers are 262 and 263.

N=Natural Numbers For Example 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19 and so on

int sum( int n){if( n==0)return 0;elsereturn n + sum(n-1);}A call to the function as:sum(n)will return the sum of n numbers.

I have found the coefficient of variation of the first natural numbers and also other functions.

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