Program to calculate sum of an Infinite Arithmetic-Geometric Sequence

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: 0.666667

Input: 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 N^th term of the Arithmetic-Geometric Progression is given by:

=> T_N = [a + (N - 1) * d] * (b * r^{n - 1})

The sum of the Arithmetic-Geometric Progression is given by:

=>S_? = \frac{a}{(1 - r)} + \frac{d * r}{(1 - r)^2}

where, |r| < 1.

Below is the implementation of the above approach:

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the sum of the
// infinite AGP
void 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 Code
int main()
{
    double a = 0, d = 1, r = 0.5;
    sumOfInfiniteAGP(a, d, r);

    return 0;
}
// Correction done by Yogesh0903

import java.lang.Math;
// Java program for the above approach
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
    System.out.print(ans);
}

// Driver Code
public 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 program for the above approach

# Function to find the sum of the
# infinite AGP
def 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 Code
if __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# program for the above approach
using 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

<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.