# Print first N terms of series (0.25, 0.5, 0.75, …) in fraction representation

• Last Updated : 18 Mar, 2021

Given an integer N, the task is to print the first N terms of the series in their fraction form i.e.

1/4, 1/2, 3/4, 1, 5/4, …

The above series has values as 0.25, 0.5, 0.75, 1, 1.25, ….etc. It is an Arithmetic progression that begins with 0.25 and has a difference of 0.25.
Examples:

Input: N = 6
Output: 1/4 1/2 3/4 1 5/4 3/2
Input: N = 9
Output: 1/4 1/2 3/4 1 5/4 3/2 7/4 2 9/4

Approach: Consider the first four terms of the series as the base terms. Store the numerator elements and denominator elements separately.
Consider the first term 1/4, the fifth term is 1 + (1 * 4) / 4 which is 1/5
Similarly, consider the second term 1/2 the sixth term is 1 + (1 * 2) / 2 which is 3/2
Hence, we can consider the denominators will always be either 2, 4 or no denominator and the numerator of the term can be calculated from the denominator.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to print the required series``void` `printSeries(``int` `n)``{``    ``// Numerators for the first four numerators``    ``// of the series``    ``int` `nmtr = { 1, 1, 1, 3 };` `    ``// Denominators for the first four denominators``    ``// of the series``    ``int` `dntr = { 0, 4, 2, 4 };` `    ``for` `(``int` `i = 1; i <= n; i++) {` `        ``// If location of the term in the series is``        ``// a multiple of 4 then there will be no denominator``        ``if` `(i % 4 == 0)``            ``cout << nmtr[i % 4] + (i / 4) - 1 << ``" "``;` `        ``// Otherwise there will be denominator``        ``else` `{` `            ``// Printing the numerator and the denominator terms``            ``cout << nmtr[i % 4] + ((i / 4) * dntr[i % 4])``                 ``<< ``"/"` `<< dntr[i % 4] << ``" "``;``        ``}``    ``}``}` `// Driver code``int` `main()``{``    ``int` `n = 9;``    ``printSeries(n);``    ``return` `0;``}`

## Java

 `// Java implementation of the approach` `class` `GFG``{` `// Function to print the required series``public` `static` `void` `printSeries(``int` `n)``{``    ``// Numerators for the first four numerators``    ``// of the series``    ``int``[] nmtr = ``new` `int``[]{ ``1``, ``1``, ``1``, ``3` `};` `    ``// Denominators for the first four denominators``    ``// of the series``    ``int``[] dntr = ``new` `int``[]{ ``0``, ``4``, ``2``, ``4` `};` `    ``for` `(``int` `i = ``1``; i <= n; i++)``    ``{` `        ``// If location of the term in the series is``        ``// a multiple of 4 then there will be no denominator``        ``if` `(i % ``4` `== ``0``)``            ``System.out.print( nmtr[i % ``4``] + (i / ``4``) - ``1` `+ ``" "``);` `        ``// Otherwise there will be denominator``        ``else``        ``{` `            ``// Printing the numerator and the denominator terms``            ``System.out.print( nmtr[i % ``4``] + ((i / ``4``) * dntr[i % ``4``])``                ``+``"/"` `+ dntr[i % ``4``] +``" "``);``        ``}``    ``}``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``9``;``    ``printSeries(n);``}``}` `// This code is contributed``// by 29AjayKumar`

## Python3

 `# Python 3 implementation of the approach` `# Function to print the required series``def` `printSeries(n):``    ` `    ``# Numerators for the first four``    ``# numerators of the series``    ``nmtr ``=` `[``1``, ``1``, ``1``, ``3``]` `    ``# Denominators for the first four``    ``# denominators of the series``    ``dntr ``=` `[``0``, ``4``, ``2``, ``4``]` `    ``for` `i ``in` `range``(``1``, n ``+` `1``, ``1``):``        ` `        ``# If location of the term in the``        ``# series is a multiple of 4 then``        ``# there will be no denominator``        ``if` `(i ``%` `4` `=``=` `0``):``            ``print``(nmtr[i ``%` `4``] ``+` `int``(i ``/` `4``) ``-` `1``,``                                     ``end ``=` `" "``)` `        ``# Otherwise there will be denominator``        ``else``:``            ` `            ``# Printing the numerator and``            ``# the denominator terms``            ``print``(nmtr[i ``%` `4``] ``+` `(``int``(i ``/` `4``) ``*``                    ``dntr[i ``%` `4``]), end ``=` `"")``            ``print``(``"/"``, end ``=` `"")``            ``print``(dntr[i ``%` `4``], end ``=` `" "``)` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``n ``=` `9``    ``printSeries(n)` `# This code is contributed by``# Shashank_Sharma`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `// Function to print the required series``static` `void` `printSeries(``int` `n)``{``    ` `    ``// Numerators for the first four numerators``    ``// of the series``    ``int``[] nmtr = { 1, 1, 1, 3 };` `    ``// Denominators for the first four denominators``    ``// of the series``    ``int``[] dntr = { 0, 4, 2, 4 };` `    ``for` `(``int` `i = 1; i <= n; i++)``    ``{` `        ``// If location of the term in the series is``        ``// a multiple of 4 then there will be no denominator``        ``if` `(i % 4 == 0)``            ``Console.Write((nmtr[i % 4] + (i / 4) - 1) + ``" "``);` `        ``// Otherwise there will be denominator``        ``else``        ``{` `            ``// Printing the numerator and the denominator terms``            ``Console.Write((nmtr[i % 4] + ((i / 4) * dntr[i % 4])) +``                                        ``"/"` `+ dntr[i % 4] + ``" "``);``        ``}``    ``}``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `n = 9;``    ``printSeries(n);``}``}` `// This code is contributed``// by Akanksha Rai`

## Javascript

 ``
Output:
`1/4 1/2 3/4 1 5/4 3/2 7/4 2 9/4`

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