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In this article, you will learn and get code about how to add **n** numbers (n integer numbers, n real numbers, and n natural
numbers) in C programming. Here **n** indicates the amount/quantity. That is, how many numbers, user wants to enter and add them
as shown here.

For example, if user enters 5 as the value of **n**, then program further ask from user to enter any **5** number.
After receiving **5** numbers say **10**, **20**, **30**, **40**, and **50** from user, program sum-up all the 5
numbers and display its summation as result on output. That will be **150** (because **10+20+30+40+50** equals 150).

To add n numbers in C programming, you have to ask from user to enter the value of **n**, then ask to enter **n** numbers to
perform the addition of all the provided **n** numbers (by user). And then display the addition result as output on screen.
Let's take a look at the program.

// Add n numbers in C // -----codescracker.com----- #include<stdio.h> #include<conio.h> int main() { int i, n, num, sum=0; printf("Enter the value of n: "); scanf("%d", &n); printf("Enter %d numbers: ", n); for(i=0; i<n; i++) { scanf("%d", &num); sum = sum+num; } printf("\nSum of all %d numbers = %d", n, sum); getch(); return 0; }

This program was written under **Code::Blocks** IDE. Here is the initial snapshot of sample run:

Supply the value of **n** say **6**, then enter any 6 integer values say **10, 20, 30, 40, 50, 60** and
press **ENTER** key to see the output as shown in the second snapshot of sample run:

Some of the main steps used in above program are:

- Declare required amount of variable that is
**i**, n, num and**sum** - Initialize
**sum**with 0 - Receive the value of
**n** - Use for loop to receive desired amount of numbers
- Inside the
**for**loop, we have applied the summation operation each time when receiving the number - That is, when user enters first number say
**10**, then**sum+num**or**0+10**gets initialized to**sum** - Again when user enters second number say
**20**, then**sum+num**or**10+20**gets initialized to**sum** - Process the operation for all numbers. That is, if user enters 6 as value of
**n**, then operation goes 6 times - Print the value of
**sum**

The program given above is correct only when user supply all the numbers in integer form (without any decimal). So here is the modified version of the above program that is correct for both integer and decimal numbers.

This program is applicable to find the sum of all **n** real numbers.

// Program to Add n real numbers in C // -----codescracker.com----- #include<stdio.h> #include<conio.h> int main() { int n, i; float num, sum=0; printf("Enter the value of n: "); scanf("%d", &n); printf("Enter %d numbers: ", n); for(i=0; i<n; i++) { scanf("%f", &num); sum = sum+num; } printf("\nSum of all %d numbers = %0.2f", n, sum); getch(); return 0; }

Here is the snapshot of first sample run. In this sample run, we have provided all the numbers as integer value:

And here is the snapshot of second sample run. In this sample run, we have provided all the numbers as floating-point value:

In above program, we only done these changes:

- Declared
**num**and**sum**variable as**float**data type for real number operation - For both variable,
**%f**format specifier used here **0.2f**is used to print the value of**sum**upto 2 decimal places- Rest of the program is same as above one

As we all knows that positive numbers from **1, 2, 3, ...** are natural numbers. Therefore here we only have to ask from user,
the value of **n**, that is upto how many term, the natural number continues, and find sum of that natural number.

For example, if user enters the value of **n** as 8, then find sum of first 8 natural number, that will be **1, 2, 3, 4, 5, 6, 7**
and **8**. Let's take a look at the program:

// Program to find sum of n natural numbers in C // -----codescracker.com----- #include<stdio.h> #include<conio.h> int main() { int i, n, sum=0; printf("Enter the value of n: "); scanf("%d", &n); for(i=1; i<=n; i++) sum = sum+i; printf("\nSum of %d natural numbers = %d", n, sum); getch(); return 0; }

Here is the snapshot of sample run:

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