# Find the sum of first N terms of the series 2, 5, 8, 11, 14..

Given a positive integer **N**, the task is to find the sum of the first N terms of the series

2, 5, 8, 11, 14..

**Examples:**

Input:N = 540Output:

Input:N = 10Output:155

**Approach:**

1st term = 2

2nd term = (2 + 3) = 5

3rd term = (5 + 3) = 8

4th term = (8 + 3) = 11

.

.

Nth term = (2 + (N – 1) * 3) = 3N – 1

The sequence is formed by using the following pattern. For any value N-

Here,

a is the first term

d is the common difference

The above solution can be derived following the series of steps-

The series-

2, 5, 8, 11, …., till N terms

is in A.P. with first term of the series a = 2 and common difference d = 3

Sum of N terms of an A.P. is

**Illustration:**

Input:N = 5Output:40Explanation:

a = 2

d = 3

Below is the implementation of the above approach:

## C++

`// C++ program to implement` `// the above approach` `#include <iostream>` `using` `namespace` `std;` `// Function to return` `// Nth term of the series` `int` `nth(` `int` `n, ` `int` `a1, ` `int` `d)` `{` ` ` `return` `(a1 + (n - 1) * d);` `}` `// Function to return sum of` `// Nth term of the series` `int` `sum(` `int` `a1, ` `int` `nterm, ` `int` `n)` `{` ` ` `return` `n * (a1 + nterm) / 2;` `}` `// Driver code` `int` `main()` `{` ` ` `// Value of N` ` ` `int` `N = 5;` ` ` `// First term` ` ` `int` `a = 2;` ` ` `// Common difference` ` ` `int` `d = 3;` ` ` `// finding last term` ` ` `int` `nterm = nth(N, a, d);` ` ` `cout << sum(a, nterm, N) << endl;` ` ` `return` `0;` `}` |

## C

`// C program to implement` `// the above approach` `#include <stdio.h>` `// Function to return` `// Nth term of the series` `int` `nth(` `int` `n, ` `int` `a1, ` `int` `d)` `{` ` ` `return` `(a1 + (n - 1) * d);` `}` `// Function to return` `// sum of Nth term of the series` `int` `sum(` `int` `a1, ` `int` `nterm, ` `int` `n)` `{` ` ` `return` `n * (a1 + nterm) / 2;` `}` `// Driver code` `int` `main()` `{` ` ` `// Value of N` ` ` `int` `N = 5;` ` ` `// First term` ` ` `int` `a = 2;` ` ` `// Common difference` ` ` `int` `d = 3;` ` ` `// Finding last term` ` ` `int` `nterm = nth(N, a, d);` ` ` `printf` `(` `"%d"` `, sum(a, nterm, N));` ` ` `return` `0;` `}` |

## Java

`// Java program to implement` `// the above approach` `import` `java.io.*;` `class` `GFG {` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `// Value of N` ` ` `int` `N = ` `5` `;` ` ` `// First term` ` ` `int` `a = ` `2` `;` ` ` `// Common difference` ` ` `int` `d = ` `3` `;` ` ` `// Finding Nth term` ` ` `int` `nterm = nth(N, a, d);` ` ` `System.out.println(sum(a, nterm, N));` ` ` `}` ` ` `// Function to return` ` ` `// Nth term of the series` ` ` `public` `static` `int` `nth(` `int` `n,` ` ` `int` `a1, ` `int` `d)` ` ` `{` ` ` `return` `(a1 + (n - ` `1` `) * d);` ` ` `}` ` ` `// Function to return` ` ` `// sum of Nth term of the series` ` ` `public` `static` `int` `sum(` `int` `a1,` ` ` `int` `nterm, ` `int` `n)` ` ` `{` ` ` `return` `n * (a1 + nterm) / ` `2` `;` ` ` `}` `}` |

## Python3

`# Python code for the above approach` `# Function to return` `# Nth term of the series` `def` `nth(n, a1, d):` ` ` `return` `(a1 ` `+` `(n ` `-` `1` `) ` `*` `d);` `# Function to return sum of` `# Nth term of the series` `def` `sum` `(a1, nterm, n):` ` ` `return` `n ` `*` `(a1 ` `+` `nterm) ` `/` `2` `;` `# Driver code` `# Value of N` `N ` `=` `5` `;` `# First term` `a ` `=` `2` `;` `# Common difference` `d ` `=` `3` `;` `# finding last term` `nterm ` `=` `nth(N, a, d);` `print` `((` `int` `)(` `sum` `(a, nterm, N)))` `# This code is contributed by gfgking` |

## C#

`using` `System;` `public` `class` `GFG` `{` ` ` ` ` `// Function to return` ` ` `// Nth term of the series` ` ` `public` `static` `int` `nth(` `int` `n, ` `int` `a1, ` `int` `d)` ` ` `{` ` ` `return` `(a1 + (n - 1) * d);` ` ` `}` ` ` `// Function to return` ` ` `// sum of Nth term of the series` ` ` `public` `static` `int` `sum(` `int` `a1, ` `int` `nterm, ` `int` `n)` ` ` `{` ` ` `return` `n * (a1 + nterm) / 2;` ` ` `}` ` ` `static` `public` `void` `Main()` ` ` `{` ` ` `// Code` ` ` `// Value of N` ` ` `int` `N = 5;` ` ` `// First term` ` ` `int` `a = 2;` ` ` `// Common difference` ` ` `int` `d = 3;` ` ` `// Finding Nth term` ` ` `int` `nterm = nth(N, a, d);` ` ` `Console.Write(sum(a, nterm, N));` ` ` `}` `}` `// This code is contributed by Potta Lokesh` |

## Javascript

`<script>` ` ` `// JavaScript code for the above approach` ` ` `// Function to return` ` ` `// Nth term of the series` ` ` `function` `nth(n, a1, d) {` ` ` `return` `(a1 + (n - 1) * d);` ` ` `}` ` ` `// Function to return sum of` ` ` `// Nth term of the series` ` ` `function` `sum(a1, nterm, n) {` ` ` `return` `n * (a1 + nterm) / 2;` ` ` `}` ` ` `// Driver code` ` ` `// Value of N` ` ` `let N = 5;` ` ` `// First term` ` ` `let a = 2;` ` ` `// Common difference` ` ` `let d = 3;` ` ` `// finding last term` ` ` `let nterm = nth(N, a, d);` ` ` `document.write(sum(a, nterm, N) + ` `'<br>'` `);` ` ` `// This code is contributed by Potta Lokesh` ` ` `</script>` |

**Output:**

40

**Time complexity:** O(1) because performing constant operations

**Auxiliary Space: **O(1)