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Program to calculate sum of an Infinite Arithmetic-Geometric Sequence
  • Last Updated : 13 May, 2021

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 Nth 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++




// 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) / (1 - r * r);
 
    // Print the required sum
    cout << ans;
}
 
// Driver Code
int main()
{
    double a = 0, d = 1, r = 0.5;
    sumOfInfiniteAGP(a, d, r);
 
    return 0;
}

Java




// 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) / (1 - r * r);
 
    // 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

Python3




# 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 * r);
 
    # 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.

C#




// 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) / (1 - r * r);
 
        // 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.

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) / (1 - r * r);
 
            // 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>
Output
0.666667

Time Complexity: O(1)
Auxiliary Space: O(1)

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