# Program for sum of arithmetic series

Last Updated : 16 Feb, 2023

A series with the same common difference is known as arithmetic series. The first term of series is a and the common difference is d. The series looks like a, a + d, a + 2d, a + 3d, . . . Task is to find the sum of the series.

Examples:

```Input : a = 1
d = 2
n = 4
Output : 16
1 + 3 + 5 + 7 = 16

Input : a = 2.5
d = 1.5
n = 20
Output : 335```
Recommended Practice

A simple solution to find the sum of arithmetic series.

## C++

 `// CPP Program to find the sum of arithmetic  ` `// series. ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find sum of series. ` `float` `sumOfAP(``float` `a, ``float` `d, ``int` `n) ` `{ ` `    ``float` `sum = 0; ` `    ``for` `(``int` `i=0;i

## Java

 `// JAVA Program to find the sum of  ` `// arithmetic series. ` ` `  `class` `GFG{ ` `     `  `    ``// Function to find sum of series. ` `    ``static` `float` `sumOfAP(``float` `a, ``float` `d,  ` `                                  ``int` `n) ` `    ``{ ` `        ``float` `sum = ``0``; ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{ ` `            ``sum = sum + a; ` `            ``a = a + d; ` `        ``} ` `        ``return` `sum; ` `    ``} ` `     `  `    ``// Driver function ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `n = ``20``; ` `        ``float` `a = ``2``.5f, d = ``1``.5f; ` `        ``System.out.println(sumOfAP(a, d, n)); ` `    ``} ` `} ` ` `  `/*This code is contributed by Nikita Tiwari.*/`

## Python

 `# Python Program to find the sum of  ` `# arithmetic series. ` ` `  `# Function to find sum of series. ` `def` `sumOfAP( a, d,n) : ` `    ``sum` `=` `0` `    ``i ``=` `0` `    ``while` `i < n : ` `        ``sum` `=` `sum` `+` `a ` `        ``a ``=` `a ``+` `d ` `        ``i ``=` `i ``+` `1` `    ``return` `sum` `     `  `# Driver function ` `n ``=` `20` `a ``=` `2.5` `d ``=` `1.5` `print` `(sumOfAP(a, d, n)) ` ` `  `# This code is contributed by Nikita Tiwari. `

## C#

 `// C# Program to find the sum of  ` `// arithmetic series. ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// Function to find sum of series. ` `    ``static` `float` `sumOfAP(``float` `a, ``float` `d,  ` `                                    ``int` `n) ` `    ``{ ` `        ``float` `sum = 0; ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``sum = sum + a; ` `            ``a = a + d; ` `        ``} ` `         `  `        ``return` `sum; ` `    ``} ` `     `  `    ``// Driver function ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 20; ` `        ``float` `a = 2.5f, d = 1.5f; ` `         `  `        ``Console.Write(sumOfAP(a, d, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by parashar. `

## Javascript

Output:

`335`

Time Complexity: O(n)

Space complexity: O(1) because using constant space

### Approach 2:

An Efficient solution to find the sum of arithmetic series is to use the below formula as follows:

```Sum of arithmetic series
= ((n / 2) * (2 * a + (n - 1) * d))
Where
a - First term
d - Common difference
n - No of terms```

Example:

## C#

 `// C# efficient solution to find  ` `// sum of arithmetic series. ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``static` `float` `sumOfAP(``float` `a,  ` `                         ``float` `d,  ` `                         ``float` `n) ` `    ``{ ` `        ``float` `sum = (n / 2) *  ` `                    ``(2 * a +  ` `                    ``(n - 1) * d); ` `        ``return` `sum; ` `    ``} ` `     `  `    ``// Driver code ` `    ``static` `public` `void` `Main () ` `    ``{ ` `        ``float` `n = 20; ` `        ``float` `a = 2.5f, d = 1.5f; ` `        ``Console.WriteLine(sumOfAP(a, d, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by Ajit. `

## PHP

 ` `

## Javascript

Output

`335`

Time Complexity: O(1)

Space complexity: O(1) since using only constant variables

How does this formula work?

We can prove the formula using mathematical induction. We can easily see that the formula holds true for n = 1 and n = 2. Let this be true for n = k-1.

```Let the formula be true for n = k-1.
Sum of first k - 1 elements of arithmetic series is
= (((k-1))/ 2) * (2 * a + (k - 2) * d))
We know k-th term of arithmetic series is
= a + (k - 1)*d

Sum of first k elements =
= Sum of (k-1) numbers + k-th element
= (((k-1)/2)*(2*a + (k-2)*d)) + (a + (k-1)*d)
= [((k-1)(2a + (k-2)d) + (2a + 2kd - 2d)]/2
= ((k / 2) * (2 * a + (k - 1) * d))```

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