# Sum of series 8/10, 8/100, 8/1000, 8/10000. . . till N terms

• Last Updated : 16 Aug, 2022

Given a positive integer n, the task is to find the sum of series

8/10 + 8/100 + 8/1000 + 8/10000. . . till Nth term

Examples:

Input: n = 3
Output: 0.888

Input: n = 5
Output: 0.88888

Approach:

The total sum till nth term of the given G.P. series can be generalized as-

The above formula can be derived following the series of steps-

The given G.P. series

Here,

Thus, using the sum of G.P. formula for r<1

Substituting the values of a and r in the above equation

Illustration:

Input: n = 3
Output: 0.888
Explanation:

S_{n}=\frac{8}{9}(1-(\frac{1}{10})^{3})
= 0.888 * 0.999
= 0.888

Below is the implementation of the above problem-

## C++

 // C++ program to implement // the above approach #include using namespace std;   // Function to calculate sum of // given series till Nth term double sumOfSeries(double N) {     return (8 * ((pow(10, N) - 1) / pow(10, N))) / 9; }   // Driver code int main() {     double N = 5;     cout << sumOfSeries(N);     return 0; }

## Java

 // Java program to implement // the above approach import java.util.*; public class GFG {         // Function to calculate sum of     // given series till Nth term     static double sumOfSeries(double N)     {         return (8                 * ((Math.pow(10, N) - 1) / Math.pow(10, N)))             / 9;     }       // Driver code     public static void main(String args[])     {         double N = 5;         System.out.print(sumOfSeries(N));     } }   // This code is contributed by Samim Hossain Mondal.

## Python3

 # Python code for the above approach   # Function to calculate sum of # given series till Nth term def sumOfSeries(N):     return (8 * (((10 ** N) - 1) / (10 ** N))) / 9;   # Driver code N = 5; print(sumOfSeries(N));   # This code is contributed by gfgking

## C#

 // C# program to implement // the above approach using System; class GFG {         // Function to calculate sum of     // given series till Nth term     static double sumOfSeries(double N)     {         return (8                 * ((Math.Pow(10, N) - 1) / Math.Pow(10, N)))             / 9;     }       // Driver code     public static void Main()     {         double N = 5;         Console.WriteLine(sumOfSeries(N));     } }   // This code is contributed by ukasp.

## Javascript



Output

0.88888

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

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