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# 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 = 5
Output: 40

Input: N = 10
Output: 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 = 5
Output: 40
Explanation:
a = 2
d = 3   Below is the implementation of the above approach:

## C++

 // C++ program to implement // the above approach #include  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    // 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

 

Output:

40

Time complexity: O(1) because performing constant operations

Auxiliary Space: O(1) // since no extra array or recursion is used so the space taken by the algorithm is constant

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