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

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  • 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-

S_{N}=\frac{N}{2}(2*a+(N-1)d)

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

S_{N}=\frac{N}{2}(2*a+(N-1)d)

Illustration:

Input: N = 5
Output: 40
Explanation:
a = 2 
d = 3
S_{N}=\frac{N}{2}(2*a+(N-1)d)
S_{N}=\frac{5}{2}(2*2+(5-1)3)
S_{N}=40

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)
 


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