Given three integers A, D, and R representing the first term, common difference, and common ratio of an infinite Arithmetic-Geometric Progression, the task is to find the sum of the given infinite Arithmetic-Geometric Progression such that the absolute value of R is always less than 1.
Examples:
Input: A = 0, D = 1, R = 0.5
Output: 0.666667Input: A = 2, D = 3, R = -0.3
Output: 0.549451
Approach: An arithmetic-geometric sequence is the result of the term-by-term multiplication of the geometric progression series with the corresponding terms of an arithmetic progression series. The series is given by:
a, (a + d) * r, (a + 2 * d) * r2, (a + 3 * d) * r3, …, [a + (N ? 1) * d] * r(N ? 1).
The Nth term of the Arithmetic-Geometric Progression is given by:
=>
![T_N = [a + (N - 1) * d] * (b * r^{n - 1})](https://neveropen.co.uk/wp-content/uploads/2023/11/dominic_rubhabha_quicklatex.com-87b7f894daa8b072bc4ae02316a45992_l3.png "Rendered by QuickLaTeX.com")
The sum of the Arithmetic-Geometric Progression is given by:
=>
where, |r| < 1.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the sum of the// infinite AGPvoid sumOfInfiniteAGP(double a, double d, double r){ // Stores the sum of infinite AGP double ans = a / (1 - r) + (d * r) / pow((1-r),2); // Print the required sum cout << ans;}// Driver Codeint main(){ double a = 0, d = 1, r = 0.5; sumOfInfiniteAGP(a, d, r); return 0;}// Correction done by Yogesh0903 |
Java
import java.lang.Math;// Java program for the above approachclass GFG{// Function to find the sum of the// infinite AGPstatic void sumOfInfiniteAGP(double a, double d, double r){ // Stores the sum of infinite AGP double ans = a / (1 - r) + (d * r) / Math.pow((1-r),2); // Print the required sum System.out.print(ans);}// Driver Codepublic static void main(String[] args){ double a = 0, d = 1, r = 0.5; sumOfInfiniteAGP(a, d, r);}}// This code is contributed by 29AjayKumar// Correction done by Yogesh0903 |
Python3
# Python3 program for the above approach# Function to find the sum of the# infinite AGPdef sumOfInfiniteAGP(a, d, r): # Stores the sum of infinite AGP ans = a / (1 - r) + (d * r) / (1 - r)**2; # Print the required sum print (round(ans,6))# Driver Codeif __name__ == '__main__': a, d, r = 0, 1, 0.5 sumOfInfiniteAGP(a, d, r)# This code is contributed by mohit kumar 29.# Correction done by Yogesh0903 |
C#
// C# program for the above approachusing System;class GFG{ // Function to find the sum of the // infinite AGP static void sumOfInfiniteAGP(double a, double d, double r) { // Stores the sum of infinite AGP double ans = a / (1 - r) + (d * r) / Math.Pow((1-r),2); // Print the required sum Console.Write(ans); } // Driver Code public static void Main() { double a = 0, d = 1, r = 0.5; sumOfInfiniteAGP(a, d, r); }}// This code is contributed by ukasp.// Correction done by Yogesh0903 |
Javascript
<script> // Javascript program for the above approach // Function to find the sum of the // infinite AGP function sumOfInfiniteAGP(a, d, r) { // Stores the sum of infinite AGP let ans = a / (1 - r) + (d * r) / Math.pow((1-r),2); // Print the required sum document.write(ans) } // Driver Code let a = 0, d = 1, r = 0.5; sumOfInfiniteAGP(a, d, r); // This code is contributed by Hritik </script> |
0.666667
Time Complexity: O(1)
Auxiliary Space: O(1), since no extra space has been taken.
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