Given a positive integer N, the task is to find the Nth term of the series
5, 10, 20, 40….till N terms
Examples:
Input: N = 5
Output: 80Input: N = 3
Output: 20
Approach:
1st term = 5 * (2 ^ (1 – 1)) = 5
2nd term = 5 * (2 ^ (2 – 1)) = 10
3rd term = 5 * (2 ^ (3 – 1)) = 20
4th term = 5 * (2 ^ (4 – 1)) = 40
.
.
Nth term = 5 * (2 ^ (N – 1))
The Nth term of the given series can be generalized as-
TN = (a * (r ^ (N – 1))
The following steps can be followed to derive the formula-
The series 5, 10, 20, 40….till N terms
is in G.P. with
first term a = 5
common ratio r = 2 because each term is double the one before it.
The Nth term of a G.P. is
TN = (a * (r ^ (N – 1))
Illustration:
Input: N = 5
Output: 80
Explanation:
TN = (a * (r ^ (N – 1))
= (5 * (2 ^ (5 – 1))
= (5 * 16)
= 80
Below is the implementation of the above approach-
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate nth termint nTerm(int a, int r, int n){ return a * pow(r, n - 1);}// Driver codeint main(){ // Value of N int N = 5; // First term of the series int a = 5; // Common ratio int r = 2; cout << nTerm(a, r, N); return 0;} |
C
// C program to implement// the above approach#include <math.h>#include <stdio.h>// Function to calculate nth termint nTerm(int a, int r, int n){ return a * pow(r, n - 1);}// Driver codeint main(){ // Value of N int N = 5; // First term int a = 5; // Common ratio int r = 2; printf("%d", nTerm(a, r, n)); return 0;} |
Java
// Java program to implement// the above approachimport java.io.*;class GFG { // Driver code public static void main(String[] args) { // Value of N int N = 5; // First term int a = 5; // Common ratio int r = 2; System.out.println(nTerm(a, r, N)); } // Function to calculate nth term public static int nTerm(int a, int r, int n) { return a * ((int)Math.pow(r, n - 1)); }} |
Python3
# python3 program to implement# the above approach# Function to calculate nth termdef nTerm(a, r, n): return a * pow(r, n - 1)# Driver codeif __name__ == "__main__": # Value of N N = 5 # First term of the series a = 5 # Common ratio r = 2 print(nTerm(a, r, N))# This code is contributed by rakeshsahni |
C#
using System;public class GFG{ // Function to calculate nth term public static int nTerm(int a, int r, int n) { return a * ((int)Math.Pow(r, n - 1)); } static public void Main() { // Code // Value of N int N = 5; // First term int a = 5; // Common ratio int r = 2; Console.Write(nTerm(a, r, N)); }}// This code is contributed by Potta Lokesh |
Javascript
<script> // JavaScript code for the above approach // Function to calculate nth term function nTerm(a, r, n) { return a * Math.pow(r, n - 1); } // Driver code // Value of N let N = 5; // First term of the series let a = 5; // Common ratio let r = 2; document.write(nTerm(a, r, N)); // This code is contributed by Potta Lokesh </script> |
80
Time complexity: O(logrn) because it is using inbuilt pow function
Auxiliary Space: O(1) // since no extra array is used so the space taken by the algorithm is constant
