Given a number N. The task is to write a program to find the Nth term in the below series:
0, 2, 4, 8, 12, 18…
Examples:
Input: 3
Output: 4
For N = 3
Nth term = ( 3 + ( 3 - 1 ) * 3 ) / 2
= 4
Input: 5
Output: 12
On observing carefully, the Nth term in the above series can be generalized as:
Nth term = ( N + ( N - 1 ) * N ) / 2
Below is the implementation of the above approach:
C++
// CPP program to find N-th term of the series:// 0, 2, 4, 8, 12, 18...#include <iostream>using namespace std;// Calculate Nth term of seriesint nthTerm(int N){ return (N + N * (N - 1)) / 2;}// Driver Functionint main(){ int N = 5; cout << nthTerm(N); return 0;} |
Java
// Java program to find N-th term of the series:// 0, 2, 4, 8, 12, 18...import java.io.*;// Main class for main methodclass GFG { public static int nthTerm(int N) { // By using above formula return (N + (N - 1) * N) / 2; } // Driver code public static void main(String[] args) { int N = 5; // 5th term is 12 System.out.println(nthTerm(N)); }} |
Python 3
# Python 3 program to find N-th term of the series: # 0, 2, 4, 8, 12, 18.# Calculate Nth term of seriesdef nthTerm(N) : return (N + N * (N - 1)) // 2# Driver Codeif __name__ == "__main__" : N = 5 print(nthTerm(N))# This code is contributed by ANKITRAI1 |
C#
// C# program to find N-th term of the series:// 0, 2, 4, 8, 12, 18...using System;class gfg{ // Calculate Nth term of series public int nthTerm(int N) { int n = ((N + N * (N - 1)) / 2); return n; } //Driver program static void Main(string[] args) { gfg p = new gfg(); int a = p.nthTerm(5); Console.WriteLine(a); Console.Read(); }}//This code is contributed by SoumikMondal |
PHP
<?php// PHP program to find // N-th term of the series:// 0, 2, 4, 8, 12, 18...// Calculate Nth term of seriesfunction nthTerm($N){ return (int)(($N + $N * ($N - 1)) / 2);}// Driver Code$N = 5;echo nthTerm($N);// This code is contributed by mits?> |
Javascript
<script>// JavaScript program to find N-th term of the series:// 0, 2, 4, 8, 12, 18...// Calculate Nth term of seriesfunction nthTerm(N){ return parseInt((N + N * (N - 1)) / 2);}// Driver Function let N = 5; document.write(nthTerm(N)); // This code contributed by Rajput-Ji </script> |
Output:
12
Time Complexity: O(1)
Auxiliary space: O(1) as it is using constant variables
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