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++
// 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
// 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
# 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#
// 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.
PHP
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++
// 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
// 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
# 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#
// 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.
PHP
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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