Program for N-th term of Arithmetic Progression series

Given first term (a), common difference (d) and a integer N of the Arithmetic Progression series, the task is to find N^thterm of the series.
Examples :

Input : a = 2 d = 1 N = 5
Output :
The 5th term of the series is : 6

Input : a = 5 d = 2 N = 10
Output :
The 10th term of the series is : 23

Approach:

We know the Arithmetic Progression series is like = 2, 5, 8, 11, 14 …. …
In this series 2 is the stating term of the series .
Common difference = 5 – 2 = 3 (Difference common in the series).
so we can write the series as :
t₁ = a₁
t₂ = a₁ + (2-1) * d
t₃ = a₁ + (3-1) * d
.
.
.
t_N = a₁ + (N-1) * d

To find the N^th term in the Arithmetic Progression series we use the simple formula .

T_N = a₁ + (N-1) * d

C++

// CPP Program to find nth term of
// Arithmetic progression
#include <bits/stdc++.h>
using namespace std;
 
int Nth_of_AP(int a, int d, int N)
{
    // using formula to find the
    // Nth term t(n) = a(1) + (n-1)*d
    return (a + (N - 1) * d);
     
}
 
// Driver code
int main()
{
    // starting number
    int a = 2;
     
    // Common difference
    int d = 1;
     
    // N th term to be find
    int N = 5;
     
    // Display the output
    cout << "The "<< N
         <<"th term of the series is : "
         << Nth_of_AP(a,d,N);
 
    return 0;
}

Java

// Java program to find nth term
// of Arithmetic progression
import java.io.*;
import java.lang.*;
 
class GFG
{
    public static int Nth_of_AP(int a,
                                int d,
                                int N)
    {
        // using formula to find the Nth
        // term t(n) = a(1) + (n-1)*d
        return ( a + (N - 1) * d );
    }
 
    // Driver code
    public static void main(String[] args)
    {
        // starting number
        int a = 2;
         
        // Common difference
        int d = 1;
         
        // N th term to be find
        int N = 5;
 
        // Display the output
        System.out.print("The "+ N +
                         "th term of the series is : " +
                          Nth_of_AP(a, d, N));
    }
}

Python3

# Python 3 Program to
# find nth term of
# Arithmetic progression
 
def Nth_of_AP(a, d, N) :
 
    # using formula to find the
    # Nth term t(n) = a(1) + (n-1)*d
    return (a + (N - 1) * d)
      
  
# Driver code
a = 2  # starting number
d = 1  # Common difference
N = 5  # N th term to be find
  
# Display the output
print( "The ", N ,"th term of the series is : ",
       Nth_of_AP(a, d, N))
 
 
 
# This code is contributed
# by Nikita Tiwari.

Javascript

<script>
 
// JavaScript Program to find nth term of 
// Arithmetic progression
 
    function Nth_of_AP(a, d, N)
    { 
        // using formula to find the 
        // Nth term t(n) = a(1) + (n-1)*d
        return (a + (N - 1) * d);
       
    }
   
    // Driver code
 
    // starting number
    let a = 2; 
       
    // Common difference
    let d = 1; 
       
    // N th term to be find
    let N = 5; 
       
    // Display the output
    document.write("The "+ N + "th term of the series is : "
    + Nth_of_AP(a,d,N));
   
 
// This code is contributed by Mayank Tyagi
 
</script>

C#

// C# program to find nth term
// of Arithmetic progression
using System;
 
class GFG
{
    public static int Nth_of_AP(int a,
                                int d,
                                int N)
    {
         
        // using formula to find the Nth
        // term t(n) = a(1) + (n-1)*d
        return ( a + (N - 1) * d );
    }
 
    // Driver code
    public static void Main()
    {
        // starting number
        int a = 2;
         
        // Common difference
        int d = 1;
         
        // N th term to be find
        int N = 5;
 
        // Display the output
        Console.WriteLine("The "+ N +
                          "th term of the series is : " +
                           Nth_of_AP(a, d, N));
    }
}
 
// This code is contributed by vt_m.

PHP

<?php
// PHP Program to find nth term of
// Arithmetic progression
 
function Nth_of_AP($a, $d, $N)
{
    // using formula to find the
    // Nth term t(n) = a(1) + (n-1)*d
    return ($a + ($N - 1) * $d);
     
}
 
// Driver code
 
// starting number
$a = 2;
 
// Common difference
$d = 1;
 
// N th term to be find
$N = 5;
     
// Display the output
echo("The " . $N . "th term of the series is : " .
                           Nth_of_AP($a, $d, $N));
 
// This code is contributed by Ajit.
?>

Output :

The 5th term of the series is : 6

Time Complexity: O(1), the code will run in a constant time.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Approach 2: Using a loop to iterate over each term in the series until the N-th term is reached.

C++

#include <iostream>
using namespace std;
 
int main()
{
    int a = 2;
    int d = 1;
    int n = 5;
     
    int nthTerm = a;
    for (int i = 1; i < n; i++)
    {
        nthTerm += d;
    }
    cout << "The " << n << "th term of the series is: " << nthTerm << endl;
    return 0;
}

Java

import java.util.*;
 
public class Main {
    public static void main(String[] args) {
        int a = 2;
        int d = 1;
        int n = 5;
 
        int nthTerm = a;
        for (int i = 1; i < n; i++) {
            nthTerm += d;
        }
        System.out.println("The " + n + "th term of the series is: " + nthTerm);
    }
}

Python3

a = 2
d = 1
n = 5
 
nthTerm = a
for i in range(1, n):
    nthTerm += d
 
print("The", n, "th term of the series is:", nthTerm)
 
# This code is contributed by shivhack999

Javascript

let a = 2;
let d = 1;
let n = 5;
 
let nthTerm = a;
for (let i = 1; i < n; i++) {
    nthTerm += d;
}
console.log("The " + n + "th term of the series is: " + nthTerm);

C#

using System;
 
public class MainClass
{
    public static void Main(string[] args)
    {
        int a = 2;
        int d = 1;
        int n = 5;
 
        int nthTerm = a;
        for (int i = 1; i < n; i++)
        {
            nthTerm += d;
        }
        Console.WriteLine("The " + n + "th term of the series is: " + nthTerm);
    }
}