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Minimize Nth term of an Arithmetic progression (AP)
• Last Updated : 14 Apr, 2021

Given two integers A, B which are any two terms of an Arithmetic Progression series, and an integer N, the task is to minimize the Nth term of that arithmetic progression.

Note: All the elements of an AP series must be positive.

Examples:

Input: A = 1, B = 6, N = 3
Output: 11
Explanations: First three terms of the given arithmetic progression are: {1, 6, 11}.
Therefore, the required output is 11.

Input: A = 20, B = 50, N = 5
Output: 50
Explanations: First Five terms of the given arithmetic progression are: {10, 20, 30, 40, 50}.
Therefore, the required output is 50.

Approach: The problem can be solved by placing A and B at all possible positions in an Arithmetic Progression and check which generates the least possible Nth term. Considering the positions of A and B to be i and j respectively, then the following calculations need to be made:

Common Difference(D) of an AP = (B – A)/(j – i)
First term of an AP = A – (i – 1) * D
Nth term of an AP = First Term + (N – 1)* D

Finally, return the smallest Nth term obtained.
Follow the steps below to solve the problem:

1. Initialize a variable, say res to store the smallest possible value of the required Nth term of arithmetic progression.
2. Using two nested loops, place A and B at all possible positions in an AP, and calculate the Nth term using above calculations. Keep updating res to store the least value of Nth term obtain.
3. Finally, print the value of res as the required answer.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach` `#include ``using` `namespace` `std;` `// Function to find the smallest``// Nth term of an AP possible``int` `smallestNth(``int` `A, ``int` `B, ``int` `N)``{``    ``// Stores the smallest Nth term``    ``int` `res = INT_MAX;` `    ``for` `(``int` `i = 1; i < N; i++) {``        ``for` `(``int` `j = N; j > i; j--) {` `            ``// Check if common difference``            ``// of AP is an integer``            ``if` `((B - A) % (j - i) == 0) {` `                ``// Store the common``                ``// difference``                ``int` `D = (B - A) / (j - i);` `                ``// Store the First Term``                ``// of that AP``                ``int` `FirstTerm = A - (i - 1) * D;` `                ``// Store the Nth term of``                ``// that AP``                ``int` `NthTerm = FirstTerm + (N - 1) * D;` `                ``// Check if all elements of``                ``// an AP are positive``                ``if` `(FirstTerm > 0)``                    ``res = min(res, NthTerm);``            ``}``        ``}``    ``}` `    ``// Return the least``    ``// Nth term obtained``    ``return` `res;``}` `// Driver Code``int` `main()``{``    ``int` `N = 3;``    ``int` `A = 1;``    ``int` `B = 6;``    ``cout << smallestNth(A, B, N);``}`

## Java

 `// Java program to implement``// the above approach``import` `java.io.*;` `class` `GFG{`` ` `// Function to find the smallest``// Nth term of an AP possible``static` `int` `smallestNth(``int` `A, ``int` `B, ``int` `N)``{` `    ``// Stores the smallest Nth term``    ``int` `res = Integer.MAX_VALUE;`` ` `    ``for``(``int` `i = ``1``; i < N; i++)``    ``{``        ``for``(``int` `j = N; j > i; j--)``        ``{``            ` `            ``// Check if common difference``            ``// of AP is an integer``            ``if` `((B - A) % (j - i) == ``0``)``            ``{``                ` `                ``// Store the common``                ``// difference``                ``int` `D = (B - A) / (j - i);`` ` `                ``// Store the First Term``                ``// of that AP``                ``int` `FirstTerm = A - (i - ``1``) * D;`` ` `                ``// Store the Nth term of``                ``// that AP``                ``int` `NthTerm = FirstTerm + (N - ``1``) * D;`` ` `                ``// Check if all elements of``                ``// an AP are positive``                ``if` `(FirstTerm > ``0``)``                    ``res = Math.min(res, NthTerm);``            ``}``        ``}``    ``}`` ` `    ``// Return the least``    ``// Nth term obtained``    ``return` `res;``}`` ` `// Driver Code``public` `static` `void` `main (String[] args)``{``    ``int` `N = ``3``;``    ``int` `A = ``1``;``    ``int` `B = ``6``;`` ` `    ``System.out.print(smallestNth(A, B, N));``}``}` `// This code is contributed by code_hunt`

## Python3

 `# Python3 program to implement``# the above approach``import` `sys` `# Function to find the smallest``# Nth term of an AP possible``def` `smallestNth(A, B, N):``    ` `    ``# Stores the smallest Nth term``    ``res ``=` `sys.maxsize` `    ``for` `i ``in` `range``(``1``, N):``        ``for` `j ``in` `range``(N, i, ``-``1``):` `            ``# Check if common difference``            ``# of AP is an integer``            ``if` `((B ``-` `A) ``%` `(j ``-` `i) ``=``=` `0``):` `                ``# Store the common``                ``# difference``                ``D ``=` `(B ``-` `A) ``/``/` `(j ``-` `i)` `                ``# Store the First Term``                ``# of that AP``                ``FirstTerm ``=` `A ``-` `(i ``-` `1``) ``*` `D` `                ``# Store the Nth term of``                ``# that AP``                ``NthTerm ``=` `FirstTerm ``+` `(N ``-` `1``) ``*` `D` `                ``# Check if all elements of``                ``# an AP are positive``                ``if` `(FirstTerm > ``0``):``                    ``res ``=` `min``(res, NthTerm)` `    ``# Return the least``    ``# Nth term obtained``    ``return` `res` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``N ``=` `3``    ``A ``=` `1``    ``B ``=` `6``    ` `    ``print``(smallestNth(A, B, N))``    ` `# This code is contributed by mohit kumar 29`

## C#

 `// C# program to implement``// the above approach``using` `System;` `class` `GFG{`` ` `// Function to find the smallest``// Nth term of an AP possible``static` `int` `smallestNth(``int` `A, ``int` `B, ``int` `N)``{``    ` `    ``// Stores the smallest Nth term``    ``int` `res = Int32.MaxValue;`` ` `    ``for``(``int` `i = 1; i < N; i++)``    ``{``        ``for``(``int` `j = N; j > i; j--)``        ``{``            ` `            ``// Check if common difference``            ``// of AP is an integer``            ``if` `((B - A) % (j - i) == 0)``            ``{``                ` `                ``// Store the common``                ``// difference``                ``int` `D = (B - A) / (j - i);`` ` `                ``// Store the First Term``                ``// of that AP``                ``int` `FirstTerm = A - (i - 1) * D;`` ` `                ``// Store the Nth term of``                ``// that AP``                ``int` `NthTerm = FirstTerm + (N - 1) * D;`` ` `                ``// Check if all elements of``                ``// an AP are positive``                ``if` `(FirstTerm > 0)``                    ``res = Math.Min(res, NthTerm);``            ``}``        ``}``    ``}`` ` `    ``// Return the least``    ``// Nth term obtained``    ``return` `res;``}`` ` `// Driver Code``public` `static` `void` `Main ()``{``    ``int` `N = 3;``    ``int` `A = 1;``    ``int` `B = 6;``    ` `    ``Console.Write(smallestNth(A, B, N));``}``}` `// This code is contributed by code_hunt`

## Javascript

 ``
Output:
`11`

Time Complexity: O(N2)
Auxiliary Space: O(1)

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