Program for sum of geometric series

A Geometric series is a series with a constant ratio between successive terms. The first term of the series is denoted by a and common ratio is denoted by r. The series looks like this :- a, ar, ar², ar³, ar⁴, . . .. The task is to find the sum of such a series.

Examples :

Input : a = 1
        r = 0.5
        n = 10
Output : 1.99805

Input : a = 2
        r = 2
        n = 15
Output : 65534

A Simple solution to calculate the sum of geometric series.

C++

// A naive solution for calculating sum of
// geometric series.
#include
using namespace std;

// function to calculate sum of
// geometric series
float sumOfGP(float a, float r, int n)
{
float sum = 0;
for (int i = 0; i < n; i++) { sum = sum + a; a = a * r; } return sum; } // driver function int main() { int a = 1; // first term float r = 0.5; // common ratio int n = 10; // number of terms cout << sumOfGP(a, r, n) << endl; return 0; }

Java

// A naive solution for calculating sum of
// geometric series.
import java.io.*;

class GFG{

// function to calculate sum of
// geometric series
static float sumOfGP(float a, float r, int n)
{
float sum = 0;
for (int i = 0; i < n; i++) { sum = sum + a; a = a * r; } return sum; } // driver function public static void main(String args[]) { int a = 1; // first term float r = (float)(1/2.0) ;// common ratio int n = 10 ; // number of terms System.out.printf("%.5f",(sumOfGP(a, r, n))); } } //This code is contributed by Nikita Tiwari

Python

# A naive solution for calculating sum of
# geometric series.

# function to calculate sum of
# geometric series
def sumOfGP(a, r, n) :

sum = 0
i = 0
while i < n : sum = sum + a a = a * r i = i + 1 return sum #driver function a = 1 # first term r = (float)(1/2.0) # common ratio n = 10 # number of terms print("%.5f" %sumOfGP(a, r, n)), # This code is contributed by Nikita Tiwari

C#

// A naive solution for calculating
// sum of geometric series.
using System;

class GFG {

// function to calculate
// sum of geometric series
static float sumOfGP(float a,
float r,
int n)
{
float sum = 0;
for (int i = 0; i < n; i++) { sum = sum + a; a = a * r; } return sum; } // Driver Code static public void Main () { // first term int a = 1; // common ratio float r = (float)(1/2.0) ; // number of terms int n = 10 ; Console.WriteLine((sumOfGP(a, r, n))); } } // This code is contributed by Ajit.

Output :

1.99805

Time Complexity: O(n).

An Efficient solution to solve the sum of geometric series where first term is a and common ration is r
is by the formula :-
sum of series = a(1 – rⁿ)/(1 – r).
Where r = T2/T1 = T3/T2 = T4/T3 . . .
and T1, T2, T3, T4 . . . ,Tn are the first, second, third, . . . ,nth terms respectively.
For example – The series is 2, 4, 8, 16, 32, 64, . . . upto 15 elements. In the above series, find the sum of first 15 elements where
first term a = 2 and common ration r = 4/2 = 2 or = 8/4 = 2
Then,
sum = 2 * (1 – 2¹⁵) / (1 – 2).
sum = 65534

C++

// An Efficient solution to solve sum of
// geometric series.
#include
using namespace std;

// function to calculate sum of
// geometric series
float sumOfGP(float a, float r, int n)
{
// calculating and storing sum
return (a * (1 – pow(r, n))) / (1 – r);
}

// driver code
int main()
{
float a = 2; // first term
float r = 2; // common ratio
int n = 15; // number of terms

cout << sumOfGP(a, r, n); return 0; }

Java

// An Efficient solution to solve sum of
// geometric series.
import java.math.*;

class GFG{

// function to calculate sum of
// geometric series
static float sumOfGP(float a, float r, int n)
{
// calculating and storing sum
return (a * (1 – (int)(Math.pow(r, n)))) /
(1 – r);
}

// driver code
public static void main(String args[])
{
float a = 2; // first term
float r = 2; // common ratio
int n = 15; // number of terms

System.out.println((int)(sumOfGP(a, r, n)));
}
}

//This code is contributed by Nikita Tiwari.

Python

# An Efficient solution to solve sum of
# geometric series.

# function to calculate sum of
# geometric series
def sumOfGP( a, r, n) :

# calculating and storing sum
return (a * (1 – pow(r, n))) / (1 – r)

# driver code
a = 2 # first term
r = 2 # common ratio
n = 15 # number of terms

print sumOfGP(a, r, n)

# This code is contributed by Nikita Tiwari.

C#

// C# program to An Efficient solution
// to solve sum of geometric series.
using System;

class GFG {

// function to calculate sum of
// geometric series
static float sumOfGP(float a, float r, int n)
{
// calculating and storing sum
return (a * (1 – (int)(Math.Pow(r, n)))) /
(1 – r);
}

// Driver Code
public static void Main()
{
float a = 2; // first term
float r = 2; // common ratio
int n = 15; // number of terms

Console.Write((int)(sumOfGP(a, r, n)));
}
}

// This code is contributed by Nitin Mittal.

Output :

65534

Time Complexity: Depends on implementation of pow() function in C/C++. In general, we can compute integer powers in O(Log n) time.

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