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

• Last Updated : 03 Aug, 2022

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 seriesint nth(int n, int a1, int d){    return (a1 + (n - 1) * d);} // Function to return sum of// Nth term of the seriesint sum(int a1, int nterm, int n){    return n * (a1 + nterm) / 2;} // Driver codeint 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 seriesint nth(int n, int a1, int d){    return (a1 + (n - 1) * d);} // Function to return// sum of Nth term of the seriesint sum(int a1, int nterm, int n){    return n * (a1 + nterm) / 2;} // Driver codeint 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 approachimport 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 seriesdef nth(n, a1, d):    return (a1 + (n - 1) * d); # Function to return sum of# Nth term of the seriesdef sum(a1, nterm, n):    return n * (a1 + nterm) / 2; # Driver code # Value of NN = 5; # First terma = 2; # Common differenced = 3; # finding last termnterm = 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)

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