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Program to find N-th term of the series a, b, b, c, c, c,…….

  • Difficulty Level : Medium
  • Last Updated : 24 Mar, 2021

Given a number N. The task is to write a program to find the N-th term in the below series:
 

a, b, b, c, c, c, d, d, d, d, …..

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Examples
 

Input : 12
Output : e

Input : 288
Output : x

 



The idea is to use AP sum formula to find the solution to this problem. Clearly the series is depicted as 1a, 2b’s, 3c’s, 4d’s, 5e’s and so on. Thus making it an AP.
Now we can use the AP sum formula: 
 

sum = (n/2)*(a + (n-1)*d)  

Which in this case becomes sum = (n(n+1))/2( since a = 1 and d = 1 ) where ‘sum’ here is the Nth term given.
Below is the implementation of the above approach: 
 

C++




// CPP program to find nth term of the
// given series
#include <bits/stdc++.h>
using namespace std;
 
// Function to find nth term of the
// given series
void findNthTerm(int n)
{
    // Let us find roots of equation x * (x + 1)/2 = n
    n = n * 2;
    int a = 1, b = 1, c = -1 * n;
    int d = b * b - 4 * a * c;
    double sqrt_val = sqrt(abs(d));
    int x1 = (double)(-b + sqrt_val) / (2 * a);
    int x2 = (double)(-b - sqrt_val) / (2 * a);
 
    if (x1 >= 1)
        cout << (char)('a' + x1) << endl;
    else if (x2 >= 1)
        cout << (char)('a' + x2) << endl;
}
 
// Driver program
int main()
{
    int n = 12;
    findNthTerm(n);
 
    n = 288;
    findNthTerm(n);
 
    return 0;
}

Java




// Java program to find nth
// term of the given series
import java.io.*;
 
class GFG {
 
    // Function to find nth term
    // of the given series
    static void findNthTerm(int n)
    {
        // Let us find roots of
        // equation x * (x + 1)/2 = n
        n = n * 2;
        int a = 1, b = 1, c = -1 * n;
        int d = b * b - 4 * a * c;
        double sqrt_val = Math.sqrt(Math.abs(d));
        int x1 = (int)((-b + sqrt_val) / (2 * a));
        int x2 = (int)((-b - sqrt_val) / (2 * a));
 
        if (x1 >= 1)
            System.out.println((char)('a' + x1));
        else if (x2 >= 1)
            System.out.println((char)('a' + x2));
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 12;
        findNthTerm(n);
 
        n = 288;
        findNthTerm(n);
    }
}
 
// This code has been contributed
// by anuj_67.

Python 3




# Python 3 program to find nth
# term of the given series
import math
 
# Function to find nth term
# of the given series
def findNthTerm(n):
 
    # Let us find roots of
    # equation x * (x + 1)/2 = n
    n = n * 2
    a = 1
    b = 1
    c = -1 * n
    d = b * b - 4 * a * c
    sqrt_val = math.sqrt(abs(d))
    x1 = (-b + sqrt_val) // (2 * a)
    x2 = (-b - sqrt_val) // (2 * a)
    x1 = int(x1)
    x2 = int(x2)
     
    # ASCII of 'a' is 97
    if (x1 >= 1):
        print(chr(97+x1))
    elif (x2 >= 1):
        print(chr(97+x2))
 
# Driver Code
if __name__ == "__main__":
    n = 12
    findNthTerm(n)
 
    n = 288
    findNthTerm(n)
 
# This code is contributed
# by ChitraNayal

C#




// C# program to find nth
// term of the given series
using System;
 
public class GFG {
 
    // Function to find nth term
    // of the given series
    static void findNthTerm(int n)
    {
        // Let us find roots of
        // equation x * (x + 1)/2 = n
        n = n * 2;
        int a = 1, b = 1, c = -1 * n;
        int d = b * b - 4 * a * c;
        double sqrt_val = Math.Sqrt(Math.Abs(d));
        int x1 = (int)((-b + sqrt_val) / (2 * a));
        int x2 = (int)((-b - sqrt_val) / (2 * a));
 
        if (x1 >= 1)
            Console.WriteLine((char)('a' + x1));
        else if (x2 >= 1)
            Console.WriteLine((char)('a' + x2));
    }
 
    // Driver Code
    static public void Main(String[] args)
    {
        int n = 12;
        findNthTerm(n);
 
        n = 288;
        findNthTerm(n);
    }
}
// contributed by Arnab Kundu

PHP




<?php
// PHP program to find nth
// term of the given series
 
// Function to find nth term
// of the given series
function findNthTerm($n)
{
    // Let us find roots of
    // equation x * (x + 1)/2 = n
    $n = $n * 2;
    $a = 1;
    $b = 1;
    $c = -1 * $n;
    $d = $b * $b - 4 * $a * $c;
    $sqrt_val = sqrt(abs($d));
    $x1 = (-$b + $sqrt_val) / (2 * $a);
    $x2 = (-$b - $sqrt_val) / (2 * $a);
 
    // Ascii of 'a' is 97
    if((int)$x1 >= 1)
        echo chr(97+$x1) . "\n";
    else if ((int)$x2 >= 1)
        echo chr(97+$x2), "\n";
}
 
// Driver Code
$n = 12;
findNthTerm($n);
 
$n = 288;
findNthTerm($n);
 
// This Code is contributed by mits
?>

Javascript




<script>
 
// Javascript program to find nth
// term of the given series
const str = "abcdefghijklmnopqrstuvwxyz";
 
// Function to find nth term
    // of the given series
    function findNthTerm( n) {
        // Let us find roots of
        // equation x * (x + 1)/2 = n
        n = n * 2;
        let a = 1, b = 1, c = -1 * n;
        let d = b * b - 4 * a * c;
        let sqrt_val = Math.sqrt(Math.abs(d));
        let x1 = parseInt( ((-b + sqrt_val) / (2 * a)));
        let x2 = parseInt( ((-b - sqrt_val) / (2 * a)));
 
        if (x1 >= 1)
            document.write(str[x1]+"<br/>");
        else if (x2 >= 1)
            document.write(str[x2]+"<br/>");
    }
 
    // Driver Code
      
        let n = 12;
        findNthTerm(n);
 
        n = 288;
        findNthTerm(n);
 
// This code is contributed by shikhasingrajput
 
</script>
Output: 
e
x

 




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