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

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 `

Output:

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

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