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# Program for sum of arithmetic series

• Difficulty Level : Easy
• Last Updated : 01 Apr, 2021

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

A simple solution to find 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)
An Efficient solution to find the sum of arithmetic series is to use below formula.

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

## 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)
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 geometric 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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