Encuentra la suma de los primeros N términos de la serie 2, 5, 8, 11, 14..

Dado un entero positivo N , la tarea es encontrar la suma de los primeros N términos de la serie 

2, 5, 8, 11, 14..

Ejemplos:

Entrada: N = 5
Salida: 40

Entrada : N = 10
Salida : 155

Acercarse:

1er término = 2

2do término = (2 + 3) = 5

3er término = (5 + 3) = 8

4to término = (8 + 3) = 11

.

.

N-ésimo término = (2 + (N – 1) * 3) = 3N – 1

La secuencia se forma usando el siguiente patrón. Para cualquier valor N-

Aquí, 
a es el primer término
d es la diferencia común

La solución anterior se puede derivar siguiendo la serie de pasos:

Las series-

2, 5, 8, 11, …., hasta N términos

está en AP con primer término de la serie a = 2 y diferencia común d = 3

La suma de N términos de un AP es

Ilustración:

Entrada: N = 5
Salida: 40
Explicación:
a = 2 
d = 3

A continuación se muestra la implementación del enfoque anterior:

// 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 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 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;
    }
}

# 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

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

<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>

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