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# Program for sum of arithmetic series

• Difficulty Level : Easy
• 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.`

## PHP

 ``

## 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++

 `// Efficient solution to find sum of arithmetic series.``#include``using` `namespace` `std;`` ` `float` `sumOfAP(``float` `a, ``float` `d, ``float` `n)``{``    ``float` `sum = (n / 2) * (2 * a + (n - 1) * d);``    ``return` `sum;``}`` ` `// Driver code``int` `main()``{``    ``float` `n = 20;``    ``float` `a = 2.5, d = 1.5;``    ``cout<

## Java

 `// Java Efficient solution to find ``// sum of arithmetic series.``class` `GFG``{``    ``static` `float` `sumOfAP(``float` `a, ``float` `d, ``float` `n)``    ``{``        ``float` `sum = (n / ``2``) * (``2` `* a + (n - ``1``) * d);``        ``return` `sum;``    ``}`` ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args) ``    ``{``        ``float` `n = ``20``;``        ``float` `a = ``2``.5f, d = ``1``.5f;``        ``System.out.print(sumOfAP(a, d, n));``    ``}``}`` ` `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python3 Efficient ``# solution to find sum ``# of arithmetic series.`` ` `def`  `sumOfAP(a,  d,  n):``    ``sum` `=` `(n ``/` `2``) ``*` `(``2` `*` `a ``+` `(n ``-` `1``) ``*` `d)``    ``return` `sum``     ` `# Driver code    ``n ``=` `20``a ``=` `2.5``d ``=` `1.5`` ` `print``(sumOfAP(a, d, n))`` ` `# This code is ``# contributed by sunnysingh``  `

## 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

 `// Efficient solution to find sum of arithmetic series.`` ` `function` `sumOfAP(a, d, n) {``    ``let sum = (n / 2) * (2 * a + (n - 1) * d);``    ``return` `sum;``}`` ` `// Driver code``let n = 20;``let a = 2.5, d = 1.5;``document.write(sumOfAP(a, d, n));`` ` `// This code is contributed by Ashok`

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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