# Find the Nth term of the series where each term f[i] = f[i – 1] – f[i – 2]

Given three integers X, Y and N, the task is to find the Nth term of the series f[i] = f[i – 1] – f[i – 2], i > 1 where f[0] = X and f[1] = Y.

Examples:

Input: X = 2, Y = 3, N = 3
Output: -2
The series will be 2 3 1 -2 -3 -1 2 and f[3] = -2

Input: X = 3, Y = 7, N = 8
Output: 4

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

Approach: An important observation here is that there will be atmost 6 distinct terms, before the sequence starts repeating itself. So, find the first 6 terms of the series and then the Nth term would be same as the (N % 6)th term.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the Nth term ` `// of the given series ` `int` `findNthTerm(``int` `x, ``int` `y, ``int` `n) ` `{ ` `    ``int` `f[6]; ` ` `  `    ``// First and second term of the series ` `    ``f[0] = x; ` `    ``f[1] = y; ` ` `  `    ``// Find first 6 terms ` `    ``for` `(``int` `i = 2; i <= 5; i++) ` `        ``f[i] = f[i - 1] - f[i - 2]; ` ` `  `    ``// Return the Nth term ` `    ``return` `f[n % 6]; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `x = 2, y = 3, n = 3; ` `    ``cout << findNthTerm(x, y, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.io.*;  ` ` `  `class` `GFG  ` `{  ` `     `  `    ``// Function to find the nth term of series  ` `    ``static` `int` `findNthTerm(``int` `x, ``int` `y, ``int` `n)  ` `    ``{      ` `        ``int``[] f = ``new` `int``[``6``]; ` `         `  `        ``f[``0``] = x; ` `        ``f[``1``] = y; ` `         `  `        ``// Loop to add numbers  ` `        ``for` `(``int` `i = ``2``; i <= ``5``; i++)  ` `            ``f[i] = f[i - ``1``] - f[i - ``2``];  ` `         `  `        ``return` `f[n % ``6``];  ` `    ``}  ` ` `  `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main(String args[])  ` `    ``{  ` `        ``int` `x = ``2``, y = ``3``, n = ``3``;  ` `        ``System.out.println(findNthTerm(x, y, n));  ` `    ``}  ` `}  ` ` `  `// This code is contributed by mohit kumar 29 `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the Nth term  ` `# of the given series  ` `def` `findNthTerm(x, y, n):  ` ` `  `    ``f ``=` `[``0``] ``*` `6` ` `  `    ``# First and second term of  ` `    ``# the series  ` `    ``f[``0``] ``=` `x  ` `    ``f[``1``] ``=` `y  ` ` `  `    ``# Find first 6 terms  ` `    ``for` `i ``in` `range``(``2``, ``6``):  ` `        ``f[i] ``=` `f[i ``-` `1``] ``-` `f[i ``-` `2``]  ` ` `  `    ``# Return the Nth term  ` `    ``return` `f[n ``%` `6``]  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``x, y, n ``=` `2``, ``3``, ``3` `    ``print``(findNthTerm(x, y, n)) ` ` `  `# This code is contributed by ` `# Rituraj Jain `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` `     `  `    ``// Function to find the nth term of series  ` `    ``static` `int` `findNthTerm(``int` `x, ``int` `y, ``int` `n)  ` `    ``{  ` `        ``int``[] f = ``new` `int``[6];  ` `         `  `        ``f[0] = x;  ` `        ``f[1] = y;  ` `         `  `        ``// Loop to add numbers  ` `        ``for` `(``int` `i = 2; i <= 5; i++)  ` `            ``f[i] = f[i - 1] - f[i - 2];  ` `         `  `        ``return` `f[n % 6];  ` `    ``}  ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `        ``int` `x = 2, y = 3, n = 3;  ` `        ``Console.WriteLine(findNthTerm(x, y, n));  ` `    ``}  ` `}  ` ` `  `// This code is contributed by Ryuga `

## PHP

 ` `

Output:

```-2
```

Time Complexity: O(1)

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