Minimize Nth term of an Arithmetic progression (AP)

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 N^th 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 N^th 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 N^th 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 N^th term of arithmetic progression.
2. Using two nested loops, place A and B at all possible positions in an AP, and calculate the N^th term using above calculations. Keep updating res to store the least value of N^th 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 <bits/stdc++.h>
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

<script>

// JavaScript program to implement
// the above approach

// Function to find the smallest
// Nth term of an AP possible
function smallestNth(A, B, N)
{

    // Stores the smallest Nth term
    let res = Number.MAX_VALUE;

    for(let i = 1; i < N; i++)
    {
        for(let j = N; j > i; j--)
        {

            // Check if common difference
            // of AP is an integer
            if ((B - A) % (j - i) == 0)
            {

                // Store the common
                // difference
                let D = (B - A) / (j - i);

                // Store the First Term
                // of that AP
                let FirstTerm = A - (i - 1) * D;

                // Store the Nth term of
                // that AP
                let 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
    let N = 3;
    let A = 1;
    let B = 6;

    document.write(smallestNth(A, B, N));

// This code is contributed by target_2.
</script>

Output:

11

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